A255052 -id:A255052 - cp's OEIS frontend

A013663 Decimal expansion of zeta(5).

1, 0, 3, 6, 9, 2, 7, 7, 5, 5, 1, 4, 3, 3, 6, 9, 9, 2, 6, 3, 3, 1, 3, 6, 5, 4, 8, 6, 4, 5, 7, 0, 3, 4, 1, 6, 8, 0, 5, 7, 0, 8, 0, 9, 1, 9, 5, 0, 1, 9, 1, 2, 8, 1, 1, 9, 7, 4, 1, 9, 2, 6, 7, 7, 9, 0, 3, 8, 0, 3, 5, 8, 9, 7, 8, 6, 2, 8, 1, 4, 8, 4, 5, 6, 0, 0, 4, 3, 1, 0, 6, 5, 5, 7, 1, 3, 3, 3, 3

Offset: 1

N. J. A. Sloane

In a widely distributed May 2011 email, Wadim Zudilin gave a rebuttal to v1 of Kim's 2011 preprint: "The mistake (unfixable) is on p. 6, line after eq. (3.3). 'Without loss of generality' can be shown to work only for a finite set of n_k's; as the n_k are sufficiently large (and N is fixed), the inequality for epsilon is false." In a May 2013 email, Zudilin extended his rebuttal to cover v2, concluding that Kim's argument "implies that at least one of zeta(2), zeta(3), zeta(4) and zeta(5) is irrational, which is trivial." - Jonathan Sondow, May 06 2013

General: zeta(2*s + 1) = (A000364(s)/A331839(s)) * Pi^(2*s + 1) * Product_{k >= 1} (A002145(k)^(2*s + 1) + 1)/(A002145(k)^(2*s + 1) - 1), for s >= 1. - Dimitris Valianatos, Apr 27 2020

			1/1^5 + 1/2^5 + 1/3^5 + 1/4^5 + 1/5^5 + 1/6^5 + 1/7^5 + ... =
1 + 1/32 + 1/243 + 1/1024 + 1/3125 + 1/7776 + 1/16807 + ... = 1.036927755143369926331365486457...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 262.
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.

Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Michael J. Dancs and Tian-Xiao He, An Euler-type formula for zeta(2k+1), Journal of Number Theory, Volume 118, Issue 2, June 2006, Pages 192-199.
Robert J. Harley, Zeta(3), Zeta(5), .., Zeta(99) 10000 digits (txt, 400 KB).
Yong-Cheol Kim, zeta(5) is irrational, arXiv:1105.0730 [math.CA], 2011. [Jonathan Vos Post, May 4, 2011].
Simon Plouffe, Computation of Zeta(5)
Simon Plouffe, Zeta(5), the sum(1/n**5, n=1..infinity) to 512 digits
Simon Plouffe, Other interesting computations at numberworld.org.
Zhi-Wei Sun, A curious hypergeometric series related to Riemann's zeta function, Question 486353 at MathOverflow, Jan. 21, 2025.
Chuanan Wei, Some fast convergent series for the mathematical constants zeta(4) and zeta(5), arXiv:2303.07887 [math.CO], 2023.
Wikipedia, Zeta constant
Wadim Zudilin, One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational, Russ. Math. Surv., 56 (2001), 774-776.
Index entries for constants related to zeta

Cf. A013661, A002117, A013662, A013664, A013665, A013666, A013667, A013668, A013669, A013670, A013671, A013672, A013673, A013674, A013675, A013676, A013677, A013678, A293904.
Cf. A243264, A255323.
Cf. A023872, A023873, A248882, A255050, A255052, A057528, A260404.

RealDigits[Zeta[5], 10, 100][[1]] (* Alonso del Arte, Jan 13 2012 *)

PARI
```
zeta(5) \\ Michel Marcus, Apr 17 2016
```

From Peter Bala, Dec 04 2013: (Start)
Definition: zeta(5) = Sum_{n >= 1} 1/n^5.
zeta(5) = 2^5/(2^5 - 1)*(Sum_{n even} n^5*p(n)*p(1/n)/(n^2 - 1)^6 ), where p(n) = n^2 + 3. See A013667, A013671 and A013675. (End)
zeta(5) = Sum_{n >= 1} (A010052(n)/n^(5/2)) = Sum_{n >= 1} ((floor(sqrt(n)) - floor(sqrt(n-1)))/n^(5/2)). - Mikael Aaltonen, Feb 22 2015
zeta(5) = Product_{k>=1} 1/(1 - 1/prime(k)^5). - Vaclav Kotesovec, Apr 30 2020
From Artur Jasinski, Jun 27 2020: (Start)
zeta(5) = (-1/30)*Integral_{x=0..1} log(1-x^4)^5/x^5.
zeta(5) = (1/24)*Integral_{x=0..infinity} x^4/(exp(x)-1).
zeta(5) = (2/45)*Integral_{x=0..infinity} x^4/(exp(x)+1).
zeta(5) = (1/(1488*zeta(1/2)^5))*(-5*Pi^5*zeta(1/2)^5 + 96*zeta'(1/2)^5 - 240*zeta(1/2)*zeta'(1/2)^3*zeta''(1/2) + 120*zeta(1/2)^2*zeta'(1/2)*zeta''(1/2)^2 + 80*zeta(1/2)^2*zeta'(1/2)^2*zeta'''(1/2)- 40*zeta(1/2)^3*zeta''(1/2)*zeta'''(1/2) - 20*zeta(1/2)^3*zeta'(1/2)*zeta''''(1/2)+4*zeta(1/2)^4*zeta'''''(1/2)). (End).
From Peter Bala, Oct 29 2023: (Start)
zeta(3) = (8/45)*Integral_{x >= 1} x^3*log(x)^3*(1 + log(x))*log(1 + 1/x^x) dx =  (2/45)*Integral_{x >= 1} x^4*log(x)^4*(1 + log(x))/(1 + x^x) dx.
zeta(5) = 131/128 + 26*Sum_{n >= 1} (n^2 + 2*n + 40/39)/(n*(n + 1)*(n + 2))^5.
zeta(5) =  5162893/4976640 - 1323520*Sum_{n >= 1} (n^2 + 4*n + 56288/12925)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^5. Taking 10 terms of the series gives a value for zeta(5) correct to 20 decimal places.
Conjecture: for k >= 1, there exist rational numbers A(k), B(k) and c(k) such that zeta(5) = A(k) + B(k)*Sum_{n >= 1} (n^2 + 2*k*n + c(k))/(n*(n + 1)*...*(n + 2*k))^5. A similar conjecture can be made for the constant zeta(3). (End)
zeta(5) = (694/204813)*Pi^5 - Sum_{n >= 1} (6280/3251)*(1/(n^5*(exp(4*Pi*n)-1))) + Sum_{n >= 1} (296/3251)*(1/(n^5*(exp(5*Pi*n)-1))) - Sum_{n >= 1} (1073/6502)*(1/(n^5*(exp(10*Pi*n)-1))) + Sum_{n >= 1} (37/6502)*(1/(n^5*(exp(20*Pi*n)-1))). - Simon Plouffe, Jan 06 2024
From Peter Bala, Apr 27 2025: (Start)
zeta(5) = 1/5! * Integral_{x >= 0} x^5 * exp(x)/(exp(x) - 1)^2 dx = (16/15) * 1/5! * Integral_{x >= 0} x^5 * exp(x)/(exp(x) + 1)^2 dx.
zeta(5) = 1/6! * Integral_{x >= 0} x^6 * exp(x)*(exp(x) + 1)/(exp(x) - 1)^3 dx = 1/(3^3 * 5^2) * Integral_{x >= 0} x^6 * exp(x)*(exp(x) - 1)/(exp(x) + 1)^3 dx. (End)
zeta(5) = Sum_{i, j >= 1} 1/((i^4)*j*binomial(i+j, i)). More generally, zeta(n+1) = Sum_{i, j >= 1} 1/((i^n)*j*binomial(i+j, i)) for n >= 1. - Peter Bala, Aug 07 2025

A005380 Expansion of 1 / Product_{k>=1} (1-x^k)^(k+1).

Original entry on oeis.org

1, 2, 6, 14, 33, 70, 149, 298, 591, 1132, 2139, 3948, 7199, 12894, 22836, 39894, 68982, 117948, 199852, 335426, 558429, 922112, 1511610, 2460208, 3977963, 6390942, 10206862, 16207444, 25596941, 40214896, 62868772, 97814358

Offset: 0

N. J. A. Sloane

Also, a(n) = number of partitions of the integer n where there are k+1 different kinds of part k for k = 1, 2, 3, ....
Also, a(n) = number of partitions of n objects of 2 colors. These are set partitions, the n objects are not labeled but colored, using two colors.  For each subset of size k there are k+1 different possibilities, i=0..k white and k-i black objects.
Also, a(n) = number of simple unlabeled graphs with n nodes of 2 colors whose components are complete graphs. - Geoffrey Critzer, Sep 27 2012

			We represent each summand, k, in a partition of n as k identical objects. Then we color each object. We have no regard for the order of the colored objects.
a(3) = 14 because we have:  www; wwb; wbb; bbb; ww + w; ww + b;  wb + w; wb + b; bb + w; bb + b; w + w + w; w + w + b; w + b + b; b + b + b, where the 2 colors are black b and white w. - _Geoffrey Critzer_, Sep 27 2012
a(3) = 14 because we have:  3; 3'; 3''; 3'''; 2 + 1; 2 + 1';  2' + 1; 2' + 1'; 2'' + 1; 2'' + 1'; 1 + 1 + 1; 1 + 1 + 1'; 1 + 1' + 1'; 1' + 1' + 1', where a part k of different sorts is given as k, k', k'', etc. - _Joerg Arndt_, Mar 09 2015
From _Alois P. Heinz_, Mar 09 2015: (Start)
The a(4) = 33 = 5 + 9 + 6 + 8 + 5 partitions of 4 objects of 2 colors are:
5 partitions for the integer partition of 4 = 1 + 1 + 1 + 1:
  01: {{b}, {b}, {b}, {b}}
  02: {{b}, {b}, {b}, {w}}
  03: {{b}, {b}, {w}, {w}}
  04: {{b}, {w}, {w}, {w}}
  05: {{w}, {w}, {w}, {w}}
9 partitions for the integer partition of 4 = 1 + 1 + 2:
  06: {{b}, {b}, {b,b}}
  07: {{b}, {w}, {b,b}}
  08: {{w}, {w}, {b,b}}
  09: {{b}, {b}, {w,b}}
  10: {{b}, {w}, {w,b}}
  11: {{w}, {w}, {w,b}}
  12: {{b}, {b}, {w,w}}
  13: {{b}, {w}, {w,w}}
  14: {{w}, {w}, {w,w}}
6 partitions for the integer partition of 4 = 2 + 2:
  15: {{b,b}, {b,b}}
  16: {{b,b}, {w,b}}
  17: {{b,b}, {w,w}}
  18: {{w,b}, {w,b}}
  19: {{w,b}, {w,w}}
  20: {{w,w}, {w,w}}
8 partitions for the integer partition of 4 = 1 + 3:
  21: {{b}, {b,b,b}}
  22: {{w}, {b,b,b}}
  23: {{b}, {w,b,b}}
  24: {{w}, {w,b,b}}
  25: {{b}, {w,w,b}}
  26: {{w}, {w,w,b}}
  27: {{b}, {w,w,w}}
  28: {{w}, {w,w,w}}
5 partitions for the integer partition of 4 = 4:
  29: {{b,b,b,b}}
  30: {{w,b,b,b}}
  31: {{w,w,b,b}}
  32: {{w,w,w,b}}
  33: {{w,w,w,w}}
Some see number partitions, others see set partitions, ...
(End)
It is obvious from the example of _Alois P. Heinz_ that a(n) enumerates multi-set partitions of a multi-set of n elements of two kinds. In the case that there is only one kind, this reduces to the usual case of numerical partitions. In the case that all the n elements are distinct, then this reduces to the case of set partitions. - _Michael Somos_, Mar 09 2015
There are a(3) = 14 plane partitions of 6 with trace 3; of 7 with trace 4; of 8 with trace 5; etc. See a formula above with the Stanley Exercise 7.99. - _Wolfdieter Lang_, Mar 09 2015
From _Daniel Forgues_, Mar 09 2015: (Start)
The a(3) = 14 = 4 + 6 + 4 partitions of 3 objects of 2 colors are:
4 partitions for the integer partition of 3 = 1 + 1 + 1:
  01: {{b}, {b}, {b}}
  02: {{b}, {b}, {w}}
  03: {{b}, {w}, {w}}
  04: {{w}, {w}, {w}}
6 partitions for the integer partition of 3 = 1 + 2:
  05: {{b}, {b,b}}
  06: {{w}, {b,b}}
  07: {{b}, {w,b}}
  08: {{w}, {w,b}}
  09: {{b}, {w,w}}
  10: {{w}, {w,w}}
4 partitions for the integer partition of 3 = 3:
  11: {{b,b,b}}
  12: {{w,b,b}}
  13: {{w,w,b}}
  14: {{w,w,w}}
The a(2) = 6 = 3 + 3 partitions of 2 objects of 2 colors are:
3 partitions for the integer partition of 2 = 1 + 1:
  01: {{b}, {b}}
  02: {{b}, {w}}
  03: {{w}, {w}}
3 partitions for the integer partition of 2 = 2:
  04: {{b,b}}
  05: {{w,b}}
  06: {{w,w}}
The a(1) = 2 partitions of 1 object of 2 colors are:
2 partitions for the integer partition of 1 = 1:
  01: {{b}}
  02: {{w}}
a(0) = 1: the empty partition, since empty sum is 0.
Triangle(sort of, since n_th row has p(n) = A000041 terms):
  1:  2
  2:  3, 3
  3:  4, 6, 4
  4:  5, 9, 6, 8, 5
  5:  6, ?, ?, ?, ?, ?, 6
  6:  7, ?, ?, ?, ?, ?, ?, ?, ?, ?, 7
Can we find a recurrence relation? (End)

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Exercise 7.99, p. 484 and pp. 548-549.

Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from T. D. Noe)
P. J. Cameron, Some sequences of integers, Discrete Math., 75 (1989), 89-102; also in "Graph Theory and Combinatorics 1988", ed. B. Bollobas, Annals of Discrete Math., 43 (1989), 89-102.
Carlos A. A. Florentino, Plethystic exponential calculus and permutation polynomials, arXiv:2105.13049 [math.CO], 2021. Mentions this sequence.
Vaclav Kotesovec, Graph - The asymptotic ratio
P. A. MacMahon, Memoir on symmetric functions of the roots of systems of equations, Phil. Trans. Royal Soc. London, 181 (1890), 481-536; Coll. Papers II, 32-87.
N. J. A. Sloane, Transforms
R. P. Stanley, Theory and Applications of Plane Partitions: Part 2, Studies in Appl. Math., 1 (1971), 259-279.
R. P. Stanley, The conjugate trace and trace of a plane partition, J. Combin. Theory, vol. A14 53-65 1973, esp. p. 64.

Row sums of A054225.
Column k=2 of A075196.
Cf. A000219, A219555, A217093, A255050, A255052, A089353, A298988.

mul( (1-x^i)^(-i-1),i=1..80); series(%,x,80); seriestolist(%);
# second Maple program:
with(numtheory): etr:= proc(p) local b; b:=proc(n) option remember; local d,j; if n=0 then 1 else add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n)/n fi end end: a:=etr(n-> n+1): seq(a(n), n=0..40); # Alois P. Heinz, Sep 08 2008

max = 31; f[x_] = Product[ 1/(1-x^k)^(k+1), {k, 1, max}]; CoefficientList[ Series[ f[x], {x, 0, max}], x] (* Jean-François Alcover, Nov 08 2011, after g.f. *)
etr[p_] := Module[{b}, b[n_] := b[n] = Module[{d, j}, If[n==0, 1, Sum[ Sum[ d*p[d], {d, Divisors[j]}]*b[n-j], {j, 1, n}]/n]]; b]; a = etr[#+1&]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 23 2015, after Alois P. Heinz *)

a(n)=polcoeff(prod(i=1,n,(1-x^i+x*O(x^n))^-(i+1)),n)

EULER transform of b(n) = n+1.
a(n) ~ Zeta(3)^(13/36) * exp(1/12 - Pi^4/(432*Zeta(3)) + Pi^2 * n^(1/3) / (3*2^(4/3)*Zeta(3)^(1/3)) + 3*Zeta(3)^(1/3) * n^(2/3) / 2^(2/3)) / (A * 2^(23/36) * 3^(1/2) * Pi * n^(31/36)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant and Zeta(3) = A002117 = 1.202056903... . - Vaclav Kotesovec, Mar 07 2015
a(n) = A089353(n+m, m), n >= 1, for each m >= n. a(0) =1. See the Stanley reference, Exercise 7.99. - Wolfdieter Lang, Mar 09 2015
G.f.: exp(Sum_{k>=1} (sigma_1(k) + sigma_2(k))*x^k/k). - Ilya Gutkovskiy, Aug 11 2018

Edited by Christian G. Bower, Sep 07 2002
New name from Joerg Arndt, Mar 09 2015
Restored 1995 name. - N. J. A. Sloane, Mar 09 2015

A259068 Decimal expansion of zeta'(-3) (the derivative of Riemann's zeta function at -3).

Original entry on oeis.org

0, 0, 5, 3, 7, 8, 5, 7, 6, 3, 5, 7, 7, 7, 4, 3, 0, 1, 1, 4, 4, 4, 1, 6, 9, 7, 4, 2, 1, 0, 4, 1, 3, 8, 4, 2, 8, 9, 5, 6, 6, 4, 4, 3, 9, 7, 4, 2, 2, 9, 5, 5, 0, 7, 0, 5, 9, 4, 4, 7, 0, 2, 3, 2, 2, 3, 3, 2, 4, 5, 0, 1, 9, 9, 7, 9, 2, 4, 0, 6, 9, 5, 8, 6, 0, 9, 5, 1, 0, 3, 8, 7, 0, 8, 2, 5, 6, 8, 3, 2, 6, 7, 1, 2, 2, 4, 3

Offset: 0

Jean-François Alcover, Jun 18 2015

			0.0053785763577743011444169742104138428956644397422955070594470232233245...

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 2.15.1 Generalized Glaisher constants, p. 136-137.

G. C. Greubel, Table of n, a(n) for n = 0..10000
Eric Weisstein's MathWorld, Riemann Zeta Function.
Wikipedia, Riemann Zeta Function
Index entries for constants related to zeta

Cf. A000335, A000391, A000417, A000428, A023872, A057527, A057528, A255050, A255052, A258350, A258351, A258352, A260404.

Join[{0, 0}, RealDigits[Zeta'[-3], 10, 105] // First]

zeta'(-n) = (BernoulliB(n+1)*HarmonicNumber(n))/(n+1) - log(A(n)), where A(n) is the n-th Bendersky constant, that is the n-th generalized Glaisher constant.
zeta'(-3) = -11/720 - log(A(3)), where A(3) is A243263.
Equals -11/720 + (gamma + log(2*Pi))/120 - 3*Zeta'(4)/(4*Pi^4), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jul 24 2015

A217093 Number of partitions of n objects of 3 colors.

Original entry on oeis.org

1, 3, 12, 38, 117, 330, 906, 2367, 6027, 14873, 35892, 84657, 196018, 445746, 997962, 2201438, 4792005, 10300950, 21889368, 46012119, 95746284, 197344937, 403121547, 816501180, 1640549317, 3271188702, 6475456896, 12730032791, 24861111315, 48246729411, 93065426256

Offset: 0

Geoffrey Critzer, Sep 26 2012

a(n) is also the number of unlabeled simple graphs with n nodes of 3 colors whose components are complete graphs.
Number of (integer) partitions of n into 3 sorts of part 1, 6 sorts of part 2, 10 sorts of part 3, ..., (k+2)*(k+1)/2 sorts of part k. - Joerg Arndt, Dec 07 2014
In general the g.f. 1 / prod(n>=1, (1-x^k)^m(k) ) gives the number of (integer) partitions where there are m(k) sorts of part k. - Joerg Arndt, Mar 10 2015

			We represent each summand, k, in a partition of n as k identical objects. Then we color each object. We have no regard for the order of the colored objects.
a(2) = 12 because we have: ww; wg; wb; gg; gb; bb; w + w; w + g; w + b; g + g; g + b; b + b, where the 3 colors are white w, gray g, and black b.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
S. Benvenuti, B. Feng, A. Hanany and Y. H. He, Counting BPS operators in gauge theories: Quivers, syzygies and plethystics, arXiv:hep-th/0608050, Aug 2006, p.42.
Carlos A. A. Florentino, Plethystic exponential calculus and permutation polynomials, arXiv:2105.13049 [math.CO], 2021. Mentions this sequence.
Vaclav Kotesovec, Graph - The asymptotic ratio

Cf. A005380, A120844, A253289, A255050, A255052, A255802, A255803, A255835, A255836, A363601, A363602.

Column k=3 of A075196.

with(numtheory):
a:= proc(n) option remember; `if`(n=0, 1, add(add(
      d*binomial(d+2, 2), d=divisors(j))*a(n-j), j=1..n)/n)
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 26 2012
with(numtheory):
series(exp(add(((1/2)*sigma[3](k) + (3/2)*sigma[2](k) + sigma[1](k))*x^k/k, k = 1..30)), x, 31):
seq(coeftayl(%, x = 0, n), n = 0..30); # Peter Bala, Jan 16 2025

nn=30; p=Product[1/(1- x^i)^Binomial[i+2,2],{i,1,nn}]; CoefficientList[Series[p,{x,0,nn}],x]

from functools import lru_cache
from sympy import divisors
@lru_cache(maxsize=None)
def A217093_aux(n): return sum(d*(d+1)*(d+2)>>1 for d in divisors(n,generator=True))
@lru_cache(maxsize=None)
def A217093(n): return 1 if n == 0 else (A217093_aux(n)+sum(A217093_aux(k)*A217093(n-k) for k in range(1,n)))//n # Chai Wah Wu, Mar 19 2025

G.f.: Product_{i>=1} 1/(1-x^i)^binomial(i+2,2).
EULER transform of 3, 6, 10, 15, ... .
Generally for the number of partitions of k colors the generating function is Product_{i>=1} 1/(1-x^i)^binomial(i+k-1,k-1).
a(n) ~ Pi^(1/8) * exp(1/8 + 3^4 * 5^2 * Zeta(3)^3 / (2*Pi^8) - 31*Zeta(3) / (8*Pi^2) + 5^(1/4) * Pi * n^(1/4) / 6^(3/4) - 3^(13/4) * 5^(5/4) * Zeta(3)^2 * n^(1/4) / (2^(7/4) * Pi^5) + 3^(3/2) * 5^(1/2) * Zeta(3) * n^(1/2) / (2^(1/2) * Pi^2) + 2^(7/4) * Pi * n^(3/4) / (3^(5/4) * 5^(1/4))) / (A^(3/2) * 2^(73/32) * 15^(9/32) * n^(25/32)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant and Zeta(3) = A002117 = 1.202056903... . - Vaclav Kotesovec, Mar 08 2015
G.f.: exp(Sum_{k >= 1} ((1/2)*sigma_3(k) + (3/2)*sigma_2(k) + sigma_1(k))*x^k/k) = 1 + 3*x + 12*x^2 + 38*x^3 + 117*x^4 + .... - Peter Bala, Jan 16 2025

A075196 Table T(n,k) by antidiagonals: T(n,k) = number of partitions of n balls of k colors.

Original entry on oeis.org

1, 2, 2, 3, 6, 3, 4, 12, 14, 5, 5, 20, 38, 33, 7, 6, 30, 80, 117, 70, 11, 7, 42, 145, 305, 330, 149, 15, 8, 56, 238, 660, 1072, 906, 298, 22, 9, 72, 364, 1260, 2777, 3622, 2367, 591, 30, 10, 90, 528, 2198, 6174, 11160, 11676, 6027, 1132, 42, 11, 110, 735, 3582, 12292, 28784, 42805, 36450, 14873, 2139, 56

Offset: 1

Christian G. Bower, Sep 07 2002

For k>=1, n->infinity is log(T(n,k)) ~ (1+1/k) * k^(1/(k+1)) * Zeta(k+1)^(1/(k+1)) * n^(k/(k+1)). - Vaclav Kotesovec, Mar 08 2015

			Square array T(n,k) begins:
  1,  2,   3,    4,    5, ...
  2,  6,  12,   20,   30, ...
  3, 14,  38,   80,  145, ...
  5, 33, 117,  305,  660, ...
  7, 70, 330, 1072, 2777, ...

Alois P. Heinz, Rows n = 1..141, flattened

Columns 1-10: A000041, A005380, A217093, A255050, A255052, A270239, A270240, A270241, A270242, A270243.
Rows 1-3: A000027, A002378, A162147.
Main diagonal: A075197.
Cf. A255903.

with(numtheory):
A:= proc(n, k) option remember; local d, j;
      `if`(n=0, 1, add(add(d*binomial(d+k-1, k-1),
       d=divisors(j)) *A(n-j, k), j=1..n)/n)
    end:
seq(seq(A(n, 1+d-n), n=1..d), d=1..12);  # Alois P. Heinz, Sep 26 2012

Transpose[Table[nn=6;p=Product[1/(1- x^i)^Binomial[i+n,n],{i,1,nn}];Drop[CoefficientList[Series[p,{x,0,nn}],x],1],{n,0,nn}]]//Grid  (* Geoffrey Critzer, Sep 27 2012 *)

T(n,k) = Sum_{i=0..k} C(k,i) * A255903(n,i). - Alois P. Heinz, Mar 10 2015

A255050 G.f.: Product_{j>=1} 1/(1-x^j)^binomial(j+3,3).

Original entry on oeis.org

1, 4, 20, 80, 305, 1072, 3622, 11676, 36450, 110240, 324936, 935076, 2635338, 7285560, 19795370, 52930360, 139462956, 362471020, 930186694, 2358867240, 5915606398, 14680528648, 36073675792, 87816701332, 211891552280, 506981067168, 1203337174120, 2834401172172

Offset: 0

Vaclav Kotesovec, Mar 08 2015

nonn

Number of partitions of n unlabeled objects of 4 colors. - Peter Dolland, Feb 20 2025

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
Vaclav Kotesovec, Graph - The asymptotic ratio

Column k=4 of A075196.

Cf. A000292, A005380, A217093, A255052.

with(numtheory):
a:= proc(n) option remember; local d, j; `if`(n=0, 1,
      add(add(d*binomial(d+3, 3), d=divisors(j))*a(n-j), j=1..n)/n)
    end:
seq(a(n), n=0..50); # after Alois P. Heinz

nmax=50; CoefficientList[Series[Product[1/(1-x^j)^Binomial[j+3,3],{j,1,nmax}],{x,0,nmax}],x]

G.f.: Product_{j>=1} 1/(1-x^j)^C(j+3,3).
a(n) ~ Zeta(5)^(829/3600) * exp(11/72 - Zeta(3)/(4*Pi^2) + Zeta'(-3)/6 - 121*Zeta(3)^2 / (360*Zeta(5)) - Pi^6/(1800*Zeta(5)) + 11*Pi^8*Zeta(3) / (108000*Zeta(5)^2) - Pi^16/(194400000*Zeta(5)^3) + Pi^2 * n^(1/5)/ (6*2^(2/5) * Zeta(5)^(1/5)) - 11*Pi^4 * Zeta(3) * n^(1/5) / (900*2^(2/5)*Zeta(5)^(6/5)) + Pi^12 * n^(1/5) / (1350000 * 2^(2/5) * Zeta(5)^(11/5)) + 11*Zeta(3) * n^(2/5) / (6*2^(4/5) * Zeta(5)^(2/5)) - Pi^8 * n^(2/5) / (9000 * 2^(4/5) * Zeta(5)^(7/5)) + Pi^4 * n^(3/5) / (90 * 2^(1/5) * Zeta(5)^(3/5)) + 5 * Zeta(5)^(1/5) * n^(4/5) / 2^(8/5)) / (A^(11/6) * 2^(971/1800) * 5^(1/2) * Pi * n^(2629/3600)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant, Zeta(3) = A002117 = 1.202056903..., Zeta(5) = A013663 = 1.036927755... and Zeta'(-3) = ((gamma + log(2*Pi) - 11/6)/30 - 3*Zeta'(4)/Pi^4)/4 = 0.0053785763577743... .
EULER transform of 1, 4, 10, 20, 35, 56, 84, ... (= A000292(n+1)). - Peter Dolland, Feb 20 2025

A264925 G.f.: 1 / Product_{n>=0} (1 - x^(n+5))^((n+1)(n+2)(n+3)*(n+4)/4!).

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 5, 15, 35, 70, 127, 215, 360, 605, 1080, 2003, 3890, 7570, 14715, 27960, 52255, 95705, 173295, 311060, 557400, 999032, 1795880, 3235130, 5835955, 10521060, 18931287, 33956485, 60692510, 108087835, 191883595, 339724144, 600203700, 1058605775, 1864535670, 3279862975, 5762287759, 10109925380

Offset: 0

Paul D. Hanna, Nov 28 2015

nonn

Number of partitions of n objects of 5 colors, where each part must contain at least one of each color. [Conjecture - see comment by Franklin T. Adams-Watters in A052847].

			G.f.: A(x) = 1 + x^5 + 5*x^6 + 15*x^7 + 35*x^8 + 70*x^9 + 127*x^10 + 215*x^11 + 360*x^12 +...
where
1/A(x) = (1-x^5) * (1-x^6)^5 * (1-x^7)^15 * (1-x^8)^35 * (1-x^9)^70 * (1-x^10)^126 * (1-x^11)^210 * (1-x^12)^330 * (1-x^13)^495 *...
Also,
log(A(x)) = (x/(1-x))^5 + (x^2/(1-x^2))^5/2 + (x^3/(1-x^3))^5/3 + (x^4/(1-x^4))^5/4 + (x^5/(1-x^5))^5/5 + (x^6/(1-x^6))^5/6 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..5000

Cf. A052847, A264923, A264924, A264926.

Cf. A000391, A255052.

nmax = 50; CoefficientList[Series[Product[1/(1-x^k)^((k-4)*(k-3)*(k-2)*(k-1)/24), {k,1,nmax}], {x,0,nmax}], x] (* Vaclav Kotesovec, Dec 09 2015 *)

{a(n) = my(A=1); A = prod(k=0,n, 1/(1 - x^(k+4) +x*O(x^n) )^((k+1)*(k+2)*(k+3)/3!) ); polcoeff(A,n)}
for(n=0,50,print1(a(n),", "))

{a(n) = my(A=1); A = exp( sum(k=1,n+1, (x^k/(1 - x^k))^4 /k +x*O(x^n) ) ); polcoeff(A,n)}
for(n=0,50,print1(a(n),", "))

{L(n) = sumdiv(n,d, d*(d-1)*(d-2)*(d-3)*(d-4)/4!)}
{a(n) = my(A=1); A = exp( sum(k=1,n+1, L(k) * x^k/k +x*O(x^n) ) ); polcoeff(A,n)}
for(n=0,50,print1(a(n),", "))

G.f.: exp( Sum_{n>=1} ( x^n/(1-x^n) )^5 /n ).
G.f.: exp( Sum_{n>=1} L(n) * x^n/n ), where L(n) = Sum_{d|n} d*(d-1)*(d-2)*(d-3)*(d-4)/4!.
a(n) ~ Pi^(95/288) / (2 * 3^(527/576) * 7^(239/1728) * n^(1103/1728)) * exp(-25*Zeta'(-1)/12 - log(2*Pi)/2 + 595*Zeta(3)/(48*Pi^2) - 29291*Zeta(5) / (128*Pi^4) - 2480625 * Zeta(3) * Zeta(5)^2 / (2*Pi^12) + 72930375 * Zeta(5)^3 / (2*Pi^14) - 1063324867500 * Zeta(5)^5/Pi^24 - 5*Zeta'(-3)/12 + (41 * 7^(1/6) * Pi/(768*sqrt(3)) - 2625 * sqrt(3) * 7^(1/6) * Zeta(3) * Zeta(5)/(2*Pi^7) + 540225 * sqrt(3) * 7^(1/6) * Zeta(5)^2/(16*Pi^9) - 4740474375 * sqrt(3) * 7^(1/6) * Zeta(5)^4/(4*Pi^19)) * n^(1/6) + (-25 * 7^(1/3) * Zeta(3)/(4*Pi^2) + 735 * 7^(1/3) * Zeta(5) /(8*Pi^4) - 3969000 * 7^(1/3) * Zeta(5)^3 / Pi^14) * n^(1/3) + (7*sqrt(7/3)*Pi/24 - 4725 * sqrt(21) * Zeta(5)^2 / Pi^9) * sqrt(n) - 45 * 7^(2/3) * Zeta(5)/(2*Pi^4) * n^(2/3) + 2*sqrt(3)*Pi / (5*7^(1/6)) * n^(5/6)). - Vaclav Kotesovec, Dec 09 2015

cp's OEIS Frontend

A013663 Decimal expansion of zeta(5).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A005380 Expansion of 1 / Product_{k>=1} (1-x^k)^(k+1).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A259068 Decimal expansion of zeta'(-3) (the derivative of Riemann's zeta function at -3).

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

A217093 Number of partitions of n objects of 3 colors.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A075196 Table T(n,k) by antidiagonals: T(n,k) = number of partitions of n balls of k colors.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A255050 G.f.: Product_{j>=1} 1/(1-x^j)^binomial(j+3,3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A264925 G.f.: 1 / Product_{n>=0} (1 - x^(n+5))^((n+1)*(n+2)*(n+3)*(n+4)/4!).

Views

Author

Keywords

A264925 G.f.: 1 / Product_{n>=0} (1 - x^(n+5))^((n+1)(n+2)(n+3)*(n+4)/4!).