A001318 Generalized pentagonal numbers: m*(3*m - 1)/2, m = 0, +-1, +-2, +-3, ....

0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117, 126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345, 376, 392, 425, 442, 477, 495, 532, 551, 590, 610, 651, 672, 715, 737, 782, 805, 852, 876, 925, 950, 1001, 1027, 1080, 1107, 1162, 1190, 1247, 1276, 1335

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Comments

Partial sums of A026741. - Jud McCranie; corrected by Omar E. Pol, Jul 05 2012

From R. K. Guy, Dec 28 2005: (Start)

"Conway's relation twixt the triangular and pentagonal numbers: Divide the triangular numbers by 3 (when you can exactly):

0 1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 ...

0 - 1 2 .- .5 .7 .- 12 15 .- 22 26 .- .35 .40 .- ..51 ...

.....-.-.....+..+.....-..-.....+..+......-...-.......+....

"and you get the pentagonal numbers in pairs, one of positive rank and the other negative.

"Append signs according as the pair have the same (+) or opposite (-) parity.

"Then Euler's pentagonal number theorem is easy to remember:

"p(n-0) - p(n-1) - p(n-2) + p(n-5) + p(n-7) - p(n-12) - p(n-15) ++-- = 0^n

where p(n) is the partition function, the left side terminates before the argument becomes negative and 0^n = 1 if n = 0 and = 0 if n > 0.

"E.g. p(0) = 1, p(7) = p(7-1) + p(7-2) - p(7-5) - p(7-7) + 0^7 = 11 + 7 - 2 - 1 + 0 = 15."

(End)

The sequence may be used in order to compute sigma(n), as described in Euler's article. - Thomas Baruchel, Nov 19 2003

Number of levels in the partitions of n + 1 with parts in {1,2}.

a(n) is the number of 3 X 3 matrices (symmetrical about each diagonal) M = {{a, b, c}, {b, d, b}, {c, b, a}} such that a + b + c = b + d + b = n + 2, a,b,c,d natural numbers; example: a(3) = 5 because (a,b,c,d) = (2,2,1,1), (1,2,2,1), (1,1,3,3), (3,1,1,3), (2,1,2,3). - Philippe Deléham, Apr 11 2007

Also numbers a(n) such that 24*a(n) + 1 = (6*m - 1)^2 are odd squares: 1, 25, 49, 121, 169, 289, 361, ..., m = 0, +-1, +-2, ... . - Zak Seidov, Mar 08 2008

From Matthew Vandermast, Oct 28 2008: (Start)

Numbers n for which A000326(n) is a member of A000332. Cf. A145920.

This sequence contains all members of A000332 and all nonnegative members of A145919. For values of n such that n*(3*n - 1)/2 belongs to A000332, see A145919. (End)

Starting with offset 1 = row sums of triangle A168258. - Gary W. Adamson, Nov 21 2009

Starting with offset 1 = Triangle A101688 * [1, 2, 3, ...]. - Gary W. Adamson, Nov 27 2009

Starting with offset 1 can be considered the first in an infinite set generated from A026741. Refer to the array in A175005. - Gary W. Adamson, Apr 03 2010

Vertex number of a square spiral whose edges have length A026741. The two axes of the spiral forming an "X" are A000326 and A005449. The four semi-axes forming an "X" are A049452, A049453, A033570 and the numbers >= 2 of A033568. - Omar E. Pol, Sep 08 2011

A general formula for the generalized k-gonal numbers is given by n*((k - 2)*n - k + 4)/2, n=0, +-1, +-2, ..., k >= 5. - Omar E. Pol, Sep 15 2011

a(n) is the number of 3-tuples (w,x,y) having all terms in {0,...,n} and 2*w = 2*x + y. - Clark Kimberling, Jun 04 2012

Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. - Omar E. Pol, Aug 04 2012

a(n) is the sum of the largest parts of the partitions of n+1 into exactly 2 parts. - Wesley Ivan Hurt, Jan 26 2013

Conway's relation mentioned by R. K. Guy is a relation between triangular numbers and generalized pentagonal numbers, two sequences from different families, but as triangular numbers are also generalized hexagonal numbers in this case we have a relation between two sequences from the same family. - Omar E. Pol, Feb 01 2013

Start with the sequence of all 0's. Add n to each value of a(n) and the next n - 1 terms. The result is the generalized pentagonal numbers. - Wesley Ivan Hurt, Nov 03 2014

(6k + 1) | a(4k). (3k + 1) | a(4k+1). (3k + 2) | a(4k+2). (6k + 5) | a(4k+3). - Jon Perry, Nov 04 2014

Enge, Hart and Johansson proved: "Every generalised pentagonal number c >= 5 is the sum of a smaller one and twice a smaller one, that is, there are generalised pentagonal numbers a, b < c such that c = 2a + b." (see link theorem 5). - Peter Luschny, Aug 26 2016

The Enge, et al. result for c >= 5 also holds for c >= 2 if 0 is included as a generalized pentagonal number. That is, 2 = 2*1 + 0. - Michael Somos, Jun 02 2018

Suggestion for title, where n actually matches the list and b-file: "Generalized pentagonal numbers: k(n)*(3*k(n) - 1)/2, where k(n) = A001057(n) = [0, 1, -1, 2, -2, 3, -3, ...], n >= 0" - Daniel Forgues, Jun 09 2018 & Jun 12 2018

Generalized k-gonal numbers are the partial sums of the sequence formed by the multiples of (k - 4) and the odd numbers (A005408) interleaved, with k >= 5. - Omar E. Pol, Jul 25 2018

The last digits form a symmetric cycle of length 40 [0, 1, 2, 5, ..., 5, 2, 1, 0], i.e., a(n) == a(n + 40) (mod 10) and a(n) == a(40*k - n - 1) (mod 10), 40*k > n. - Alejandro J. Becerra Jr., Aug 14 2018

Only 2, 5, and 7 are prime. All terms are of the form k*(k+1)/6, where 3 | k or 3 | k+1. For k > 6, the value divisible by 3 must have another factor d > 2, which will remain after the division by 6. - Eric Snyder, Jun 03 2022

8*a(n) is the product of two even numbers one of which is n + n mod 2. - Peter Luschny, Jul 15 2022

a(n) is the dot product of [1, 2, 3, ..., n] and repeat[1, 1/2]. a(5) = 12 = [1, 2, 3, 4, 5] dot [1, 1/2, 1, 1/2, 1] = [1 + 1 + 3 + 2 + 5]. - Gary W. Adamson, Dec 10 2022

Every nonnegative number is the sum of four terms of this sequence [S. Realis]. - N. J. A. Sloane, May 07 2023

From Peter Bala, Jan 06 2025: (Start)

The sequence terms are the exponents in the expansions of the following infinite products:

1) Product_{n >= 1} (1 - s(n)*q^n) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ..., where s(n) = (-1)^(1 + mod(n+1,3)).

2) Product_{n >= 1} (1 - q^(2*n))*(1 - q^(3*n))^2/((1 - q^n)*(1 - q^(6*n))) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ....

3) Product_{n >= 1} (1 - q^n)*(1 - q^(4*n))*(1 - q^(6*n))^5/((1 - q^(2*n))*(1 - q^(3*n))*(1 - q^(12*n)))^2 = 1 - q + q^2 - q^5 - q^7 + q^12 - q^15 + q^22 + q^26 - q^35 + ....

4) Product_{n >= 1} (1 - q^(2*n))^13/((1 - (-1)^n*q^n)*(1 - q^(4*n)))^5 = 1 - 5*q + 7*q^2 - 11*q^5 + 13*q^7 - 17*q^12 + 19*q^15 - + .... See Oliver, Theorem 1.1. (End)

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 7*x^4 + 12*x^5 + 15*x^6 + 22*x^7 + 26*x^8 + 35*x^9 + ...
		

References

• Enoch Haga, A strange sequence and a brilliant discovery, chapter 5 of Exploring prime numbers on your PC and the Internet, first revised ed., 2007 (and earlier ed.), pp. 53-70.
• Ross Honsberger, Ingenuity in Mathematics, Random House, 1970, p. 117.
• Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, (to appear), section 7.2.1.4, equation (18).
• Ivan Niven and Herbert S. Zuckerman, An Introduction to the Theory of Numbers, 2nd ed., Wiley, NY, 1966, p. 231.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000
• G. E. Andrews and J. A. Sellers, Congruences for the Fishburn Numbers, arXiv preprint arXiv:1401.5345 [math.NT], 2014.
• Paul Barry, On sequences with {-1, 0, 1} Hankel transforms, arXiv preprint arXiv:1205.2565 [math.CO], 2012. - From _N. J. A. Sloane_, Oct 18 2012
• Burkard Polster (Mathologer), The hardest "What comes next?" (Euler's pentagonal formula), Youtube video, Oct 17 2020.
• S. Cooper and M. D. Hirschhorn, Results of Hurwitz type for three squares, Discrete Math., Vol. 274, No. 1-3 (2004), pp. 9-24. See P(q).
• Stephen Eberhart, Letter to N. J. A. Sloane, Jan 19 1978.
• John Elias, Illustration of Initial Terms: Generalized Penthexagrams.
• John Elias, Illustration: Star Number Fractal.
• John Elias, Illustration: Generalized Pentagonals In Generalized-Octagonal-Hexagrams.
• Andreas Enge, William Hart, and Fredrik Johansson, Short addition sequences for theta functions, arXiv:1608.06810 [math.NT], 2016.
• Leonhard Euler, Découverte d'une loi tout extraordinaire des nombres par rapport à la somme de leurs diviseurs, Opera Omnia, Series I, Vol. 2 (1751), pp. 241-253.
• Leonhard Euler, On the remarkable properties of the pentagonal numbers, arXiv:math/0505373 [math.HO], 2005.
• Leonhard Euler, De mirabilibus proprietatibus numerorum pentagonalium, par. 2
• Leonhard Euler, Observatio de summis divisorum p. 8.
• Leonhard Euler, An observation on the sums of divisors, p. 8, arXiv:math/0411587 [math.HO], 2004.
• Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math., Vol. 3, No. 2 (2008), pp. 76-114.
• Silvia Heubach and Toufik Mansour, Counting rises, levels and drops in compositions, arXiv:math/0310197 [math.CO], 2003.
• Alfred Hoehn, Illustration of initial terms.
• Barbara H. Margolius, Permutations with inversions, J. Integ. Seq., Vol. 4 (2001), Article 01.2.4.
• Johannes W. Meijer, Euler's Ship on the Pentagonal Sea, pdf and jpg.
• Johannes W. Meijer and Manuel Nepveu, Euler's ship on the Pentagonal Sea, Acta Nova, Vol. 4, No. 1 (December 2008), pp. 176-187.
• Mircea Merca, The Lambert series factorization theorem, The Ramanujan Journal, January 2017; DOI: 10.1007/s11139-016-9856-3.
• Mircea Merca and Maxie D. Schmidt, New Factor Pairs for Factorizations of Lambert Series Generating Functions, arXiv:1706.02359 [math.CO], 2017. See Remark 2.2.
• Mircea Merca, Euler's partition function in terms of 2-adic valuation, Bol. Soc. Mat. Mex. 30, 76 (2024). See p. 3.
• Ivan Niven, Formal power series, Amer. Math. Monthly, Vol. 76, No. 8 (1969), pp. 871-889.
• Robert J. Lemke Oliver, Eta quotients and theta functions, Advances in Mathematics, Vol. 241, Jul. 2013, pp. 1-17.
• Vladimir Pletser, Congruence conditions on the number of terms in sums of consecutive squared integers equal to squared integers, arXiv:1409.7969 [math.NT], 2014.
• Vladimir Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014.
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
• S. Realis, Question 271, Nouv. Corresp. Math., 4 (1878) 27-29.
• Steven J. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5 (December 2011), pp. 339-350.
• André Weil, Two lectures on number theory, past and present, L'Enseign. Math., Vol. XX (1974), pp. 87-110; Oeuvres III, pp. 279-302.
• Eric Weisstein's World of Mathematics, Pentagonal numbers, Partition Function P.
• Eric Weisstein's World of Mathematics, Pentagonal Number Theorem.
• Wikipedia, Pentagonal number theorem.
• M. Wohlgemuth, Pentagon, Kartenhaus und Summenzerlegung.
• Keke Zhang, Generalized Catalan numbers, arXiv:2011.09593 [math.CO], 2020.
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A080995 (characteristic function), A026741 (first differences), A034828 (partial sums), A165211 (mod 2).

Cf. A000326 (pentagonal numbers), A005449 (second pentagonal numbers), A000217 (triangular numbers).

Indices of nonzero terms of A010815, i.e., the (zero-based) indices of 1-bits of the infinite binary word to which the terms of A068052 converge.

Union of A036498 and A036499.

Cf. A153384, A168258, A101688, A174739, A175005.

Cf. A074378, A057569, A057570, A007310.

Sequences of generalized k-gonal numbers: this sequence (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), A195160 (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), A303305 (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), A303814 (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).

Column 1 of A195152.

Squares in APs: A221671, A221672.

Cf. A054440, A260664, A260672.

Quadrisection: A049453(k), A033570(k), A033568(k+1), A049452(k+1), k >= 0.

Cf. A002620.

Programs

GAP

a:=[0,1,2,5];; for n in [5..60] do a[n]:=2*a[n-2]-a[n-4]+3; od; a; # Muniru A Asiru, Aug 16 2018

Haskell

a001318 n = a001318_list !! n
a001318_list = scanl1 (+) a026741_list -- Reinhard Zumkeller, Nov 15 2015

Magma

[(6*n^2 + 6*n + 1 - (2*n + 1)*(-1)^n)/16 : n in [0..50]]; // Wesley Ivan Hurt, Nov 03 2014

Magma

[(3*n^2 + 2*n + (n mod 2) * (2*n + 1)) div 8: n in [0..70]]; // Vincenzo Librandi, Nov 04 2014

Maple

A001318 := -(1+z+z**2)/(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation; gives sequence without initial zero
A001318 := proc(n) (6*n^2+6*n+1)/16-(2*n+1)*(-1)^n/16 ; end proc: # R. J. Mathar, Mar 27 2011

Mathematica

Table[n*(n+1)/6, {n, Select[Range[0, 100], Mod[#, 3] != 1 &]}]
Select[Accumulate[Range[0,200]]/3,IntegerQ] (* Harvey P. Dale, Oct 12 2014 *)
CoefficientList[Series[x (1 + x + x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 70}], x] (* Vincenzo Librandi, Nov 04 2014 *)
LinearRecurrence[{1,2,-2,-1,1},{0,1,2,5,7},70] (* Harvey P. Dale, Jun 05 2017 *)
a[ n_] := With[{m = Quotient[n + 1, 2]}, m (3 m + (-1)^n) / 2]; (* Michael Somos, Jun 02 2018 *)

PARI

{a(n) = (3*n^2 + 2*n + (n%2) * (2*n + 1)) / 8}; /* Michael Somos, Mar 24 2011 */

PARI

{a(n) = if( n<0, n = -1-n); polcoeff( x * (1 - x^3) / ((1 - x) * (1-x^2))^2 + x * O(x^n), n)}; /* Michael Somos, Mar 24 2011 */

PARI

{a(n) = my(m = (n+1) \ 2); m * (3*m + (-1)^n) / 2}; /* Michael Somos, Jun 02 2018 */

Python

def a(n):
    p = n % 2
    return (n + p)*(3*n + 2 - p) >> 3
print([a(n) for n in range(60)])  # Peter Luschny, Jul 15 2022

Python

def A001318(n): return n*(n+1)-(m:=n>>1)*(m+1)>>1 # Chai Wah Wu, Nov 23 2024

Sage

@CachedFunction
def A001318(n):
    if n == 0 : return 0
    inc = n//2 if is_even(n) else n
    return inc + A001318(n-1)
[A001318(n) for n in (0..59)] # Peter Luschny, Oct 13 2012

Formula

Euler: Product_{n>=1} (1 - x^n) = Sum_{n=-oo..oo} (-1)^n*x^(n*(3*n - 1)/2).

A080995(a(n)) = 1: complement of A090864; A000009(a(n)) = A051044(n). - Reinhard Zumkeller, Apr 22 2006

Euler transform of length-3 sequence [2, 2, -1]. - Michael Somos, Mar 24 2011

a(-1 - n) = a(n) for all n in Z. a(2*n) = A005449(n). a(2*n - 1) = A000326(n). - Michael Somos, Mar 24 2011. [The extension of the recurrence to negative indices satisfies the signature (1,2,-2,-1,1), but not the definition of the sequence m*(3*m -1)/2, because there is no m such that a(-1) = 0. - Klaus Purath, Jul 07 2021]

a(n) = 3 + 2*a(n-2) - a(n-4). - Ant King, Aug 23 2011

Product_{k>0} (1 - x^k) = Sum_{k>=0} (-1)^k * x^a(k). - Michael Somos, Mar 24 2011

G.f.: x*(1 + x + x^2)/((1 + x)^2*(1 - x)^3).

a(n) = n*(n + 1)/6 when n runs through numbers == 0 or 2 mod 3. - Barry E. Williams

a(n) = A008805(n-1) + A008805(n-2) + A008805(n-3), n > 2. - Ralf Stephan, Apr 26 2003

Sequence consists of the pentagonal numbers (A000326), followed by A000326(n) + n and then the next pentagonal number. - Jon Perry, Sep 11 2003

a(n) = (6*n^2 + 6*n + 1)/16 - (2*n + 1)*(-1)^n/16; a(n) = A034828(n+1) - A034828(n). - Paul Barry, May 13 2005

a(n) = Sum_{k=1..floor((n+1)/2)} (n - k + 1). - Paul Barry, Sep 07 2005

a(n) = A000217(n) - A000217(floor(n/2)). - Pierre CAMI, Dec 09 2007

If n even a(n) = a(n-1) + n/2 and if n odd a(n) = a(n-1) + n, n >= 2. - Pierre CAMI, Dec 09 2007

a(n)-a(n-1) = A026741(n) and it follows that the difference between consecutive terms is equal to n if n is odd and to n/2 if n is even. Hence this is a self-generating sequence that can be simply constructed from knowledge of the first term alone. - Ant King, Sep 26 2011

a(n) = (1/2)*ceiling(n/2)*ceiling((3*n + 1)/2). - Mircea Merca, Jul 13 2012

a(n) = (A008794(n+1) + A000217(n))/2 = A002378(n) - A085787(n). - Omar E. Pol, Jan 12 2013

a(n) = floor((n + 1)/2)*((n + 1) - (1/2)*floor((n + 1)/2) - 1/2). - Wesley Ivan Hurt, Jan 26 2013

From Oskar Wieland, Apr 10 2013: (Start)

a(n) = a(n+1) - A026741(n),

a(n) = a(n+2) - A001651(n),

a(n) = a(n+3) - A184418(n),

a(n) = a(n+4) - A007310(n),

a(n) = a(n+6) - A001651(n)*3 = a(n+6) - A016051(n),

a(n) = a(n+8) - A007310(n)*2 = a(n+8) - A091999(n),

a(n) = a(n+10)- A001651(n)*5 = a(n+10)- A072703(n),

a(n) = a(n+12)- A007310(n)*3,

a(n) = a(n+14)- A001651(n)*7. (End)

a(n) = (A007310(n+1)^2 - 1)/24. - Richard R. Forberg, May 27 2013; corrected by Zak Seidov, Mar 14 2015; further corrected by Jianing Song, Oct 24 2018

a(n) = Sum_{i = ceiling((n+1)/2)..n} i. - Wesley Ivan Hurt, Jun 08 2013

G.f.: x*G(0), where G(k) = 1 + x*(3*k + 4)/(3*k + 2 - x*(3*k + 2)*(3*k^2 + 11*k + 10)/(x*(3*k^2 + 11*k + 10) + (k + 1)*(3*k + 4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013

Sum_{n>=1} 1/a(n) = 6 - 2*Pi/sqrt(3). - Vaclav Kotesovec, Oct 05 2016

a(n) = Sum_{i=1..n} numerator(i/2) = Sum_{i=1..n} denominator(2/i). - Wesley Ivan Hurt, Feb 26 2017

a(n) = A000292(A001651(n))/A001651(n), for n>0. - Ivan N. Ianakiev, May 08 2018

a(n) = ((-5 + (-1)^n - 6n)*(-1 + (-1)^n - 6n))/96. - José de Jesús Camacho Medina, Jun 12 2018

a(n) = Sum_{k=1..n} k/gcd(k,2). - Pedro Caceres, Apr 23 2019

Quadrisection. For r = 0,1,2,3: a(r + 4*k) = 6*k^2 + sqrt(24*a(r) + 1)*k + a(r), for k >= 1, with inputs (k = 0) {0,1,2,5}. These are the sequences A049453(k), A033570(k), A033568(k+1), A049452(k+1), for k >= 0, respectively. - Wolfdieter Lang, Feb 12 2021

a(n) = a(n-4) + sqrt(24*a(n-2) + 1), n >= 4. - Klaus Purath, Jul 07 2021

Sum_{n>=1} (-1)^(n+1)/a(n) = 6*(log(3)-1). - Amiram Eldar, Feb 28 2022

a(n) = A002620(n) + A008805(n-1). Gary W. Adamson, Dec 10 2022

E.g.f.: (x*(7 + 3*x)*cosh(x) + (1 + 5*x + 3*x^2)*sinh(x))/8. - Stefano Spezia, Aug 01 2024