Pythagoreans Figurative Numbers: The Beginning of Number Theory and Summation of Series ()
1. Introduction
Pythagoras of Samos (around 582-481 BC, Greece) and his several followers, especially, Hypsicles of Alexandria (around 190-120 BC, Greece), Plutarch of Chaeronea (around 46-120, Greece), Nicomachus of Gerasa (around 60-120, Jordan-Israel), and Theon of Smyrna (70-135, Greece) portrayed natural numbers in orderly geometrical configuration of points/dots/pebbles and labeled them as figurative numbers. From these arrangements, they deduced some astonishing number-theoretic results. This was indeed the beginning of the number theory, and an attempt to relate geometry with arithmetic. Nicomachus in his book, see [1], originally written about 100 A.D., collected earlier works of Pythagoreans on natural numbers, and presented cubic figurative numbers (solid hennumbers). Thus, figurate numbers had been studied by the ancient Greeks for polygonal numbers, pyramidal numbers, and cubes. The connection between regular geometric figures and the corresponding sequences of figurative numbers was profoundly significant in Plato’s science, after Plato of Athens (around 427-347 BC, Greece), for example in his work Timaeus. The study of figurative numbers was further advanced by Diophantus of Alexandria (about 250, Greece). His main interest was in figurate numbers based on the Platonic solids (tetrahedron, cube, octahedron, dodecahedron, and icosahedron), which he documented in De solidorum elementis. However, this treatise was lost, and rediscovered only in 1860. Dicuilus (flourished 825, Ireland) wrote Astronomical Treatise in Latin about 814-816, which contains a chapter on triangular and square numbers, see Ross and Knott [2]. After Diophantus’s work, several prominent mathematicians took interest in figurative numbers. The long list includes: Leonardo of Pisa/Fibonacci (around 1170-1250, Italy), Michael Stifel (1486-1567, Germany), Gerolamo Cardano (1501-1576, Italy), Johann Faulhaber (1580-1635, German), Claude Gaspard Bachet de Meziriac (1581-1638, France), René Descartes (1596-1650, France), Pierre de Fermat (1601-1665, France), John Pell (1611-1685, England). In 1665, Blaise Pascal (1623-1662, France) wrote the Traité du triangle arithmétique, avec quelques autres petits traitez sur la mesme matiére which contains some details of figurate numbers. Work of Leonhard Euler (1707-1783, Switzerland) and Joseph Louis Lagrange (1736-1813, France) on figurate numbers opened new avenues in number theory. Octahedral numbers were extensively examined by Friedrich Wilhelm Marpurg (1718-1795, German) in 1774, and Georg Simon Klügel (1739-1812, Germany) in 1808. The Pythagoreans could not have anticipated that figurative numbers would engage after 2000 years leading scholars such as Adrien-Marie Legendre (1752-1833, France), Karl Friedrich Gauss (1777-1855, Germany), Augustin-Louis Cauchy (1789-1857, France), Carl Guslov Jacob Jacobi (1804-1851, Germany), and Waclaw Franciszek Sierpiński (1882-1969, Poland). In 2011, Michel Marie Deza (1939-2016, Russia-France) and Elena Deza (Russia) in their book [3] had given an extensive information about figurative numbers.
In this article we shall systematically discuss most popular polygonal, centered polygonal, three dimensional numbers (including pyramidal numbers), and four dimensional figurative numbers. We shall begin with triangular numbers and end this article with pentatope numbers. For each type of polygonal figurative numbers, we shall provide definition in terms of a sequence, possible sketch, explicit formula, possible relations within the class of numbers through simple recurrence relations, properties of these numbers, generating function, sum of first finite numbers, sum of all their inverses, and relations with other types of polygonal figurative numbers. For each other type of figurative numbers mainly we shall furnish definition in terms of a sequence, possible sketch, explicit formula, generating function, sum of first finite numbers, and sum of all their inverses. The study of figurative numbers is interesting in its own sack, and often these numbers occur in real world situations. We sincerely hope after reading this article it will be possible to find new representations, patterns, relations with other types of popular numbers which are not discussed here, extensions, and real applications.
2. Triangular Numbers
In this arrangement rows contain
dots (see Figure 1).
From Figure 1 it follows that each new triangular number is obtained from the previous triangular number by adding another row containing one more dot than the previous row added, and hence
is the sum of the first n positive integers, i.e.,
(1)
i.e., the differences between successive triangular numbers produce the sequence of natural numbers. To find the sum in (1) we shall discuss two methods which are innovative.
Method 1. Since
An addition of these two arrangements immediately gives
and hence
(2)
Thus, it immediately follows that
,
,
,
,
,
,
,
. This method was first employed by Gauss. The story is his elementary school teacher asked the class to add up the numbers from 1 to 100, expecting to keep them busy for a long time. Young Gauss found the Formula (2) instantly and wrote down the correct answer 5050.
Method 2. From Figure 2 Proof without words of (2) is immediate, see Alsina and Nelsen [4]. However, a needless explanation is a “stairstep” configuration made up of one block plus two blocks plus three blocks, etc, replicated it as the shaded section in Figure 2, and fit them together to form an
rectangular array. Because the rectangle is made of two identical stairsteps (each
representing
) and the rectangle’s area is the product of base and height, that is,
, then the stairstep’s area must be half of the rectangle’s, and hence (2) holds.
To prove (2) the Principle of mathematical induction is routinely used. The relation (1) is a special case of an arithmetic progression of the finite sequence
where
, or
, i.e.,
(3)
For this, following the Method 1, it immediately follows that
(4)
Thus, the mean value of the series is
, which is similar as in discrete uniform distribution. For
, (3) reduces to (1), and (4) becomes same as (2). From (4), it is also clear that
(5)
Ancient Indian Sulbas (see Agarwal and Sen [5] ) contain several examples of arithmetic progression. Aryabhata (born 2765 BC) besides giving the Formula (4) also obtained n in terms of S, namely,
(6)
He also provided elegant results for the summation of series of squares and cubes. In Rhind Papyruses (about 1850 and 1650 BC) out of 87 problems two problems deal with arithmetical progressions and seem to indicate that Egyptian scriber Ahmes (around 1680-1620 BC) knew how to sum such series. For example, Problem 40 concerns an arithmetic progression of five terms. It states: divide 100 loaves among 5 men so that the sum of the three largest shares is 7 times the sum of the two smallest (
,
,
,
). There is a discussion of arithmetical progression in the works of Archimedes of Syracuse (287-212 BC, Greece), Hypsicles, Brahmagupta (born 30 BC, India), Diophantus, Zhang Qiujian (around 430-490, China), Bhaskara II or Bhaskaracharya (working 486, India), Alcuin of York (around 735-804, England), Dicuil, Fibonacci, Johannes de Sacrobosco (around 1195-1256, England), Levi ben Gershon (1288-1344, France). Abraham De Moivre (1667-1754. England) predicted the day of his own death. He found that he slept 15 minutes longer each night, and summing the arithmetic progression, calculated that he would die on November 27, 1754, the day that he would sleep all 24 hours. Peter Gustav Lejeune Dirichlet (1805-1859, Germany) showed that there are infinitely many primes in the arithmetic progression
, where a and b are relatively prime. Enrico Bombieri (born 1940, Italy) is known for the distribution of prime numbers in arithmetic progressions. Terence Chi-Shen Tao (born 1975, Australia-USA) showed that there exist arbitrarily long arithmetic progressions of prime numbers.
The following equalities between triangular numbers can be proved rather easily.
Instead of adding the above finite arithmetic series
, we can multiply its terms which in terms of Gamma function
can be written as
(7)
provided a/d is nonpositive.
• The triangular number
solves the handshake problem of counting the number of handshakes if each person in a room with
people shakes hands once with each person. Similarly a fully connected network of
computing devices requires
connections. The triangular number
also provides the number of games played by
teams in a Round-Robin Tournament in which each team plays every other team exactly once and no ties are allowed. Further, the triangular number
is the number of ordered pairs
, where
. For an
sided-polygon, the number of diagonals is
. From Figure 1, it follows that the number of line segments between closest pairs of dots in the triangles is
, or recursively,
,
. Thus, for example,
. A problem of Christoff Rudolff (1499-1545, Poland) reads: I am owed 3240 florins. The debtor pays me 1 florin the first day, 2 the second day, 3 the third day, and so on. How many days it takes to pay off the debt (80 days). For the Pythagoreans the fourth triangular number
(decade) was most significant of all: it contains in itself the first four integers, one, two, three, and four
, it was considered to be a symbol of “perfection”, being the sum of 1 (a point), 2 (a line), 3 (a plane) and 4 (a solid); it is the smallest integer n for which there are just as many primes between 1 and n as nonprimes, and it gives rise to the tetraktys (see Figure 1 and its alternative form Figure 3). To them, the tetraktys was the sum of the divine influences that hold the universe together, or the sum of all the manifest laws of nature. They recognized tetraktys as fate, the universe, the heaven, and even God. Pythagoras also called the Deity a Tetrad or Tetracyts, meaning the “four sacred letters”. These letters originated from the four sacrad letters JHVH, in which the ancient Jews called God our Father, the name “Jehovah”. The tetraktys was so revered by the members of the brotherhood that they shared the following oath and their most jealously guarded secret, “I swear by him who has transmitted to our minds the holy tetraktys, the roots and source of ever-flowing nature”. For Plato (Plato, meaning broad, is a nickname, his real name was Aristocles, he died at a wedding feast) number ten was the archetypal pattern of the universe. According to Eric Temple Bell (1883-1960, UK-USA), see [6], “Pythagoras asked a merchant if he could count. On the merchants’s replying that he could, Pythagoras told him to go ahead. One, two, three, four ..., he began, when Pythagoras shouted Stop! What you name four is really what you would call ten. The fourth number is not four, but decade, our tetractys and inviolable oath by which we swear”. Inadvertently, the tetractys occurs in the following: the arrangement of bowling pins in ten-pin bowling, the baryon decuplet, an archbishop’s coat of arms, the “Christmas Tree” formation in association football, a Chinese checkers board, and the list continues. The number
gives the number and arrangement of balls in Billiards. The 36th triangular number, i.e., is 666 (The Beast of Revelation-Christians often seems to have difficulties with numbers). The 666th triangular number, i.e.,
is 222111. On triangular numbers an interesting article is due to Fearnehough [7].
Figure 3. Alternative form of tetraktys.
• No triangular number has as its last digit (unit digit) 2, 4, 7 or 9. For this, let
, then
; here
. Thus, it follows that
. This relation gives only choices for k as
and 8.
• We shall show that for an integer
,
repeats every k steps if k is odd, and every 2k steps if k is even, i.e., if
is the smallest positive integer such that for all integers n
(8)
then
if k is odd, and
if k is even. For this, note that
and hence if (8) holds, then
For
and
the above equation respectively gives
Combining these two relations, we find
and hence
(9)
Now if k is odd, then in view of
is an integer, we have
This implies that
, because
is the smallest integer for which (8) holds. But, then from (9) it follows that
.
If k is even, then
is odd, and so
. Thus,
, but
and so 2k satisfies (8). This implies that
, which again from (9) gives
.
For example, for
, we have
and for
,
• Triangular numbers and binomial coefficients are related by the relation
Thus, triangular numbers are associated with Pascal’s triangle
For the origin of Pascal triangle see Agarwal and Sen [5].
• The only triangular numbers which are the product of three consecutive integers are 6, 120, 210, 990, 185,136, 258, 474, 216, see Guy [8].
• A number is called palindromic if it is identical with its reverse, i.e., reading the same forward as well as backward. There are 28 palindromic triangular numbers less than 1010, namely, 1, 3, 6, 55, 66, 171, 595, 666, 3003, 5995, 8778, 15,051, 66,066, 617,716, 828,828, 1,269,621, 1,680,861, 3,544,453, 5,073,705, 5,676,765, 6,295,926, 351,335,153, 61,477,416, 178,727,871, 1,264,114,621, 1,634,004,361, 5,289,009,825, 6,172,882,716. The largest known palindromic triangular numbers containing only odd digits and even digits are
and
. It is known, see Trigg [9], that an infinity of palindromic triangular numbers exist in several different bases, for example, three, five, and nine; however, no infinite sequence of such numbers has been found in base ten.
• Let m be a given natural number, then it is n-th triangular number, i.e.,
if and only if
. This means if and only if
is a perfect square.
• If n is a triangular number, then
and
are also triangular numbers. This result of 1775 is due to Euler. Indeed, if
, then
,
and
. An extension of Euler’s result is the identity
i.e.,
• From the identity
it follows that if n is the sum of two triangular numbers, then
is a sum of two squares.
• Differentiating the expansion
twice, we get
(10)
and hence
Hence,
is the generating function of all triangular numbers. In 1995, Sloane and Plouffe [10] have shown that
• To find the sum of the first n triangular numbers, we need an expression for
(a general reference for the summation of series is Davis [11] ). For this, we begin with Pascal’s identity
and hence
which in view of (2) gives
(11)
Archimedes as proposition 10 in his text On Spirals stated the formula
(12)
from which (11) is immediate. It is believed that he obtained (12) by letting k the successive values
in the relation
and adding the resulting
equations, together with the identity
, to arrive at
(13)
Next, letting
in the formula
and adding n equations to get
(14)
From (13) and (14), the Formula (12) follows.
Another proof of (11) is given by Fibonacci. He begins with the identity
and takes
to get the set of equations
On adding these n equations and cancelling the common terms, (11) follows.
Now from (2) and (11), we have
(15)
Relation (15) is due to Aryabhata.
For an alternative proof of (15), we note that
and hence in view of (11), we have
From (15) it follows that
which in particular for
gives
, i.e., three successive triangular numbers whose sum is a perfect square. Similarly, we have
.
From (15), we also have
, which means
divides
if
,
.
• The reciprocal of the
-th triangular number is related to the integral
• The sum of reciprocals of the first n triangular numbers is
(16)
and hence
(17)
Jacob Bernoulli (1654-1705, Switzerland) in 1689 summed numerous convergent series, the above is one of the examples. In the literature this procedure is now called telescoping, also see Lesko [12].
• Pythagoras theorem states that if a and b are the lengths of the two legs of a right triangle and c is the length of the hypothenuse, then the sum of the areas of the two squares on the legs equals the area of the square on the hypotenuse, i.e.,
(18)
A set of three positive integers a, b and c which satisfy (18) is called Pythagorean triple and written as ordered triple
. A Pythagorean triangle
is said to be primitive if
have no common divisor other than 1. For the origin, patterns, extensions, astonishing directions, and open problems, of Pythagoras theorem and his triples, see Agarwal [13] [14], and an interesting article of Beauregard and Suryanarayan [15]. There are Pythagorean triples (not necessarily primitive) each side of which is a triangular number, for example,
. It is not known whether infinitively many such triples exist.
• A number is called perfect if and only if it is equal to the sum of its positive divisors, excluding itself. For example,
is perfect, because
. The numbers
are also perfect that Pythagoreans discovered. For mystical reasons, such numbers have been given considerable attention in the past. Especially, Pythagoreans praised the number six eulogistically, concluding that the universe is harmonized by it and from it comes wholeness, permanence, as well as perfect health. In fact, Plato asserted that the creation is perfect because the number 6 is perfect. They also realized that like squares, six equilateral triangles (see Figure 4) meeting at a point (add up to 360˚) leave no space in tilling a floor.
Till very recently only 51 even perfect numbers of the form
have been discovered. It is not known whether there are any odd perfect numbers, and if there exist infinitely many perfect numbers. The following result due
to Euclid of Alexandria (around 325-265 BC, Greece) and Euler states that an even number is perfect if and only if it has the form
, where
is a prime number (known as Pére Marin Mersenne’s, 1588-1648, France, prime number). In 1575 it was observed that
, i.e., every known perfect number is also a triangular number.
• Fermat numbers are defined as
. First few Fermat’s numbers are 3, 5, 17, 257, 65537. We shall show that for
, Fermat number
is never a triangular number, i.e., there is no integer m which satisfies
. This means the discriminant of the equation
is not an integer. Suppose to contrary that there exists an integer p such that
, but then
, which implies that there exist integers r and s such that
and
. Hence, we have
for which the only solution is
. This means,
, or
, which is true only for
.
• We shall find all square triangular numbers, i.e., all positive integers n and the corresponding m so that
. This equation can be written as, so called Pell’s Equation (for its origin, see Agarwal [5] )
, where
and
. We note that if
is an integer solution of
, then
defined by the recurrence relations
(19)
satisfy
and hence
. From this observation we conclude that if
is an integer solution of
, then so is
. Since
is a solution of
(its fundamental solution is
), it follows that the iterative scheme
(20)
gives all solutions of
. System (20) can be written as
(21)
Now in (21) using the substitution
we get
(22)
Clearly, (22) generates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (22) can be computed (for details see Agarwal [16] [17] ) rather easily, and appears as
(23)
This result is originally due to Euler which he obtained in 1730. While compare to the explicit solution (23) the computation of
from the recurrence relations (22) is very simple, the following interesting relation follows from (23) by direct substitution
(24)
Hence the difference between two consecutive square triangular numbers is the square root of another square triangular number.
Now we note that the system (19) can be written as
and its (integer) solution is
(25)
From this, and simple calculations the following relations follow
It is apparent that if
is a solution of
, then
is a solution of
. Now, if n is even, we have
(26)
and, when n is odd,
(27)
and hence the right side is a perfect square for
. Therefore, the product of
consecutive triangular numbers is a perfect square for each
and
. In particular, for
,
, from (26) we have
and for
, from (27), we find
Similarly, if n is even, we have
(28)
and hence the right side is a perfect square for
(which is always even). Therefore, two times the product of
consecutive triangular numbers is a perfect square for each
and
. In particular, for
,
, from (28) we have
and for
, we find
From the equality
it follows that if the triangular number
is square, then
is also square. Since
is square, it follows that there are infinite number of square triangular numbers. This clever observation was reported in 1662, see Pietenpol et al. [18]. From this, the first four square triangular numbers, we get are
and
.
• There are infinitely many triangular numbers that are simultaneously expressible as the sum of two cubes and the difference of two cubes. For this, Burton [19] begins with the identity
and observed that if k is odd then this equality can be written as
which is the same as
For
and 5 this gives
• In 1844, Eugéne Charles Catalan (1814-1894) conjectured that 8 and 9 are the only numbers which differ by 1 and are both exact powers
,
. This conjecture was proved by Preda Mihăilescu (Born 1955, Romania) after one hundred and fifty-eight years, and published two years later in [20]. Thus the only solution in natural numbers of the Diophantine equation
for
,
is
,
,
,
. Now since
can be written as
, the only solution of this equation is
,
, i.e.,
is the only cubic triangular number.
• In 2001, Bennett [21] proved that if a, b and n are positive integers with
, then the equation
, possesses at most one solution in positive integers x and y. This result is directly applicable to show that for the equation
the only solution is
. For this, first we note that integers
and
are coprime, i.e., they do not have any common factor except 1. We also recall that if the product of coprime numbers is a p-th power, then both are also of p-the power. Now let n be even, i.e.,
, then the equation
is the same as
. Thus, it follows that
and
, and hence
, which has only one solution, namely,
which gives
, and hence
and so
is the solution of
, but we are not interested in this solution. Now we assume that n is odd, i.e.,
, then the equation
is the same as
. Thus, we must have
and
, which gives
. The only solution of this equation is
, and hence again
and so
is the undesirable solution of
.
• Startling generating function of all square triangular numbers is recorded by Plouffe [22] as
(29)
3. Square Numbers Sn
In this arrangement rows as well as columns contain
dots, (see Figure 5).
From Figure 5 it is clear that a square made up of
dots on a side can be divided into a smaller square of side n and an L, shaped border (a gnomon), which has
dots (called
th gnomonic number and denoted as
), and hence
(30)
i.e., the differences between successive nested squares produce the sequence of odd numbers. From (30) it follows that
and hence
(31)
An alternative proof of (31) is as follows
An addition of these two arrangements immediately gives
Figure 6 provides proof of (31) without words. Here odd integers, one block, three blocks, five blocks, and so on, arranged in a special way. We begin with a single block in the lower left corner; three shaded blocks surrounded it to form a
square; five unshaded blocks surround these to form a
square; with the next seven shaded blocks we have a
square; and so on. The diagram makes clear that the sum of consecutive odd integers will always yield a (geometric) square.
Comparing Figure 1 and Figure 5 or Figure 2 and Figure 6, it is clear that n-th square number is equal to the n-th triangular number increased by its predecessor, i.e.,
(32)
Indeed, we have
An addition of these two arrangements in view of (31) gives
Of course, directly from (1), (2), and (32), we also have
or simply from (1) and (2),
From (32), we find the identities
and
It also follows that
(33)
We also have equalities
(34)
(35)
and
(36)
which is the same as
and, in particular, for
reduces to
The following equality is of exceptional merit
(37)
which, in particular, for
gives
.
• Relation (30) reveals that every odd integer
is the difference of two consecutive square numbers
and
. Relation (32) shows that every square integer
is a sum of two consecutive triangular numbers
and
, whereas (33) displays it is the difference of 2n-th and two times n-th triangular numbers.
• From the equalities
it follows that there are infinite triples of consecutive numbers which can be written as the sum of two squares.
• No square number has as its last digit (unit digit) 2, 3, 7 or 8.
• From (10) it follows that
and hence
(38)
Therefore,
is the generating function of all square numbers. From (38) it also follows that the generating function for all gnomonic numbers is
• The sum of the first n square numbers is given in (11). For the exact sum of the reciprocals of the first n square numbers no formula exists; however, the problem of summing the reciprocals of all square numbers has a long history and in the literature it is known as the Basel problem. Euler in 1748 considered
which has roots at
. Then, he wrote this function in terms of infinite product
which on equating the coefficients of
, gives
and hence
(39)
The above demonstration of Euler is based on manipulations that were not justified at the time, and it was not until 1741 that he was able to produce a truly rigorous proof. Now in the literature for (39) several different proofs are known, e.g., for a recent elementary, but clever demonstration, see Murty [23].
• The following result provides a characterization of all Pythagorean triples, i.e., solutions of (18): Let u and v be any two positive integers, with
, then the three numbers
(40)
form a Pythagorean triple. If in addition u and v are of opposite parity-one even and the other odd-and they are coprime, i.e., that they do not have any common factor other than 1, then
is a primitive Pythagorean triple. The converse, i.e., any Pythagorean triple is necessarily of the form (40) also holds. For the proof and history of this result see, Agarwal [14]. From (18), (32), and (40) the following relations hold
(41)
The relation (30) can be written as
. With the help of this relation we can find Pythagorean triples
. For this, we let
, (and hence m is odd), then
,
. Thus, it follows that
(42)
For
Equation (42) gives solutions of (18):
Similar to (42) for m even we also have the relation
(43)
For
Equation (43) gives solutions of (18):
In (40), letting
and
, from (18) and (32), we get the relations
which is the same as
• In 1875, Francois Edouard Anatole Lucas (1842-1891, French) challenged the mathematical community to prove that the only solution of the equation
with
is when
and
. In the literature this has been termed as the cannonball problem, in fact, it can be visualized as the problem of taking a square arrangement of cannonballs on the ground and building a square pyramid out of them. It was only in 1918, George Neville Watson (1886-1965, Britain) used elliptic functions to provide correct (filling gaps in earlier attempts) proof of Lucas assertion. Simplified proofs of this result are available, e.g., in Ma [24] and Anglin [25].
4. Rectangular (Oblong, Pronic, Heteromecic) Numbers Rn
In this arrangement rows contain
whereas columns contain n dots, see Figure 7.
From Figure 7 it is clear that the ratio
of the sides of rectangles depends on n. Further, we have
(44)
i.e., we add successive even numbers, or two times triangular numbers. It also follows that rectangular number
is made from
by adding an L-shaped border (a gnomon), with
dots, i.e.,
(45)
i.e., the differences between successive nested rectangular numbers produce the sequence of even numbers.
Thus the odd numbers generate a limited number of forms, namely, squares, while the even ones generate a multiplicity of rectangles which are not similar. From this the Pythagoreans deduced the following correspondence:
We also have the relations
(46)
(47)
From (31) and (46) it follows that
• Relation (44) shows that the product of two consecutive positive integers n and
is the same as two times n-th triangular numbers. According to historians with this relation Pythagoreans’ enthusiasm was endless. Relation (45) reveals that every even integer 2n is the difference of two consecutive rectangular numbers
and
. Relation (46) displays that every positive integer n is the difference of n-th and
-th triangular numbers. Relation (47) is due to Plutarch), it says an integer n is a triangular number if and only if
is a perfect odd square.
• Let m be a given natural number, then it is n-th rectangular number, i.e.,
if and only if
.
• From (10) it is clear that
is the generating function of all rectangular numbers.
• From (15)-(17) and (44) it is clear that
(48)
• There is no rectangular number which is also a perfect square, in fact, the equation
has no solutions (the product of two consecutive integers cannot be a prefect square).
• To find all rectangular numbers which are also triangular numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
) where
and
. For this, corresponding to (22) the system is
(49)
This system genetrates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (49) can be written as
• Fibonacci numbers denoted as
are defined by the recurrence relation
or the closed from expression
First few of these numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144. For the origin of Fibonacci numbers, see Agarwal and Sen [5]. Lucas numbers denoted by
are defined by the same recurrence relation as Fibonacci numbers except first two numbers as
or the closed from expression
First few of these numbers are 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349. Clearly, Fibonacci numbers 1, 3, 21, 55 are also triangular numbers
. In 1989, Luo [26] had used (47) to show that these are the only Fibonacci numbers which are also triangular. This conjecture was made by Verner Emil Hoggatt Jr. (1921-1980, USA) in 1971. Similarly, only Lucas numbers which are also triangular are 1, 3, 5778, i.e.,
. From the above explicit expressions the following relations can be obtained easily
and
.
5. Pentagonal Numbers Pn
The pentagonal numbers are defined by the sequence
, i.e., beginning with 5 each number is formed from the previous one in the sequence by adding the next number in the related sequence
. Thus,
,
,
, and so on (see Figure 8 and Figure 9).
Thus, n-th pentagonal number is defined as
(50)
Comparing (50) with (3), we have
and hence from (4) it follows that
(51)
It is interesting to note that
is the sum of n integers starting from n, i.e.,
(52)
whose sum from (4) is the same as in (51).
Note that from (50), we have
From (32) and (51), we also have
(53)
• Relation (51) shows that pentagonal number
is the one-third of the
-th triangular number, whereas relation (53) reveals that it is the sum of n-th triangular number and two times of
-th triangular number, and it is the difference of
-th triangular number and
-th triangular number.
• Let m be a given natural number, then it is n-th pentagonal number, i.e.,
if and only if
.
• As in (38), we have
and hence
is the generating function of all pentagonal numbers.
• From (2), (11) and (51) it is easy to find the sum of the first n pentagonal numbers
(54)
• To find the sum of the reciprocals of all pentagonal numbers, we begin with the series
and note that
Now since
, we have
and hence
which immediately gives
(55)
• To find all square pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(56)
This system genetrates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (56) can be written as
• To find all pentagonal numbers which are also triangular numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
) where
and
. For this, corresponding to (22) the system is
(57)
This system generates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (57) can be written as
• To find all pentagonal numbers which are also rectangular numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
) where
and
. For this, corresponding to (22) the system is
(58)
This system genetrates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (58) can be written as
6. Hexagonal Numbers Hn
The hexagonal numbers are defined by the sequence
, i.e., beginning with 6 each number is formed from the previous one in the sequence by adding the next number in the related sequence
. Thus,
,
,
, and so on (see Figure 10).
Thus, n-th hexagonal number is defined as
(59)
Comparing (59) with (3), we have
and hence from (4) it follows that
(60)
• From (60) it is clear that
, i.e., alternating triangular numbers are hexagonal numbers.
• Let m be a given natural number, then it is n-th hexagonal number, i.e.,
if and only if
.
• As in (38), we have
and hence
is the generating function of all hexagonal numbers.
• From (2), (11), and (60) it is easy to find the sum of the first n hexagonal numbers
(61)
• To find the sum of the reciprocals of all hexagonal numbers, as for pentagonal numbers we begin with the series
, and get
(62)
• To find all square hexagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(63)
This system generates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (63) appears as
here,
and
are as in (25).
• To find all hexagonal numbers which are also rectangular numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
) where
and
. For this, corresponding to (22) the system is
(64)
This system generates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (64) can be written as
• To find all hexagonal numbers which are also pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
) where
and
. For this, corresponding to (22) the system is
(65)
This system generates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (65) can be written as
7. Generalized Pentagonal Numbers (Centered Hexagonal Numbers, Hex Numbers) (GP)n
The generalized pentagonal numbers are defined by the sequence
, i.e., beginning with 7 each number is formed from the previous one in the sequence by adding the next number in the related sequence
. Thus,
,
,
, and so on (see Figure 11). These numbers are also called centered hexagonal numbers as these represent hexagons with a dot in the center and all other dots surrounding the center dot in a hexagonal lattice. These numbers have practical applications in materials logistics management, for example, in packing round items into larger round containers, such as Vienna sausages into round cans, or combining individual wire strands into a cable.
Thus, n-th generalized pentagonal number is defined as
(66)
Hence, from (2) it follows that
(67)
• Incidentally,
occurs in uds baryon octet, whereas
makes a part of a Chinese checkers board.
• Since
, generalized pentagonal numbers are differences of two consecutive cubes, so that the
are the gnomon of the cubes.
• Clearly,
.
• Let m be a given natural number, then it is n-th generalized pentagonal number, i.e.,
if and only if
.
• From (10) and (67), we have
and hence
is the generating function of all generalized pentagonal numbers.
Figure 11. Generalized pentagonal numbers (centered hexagonal numbers).
• From (15) and (67) it is easy to find the sum of the first n generalized pentagonal numbers
(68)
• Since from (32) and (46), we have
from (68) it follows that
(69)
Thus the equation
has an infinite number of integer solutions. In fact, for each
equations
and
have infinite number of solutions (see Agarwal [14] ).
• To find the sum of the reciprocals of all generalized pentagonal numbers we need the following well-known result, e.g., see Andrews et al. [27], page 536, and Efthimiou [28]
(70)
Now from (70), we have
and hence
(71)
• To find all square generalized pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(72)
This system genetrates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (72) appears as
• To find all generalized pentagonal numbers which are also triangular numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(73)
This system genetrates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (73) can be written as
• There is no generalized pentagonal number which is also a rectangular number, in fact, the equation
has no solutions. For this, we note that this equation can be written as Pell’s equation
, where
and
. Now reducing this equation to
gives
, which is impossible since all squares
are either 0 or 1
.
• To find all generalized pentagonal numbers which are also pentagonal numbers, we need to find integer solutions of the equation
. This equation can also be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(74)
This system genetrates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (74) can be written as
• To find all generalized pentagonal numbers which are also hexagonal numbers, we need to find integer solutions of the equation
. This equation can also be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(75)
This system generates all (infinite) solutions
of the equation
. First few of these solutions are
For
, explicit solution of the system (75) can be written as
8. Heptagonal Numbers (Heptagon Numbers) (HEP)n
These numbers are defined by the sequence
, i.e., beginning with 7 each number is formed from the previous one in the sequence by adding the next number in the related sequence
. Thus,
,
,
, and so on (see Figure 12).
Thus,
-th heptagonal number is defined as
(76)
Comparing (76) with (3), we have
, and hence from (4) it follows that
(77)
• For all integers
it follows that
and
are odd, whereas
and
are even.
• From (77) the following equality holds
• Let m be a given natural number, then it is n-th heptagonal number, i.e.,
if and only if
.
• From (10) and (77), we have
and hence
is the generating function of all heptagonal numbers.
• In view of (15) and (77), we have
(78)
• The sum of reciprocals of all heptagonal numbers is (see https://en.wikipedia.org/wiki/Heptagonal_number)
(79)
• To find all square heptagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solutions are
and
), where
and
. For
, corresponding to (22) the system is
(80)
This system genetrates infinite number of solutions
of the equation
. First four of these solutions are
For
recurrence relations remain the same as in (80) with
and
. This leads to another set of infinite number of solutions
of the equation
. First four of these solutions are
For
also recurrence relations remain the same as in (80) with
and
. This leads to further set of infinite number of solutions
of the equation
. First four of these solutions are
• To find all heptagonal numbers which are also triangular numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solutions are
and
), where
and
. For
corresponding to (22) the system is
(81)
This system genetrates infinite number of solutions
of the equation
. First four of these solutions are
For
recurrence relations remain the same as in (81) with
and
. This leads to another set of infinite number of solutions
of the equation
. First four of these solutions are
• To find all heptagonal numbers which are also rectangular numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(82)
This system genetrates infinite number of solutions
of the equation
. First four of these solutions are
To find all heptagonal numbers which are also pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(83)
This system genetrates infinite number of solutions
of the equation
. First few of these solutions are
• To find all heptagonal numbers which are also hexagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(84)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
• To find all heptagonal numbers which are also generalized pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(85)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
9. Octagonal Numbers On
These numbers are defined by the sequence
, i.e., beginning with 8 each number is formed from the previous one in the sequence by adding the next number in the related sequence
. Thus,
,
,
, and so on (see Figure 13).
Thus, n-th octagonal number is defined as
(86)
Comparing (86) with (3), we have
, and hence from (4) it follows that
(87)
• For all integers
it follows that
are odd, whereas
are even (in fact divisible by 4).
• Let m be a given natural number, then it is n-th octagonal number, i.e.,
if and only if
.
• From (10) and (87), we have
and hence
is the generating function of all octagonal numbers.
• In view of (15) and (87), we have
(88)
• To find the sum of the reciprocals of all octagonal numbers, following Downey [29] we begin with the series
and note that
Thus, we have
Now since
it follows that
(89)
• To find all square octagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(90)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
• To find all octagonal numbers which are also triangular numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solutions are
and
), where
and
. For
corresponding to (22) the system is
(91)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
For
recurrence relations remain the same as in (91) with
and
. This leads to another set of infinite number of solutions
of the equation
. First few of these solutions are
There is no octagonal number which is also a rectangular number, in fact, the equation
has no solutions. For this, we note that this equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, Pell’s equation all solutions can be generated by the system (corresponding to (21))
(92)
Now an explicit solution of the second equation of (92) can be written as
Next, if
, then
, and hence it follows that
. We note that
and
. Thus, from the second equation of (92) mathematical induction immediately gives
and
for all
. In conclusion
or
. Finally, reducing the relation
to
gives
. Hence, in view of
, we conclude that
for all integers k, and therefore, the equation
has no solution.
• To find all octagonal numbers which are also pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solutions are
and
), where
and
. For
corresponding to (22) the system is
(93)
This system genetrates infinite number of solutions
of the equation
. First four of these solutions are
For
recurrence relations remain the same as in (93) with
and
. This leads to another set of infinite number of solutions
of the equation
. First four of these solutions are
• To find all octagonal numbers which are also hexagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solutions are
and
), where
and
. For
corresponding to (22) the system is
(94)
This system genetrates infinite number of solutions
of the equation
. First few of these solutions are
With
the system corresponding to (21) is
(95)
Now note that
and
. Thus, from the first equation of (95) mathemtical induction immediately gives
for all
. Next reducing the relation
to
gives
. Hence, in view of
, we conclude that
for all integers k, and therefore, the equation
has no solution.
• To find all octagonal numbers which are also generalized pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
, where
and
. For the equation
the only meaningful integer solution is
and it gives
.
• To find all octagonal numbers which are also heptagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(96)
This system genetrates infinite number of solutions
of the equation
. First few of these solutions are
10. Nonagonal Numbers Nn
These numbers are defined by the sequence
, i.e., beginning with 9 each number is formed from the previous one in the sequence by adding the next number in the related sequence
. Thus,
,
,
, and so on (see Figure 14).
Thus, n-th nonagonal number is defined as
(97)
Comparing (97) with (3), we have
, and hence from (4) it follows that
(98)
• For all integers
it follows that
are odd, whereas
are even.
• Let m be a given natural number, then it is n-th nonagonal number, i.e.,
if and only if
.
• From (10) and (98), we have
and hence
is the generating function of all nonagonal numbers.
• In view of (15) and (98), we have
(99)
• The sum of reciprocals of all nonagonal numbers is
(100)
here,
is the digamma function defined as the logarithmic derivative of the gamma function
, i.e.,
, and
is the Euler-Mascheroni constant.
• To find all square nonagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution are
and
), where
and
. For
corresponding to (22) the system is
(101)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
For
recurrence relations remain the same as in (101) with
and
. This leads to another set of infinite number of solutions
of the equation
. First few of these solutions are
• To find all nonagonal numbers which are also triangular numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solutions are
and
), where
and
. For
corresponding to (22) the system is
(102)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
For
and
there are no integer solutions.
• There is no nonagonal number which is also a rectangular number, in fact, the equation
has no solutions.
• To find all nonagonal numbers which are also pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solutions are
and
), where
and
. For
corresponding to (22) the system is
(103)
This system generates infinite number of solutions
of the equation
. First four of these solutions are
For
recurrence relations remain the same as in (103) with
and
. This leads to another set of infinite number of solutions
of the equation
. First four of these solutions are
• To find all nonagonal numbers which are also hexagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solutions are
and
), where
and
. For
corresponding to (20) the system is
This system gives first four integer solutions
of the equation
rather easily, which appear as
. Now the following system generates infinite number of solutions
(104)
Similarly, the following system generates infinite number of solutions
(105)
The first eight solutions
are
With
and
there are no integer solutions of the required equation.
• To find all nonagonal numbers which are also generalized pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solutions are
and
), where
and
. For
, corresponding to (22) the system is
(106)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
For
recurrence relations remain the same as in (106) with
and
. This leads to another set of infinite number of solutions
of the equation
. First four of these solutions are
• To find all nonagonal numbers which are also heptagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(107)
This system generates
infinite number of solutions
of the equation
. First few of these solutions are
• To find all nonagonal numbers which are also octagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(108)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
11. Decagonal Numbers Dn
These numbers are defined by the sequence
, i.e., beginning with 10 each number is formed from the previous one in the sequence by adding the next number in the related sequence
. Thus,
,
,
and so on (see Figure 15).
Hence, n-th decagonal number is defined as
(109)
Comparing (109) with (3), we have
, and hence from (4) it follows that
(110)
• For all integers
it follows that
are odd, whereas
are even.
• Let m be a given natural number, then it is n-th decagonal number, i.e.,
if and only if
.
• From (10) and (110), we have
and hence
is the generating function of all decagonal numbers.
• In view of (15) and (110), we have
(111)
• To find the sum of the reciprocals of all decagonal numbers, as in (89) we begin with the series
and following the same steps Downey [29] obtained
(112)
• To find all square decagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
, where
and
. For the equation
the only meaningful integer solution is
and it gives
.
• To find all decagonal numbers which are also triangular numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solutions are
and
), where
and
. For
corresponding to (22) the system is
(113)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
For
recurrence relations remain the same as in (113) with
and
. This leads to another set of infinite number of solutions
of the equation
. First few of these solutions are
• There is no decagonal number which is also a rectangular number, in fact, the equation
has no solutions.
• To find all decagonal numbers which are also pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For
corresponding to (22) the system is
(114)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
• To find all decagonal numbers which are also hexagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(115)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
• To find all decagonal numbers which are also generalized pentagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(116)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
• To find all decagonal numbers which are also heptagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(117)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
• To find all decagonal numbers which are also octagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(118)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
• To find all decagonal numbers which are also nonagonal numbers, we need to find integer solutions of the equation
. This equation can be written as Pell’s equation
(its fundamental solution is
), where
and
. For this, corresponding to (22) the system is
(119)
This system generates infinite number of solutions
of the equation
. First few of these solutions are
12. Tetrakaidecagonal Numbers (TET)n
These numbers are defined by the sequence
, i.e., beginning with 14 each number is formed from the previous one in the sequence by adding the next number in the related sequence
. Thus,
,
,
, and so on (see Figure 16).
Hence, n-th tetrakaidecagonal number is defined as
(120)
Comparing (120) with (3), we have
, and hence from (4) it follows that
(121)
• For all integers
it follows that
are odd, whereas
are even.
• Let m be a given natural number, then it is n-th tetrakaidecagonal number, i.e.,
if and only if
.
• From (10) and (121), we have
and hence
is the generating function of all tetrakaidecagonal numbers.
• In view of (15) and (121), we have