A002117 -id:A002117 - cp's OEIS frontend

A253905 Decimal expansion of zeta(3)/zeta(2).

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

Offset: 0

Geoffrey Critzer, Jan 18 2015

Three positive integers b, c, m are randomly selected (with replacement) from {1, 2, ..., n}. Let P(n) be the probability that the congruence b * x == c (mod m) has a solution. zeta(3)/zeta(2) is the limit of P(n) as n goes to infinity.

			0.73076296940143849872603673130771463952801160507937...

Steven R. Finch, Mathematical Constants II, Encyclopedia of Mathematics and Its Applications, Cambridge University Press, Cambridge, 2018, p. 154.

Cf. A002117, A013661, A023900, A222056.

Drop[Flatten[RealDigits[N[Zeta[3]/Zeta[2], 75]]], -2]

zeta(3)/zeta(2) \\ Charles R Greathouse IV, Apr 20 2016

Equals A002117/A013661.
Equals Product_{p prime} (1 - 1/(p^2 + p + 1)). - Amiram Eldar, Jun 11 2023
Equals Sum_{k>=1} A023900(k)/k^3. - Amiram Eldar, Jan 25 2024

A255269 a(n) = Product_{k=1..n} k!^k.

Original entry on oeis.org

1, 4, 864, 286654464, 7132880358604800000, 993710590042385551668019200000000000, 82086865668400428790437436119503664712777728000000000000000000

Offset: 1

Vaclav Kotesovec, Feb 20 2015

nonn

Eric Weisstein's World of Mathematics, Hyperfactorial
Eric Weisstein's World of Mathematics, Superfactorial

Cf. A000178, A002109, A036740, A055462, A255268.

Table[Product[k!^k,{k,1,n}],{n,1,10}]
FoldList[Times,Table[(k!)^k,{k,10}]] (* Harvey P. Dale, Aug 16 2021 *)

a(n) = A255268(n) / A055462(n-1).

a(n) ~ sqrt(A) * exp((3 - 45*n^2 - 32*n^3 - 9*Zeta(3)/Pi^2)/72) * n^((8*n^3 + 18*n^2 + 10*n + 1)/24) * (2*Pi)^(n*(n+1)/4), where A = A074962 = 1.28242712910062263687534256886979... is the Glaisher-Kinkelin constant and Zeta(3) = A002117 = 1.2020569031595942853997... .

A062775 Number of Pythagorean triples mod n: total number of solutions to x^2 + y^2 = z^2 mod n.

Original entry on oeis.org

1, 4, 9, 24, 25, 36, 49, 96, 99, 100, 121, 216, 169, 196, 225, 448, 289, 396, 361, 600, 441, 484, 529, 864, 725, 676, 891, 1176, 841, 900, 961, 1792, 1089, 1156, 1225, 2376, 1369, 1444, 1521, 2400, 1681, 1764, 1849, 2904, 2475, 2116, 2209, 4032, 2695, 2900

Offset: 1

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 18 2001

a(n) is multiplicative and, for a prime p, a(p) = p^2. Hence a(n) = n^2 if n is squarefree.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe, terms 1001..5000 from Seiichi Manyama)
Gottfried Helms, Pythagorean triples mod n / Solution enhanced, newsgroup sci.math.research, 2003.
László Tóth, Counting solutions of quadratic congruences in several variables revisited,arXiv:1404.4214 [math.NT], 2014.
László Tóth, Counting Solutions of Quadratic Congruences in Several Variables Revisited, J. Int. Seq. 17 (2014), Article 14.11.6.
Index to sequences related to sums of squares.

Cf. A091143 (number of solutions to x^2 + y^2 = z^2 mod 2^n).
Cf. A060968, A087687, A002117, A013662.
Number of solutions to x^k + y^k = z^k mod n: this sequence (k=2), A063454 (k=3), A288099 (k=4), A288100 (k=5), A288101 (k=6), A288102 (k=7), A288103 (k=8), A288104 (k=9), A288105 (k=10).

A062775 := proc(n)
    a := 1;
    for pe in ifactors(n)[2] do
        p := op(1,pe) ;
        e := op(2,pe) ;
        if p = 2 then
            if type(e,'odd') then
                a := a*p^((3*e+1)/2)*(2^((e+1)/2)-1) ;
            else
                a := a*p^(3*e/2)*(2^(e/2+1)-1) ;
            end if;
        else
            if type(e,'odd') then
                a := a*p^((3*e-1)/2)*(p^((e+1)/2)+p^((e-1)/2)-1) ;
            else
                a := a*p^(3*e/2-1)*(p^(e/2+1)+p^(e/2)-1) ;
            end if;
        end if;
    end do:
    a ;
end proc:
seq(A062775(n),n=1..100) ; # R. J. Mathar, Jun 25 2018

Table[cnt=0; Do[If[Mod[x^2+y^2-z^2, n]==0, cnt++ ], {x, 0, n-1}, {y, 0, n-1}, {z, 0, n-1}]; cnt, {n, 50}]
f[p_, e_] := If[OddQ[e], p^(3*(e+1)/2 - 2)*(p^((e+1)/2) + p^((e-1)/2) - 1), p^(3*e/2 - 1) * (p^(e/2 + 1) + p^(e/2) - 1)]; f[2, e_] := If[OddQ[e], 2^(3*(e+1)/2 - 1)*(2^((e+1)/2) - 1), 2^(3*e/2)*(2^(e/2+1)-1)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Oct 18 2022 *)

a(n) is multiplicative. For the powers of primes p, there are four cases. For p=2, there are cases for even and odd powers: a(2^(2k-1)) = 2^(3k-1) (2^k-1) and a(2^(2k)) = 2^(3k) (2^(k+1)-1). Similarly, for odd primes p, a(p^(2k-1)) = p^(3k-2) (p^k+p^(k-1)-1) and a(p^(2k)) = p^(3k-1) (p^(k+1)+p^k-1). - T. D. Noe, Dec 22 2003
From  Gottfried Helms, May 13 2004: (Start)
If the canonical form of n is n = 2^i * 3^j * 5^k *... * p^q, then it appears that a(n) = n * f(2, i) * f(3, j) * f(5, k) * ... * f(p, q), where f(p, 1) = p for any prime p; f(2, i) = 2^i + 2^i - 2^ceiling(i/2); f(p, i) = p^i + p^(i-1) - p^floor((i-1)/2) for any odd prime p.
For example, a(7) = 49 because a(7) = 7*f(7, 1) = 7*7; a(16) = 448 because a(16) = a(2^4) = 16 * f(2, 4) = 16 * (16+16-4) = 16*28 = 448; a(12) = 216 because a(12) = a(3*2^2) = 12*f(2, 2)*f(3, 1) = 12*(4+4-2)*3 = 216. (End)
Sum_{k=1..n} a(k) ~ c * n^3, where c = (16/45) * Product_{p prime} (1 + 1/(p^3 + p^2 + p)) = (16/45)*zeta(3)/zeta(4) = 0.39488943478263044166... . - Amiram Eldar, Oct 18 2022, Nov 30 2023

More terms from Sascha Kurz, Mar 25 2002

A067558 Sum of squares of proper divisors of n.

Original entry on oeis.org

0, 1, 1, 5, 1, 14, 1, 21, 10, 30, 1, 66, 1, 54, 35, 85, 1, 131, 1, 146, 59, 126, 1, 274, 26, 174, 91, 266, 1, 400, 1, 341, 131, 294, 75, 615, 1, 366, 179, 610, 1, 736, 1, 626, 341, 534, 1, 1106, 50, 755, 299, 866, 1, 1184, 147, 1114, 371, 846, 1, 1860, 1, 966, 581, 1365

Offset: 1

Reinhard Zumkeller, Jan 29 2002

			a(12) = 1^2 + 2^2 + 3^2 + 4^2 + 6^2 = 1 + 4 + 9 + 16 + 36 = 66.

T. D. Noe, Table of n, a(n) for n=1..10000

Cf. A001157, A002117, A032741.

Table[DivisorSigma[2, n] - n^2, {n, 1, 64}] (* Jean-François Alcover, Mar 01 2019 *)

a(n)=sigma(n,2)-n^2 \\ Charles R Greathouse IV, Dec 07 2011

a(n) = A001157(n) - n^2.
a(n) = 1 if and only if n is prime.
Dirichlet g.f.: zeta(s-2)*(zeta(s) - 1). - Ilya Gutkovskiy, Sep 08 2016
Sum_{k=1..n} a(k) ~ (zeta(3)-1) * n^3 / 3. - Amiram Eldar, Dec 31 2024

A152651 Decimal expansion of 3Zeta(5) - Zeta(3)Pi^2/6.

Original entry on oeis.org

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

Offset: 1

R. J. Mathar, Dec 10 2008

A division by 2 is missing in Mezo's penultimate formula on page 4.

			Equals 1.1334789151328136607970110178859769320890912918456042272...

David Borwein and J. M. Borwein, On an intriguing integral and some series related to zeta(4), Proc. Am. Math. Soc. 123 (1995) 1191-1198.
Istvan Mezo, Summation of Hyperharmonic Numbers, arXiv:0811.0042 [math.CO], 2008.

RealDigits[3*Zeta[5]-Zeta[3]*Pi^2/6,10,120][[1]] (* Harvey P. Dale, Apr 29 2019 *)

3*zeta(5) - zeta(3)*Pi^2/6 \\ Michel Marcus, Jul 07 2015

Equals Sum_(j >= 1) H(j)/j^4 = where H(j) = A001008(j)/A002805(j).

Equals 3*A013663 - A002117*A013661.

A255835 G.f.: Product_{k>=1} (1+x^k)^(2*k-1).

Original entry on oeis.org

1, 1, 3, 8, 15, 34, 67, 133, 255, 486, 901, 1649, 2984, 5312, 9373, 16342, 28221, 48283, 81928, 137858, 230278, 381919, 629156, 1029933, 1675856, 2711288, 4362575, 6983196, 11122327, 17630798, 27820283, 43706461, 68375137, 106534093, 165340844, 255643289

Offset: 0

Vaclav Kotesovec, Mar 07 2015

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000

Cf. A026007, A219555, A052812, A253289, A255834.

nmax=50; CoefficientList[Series[Product[(1+x^k)^(2*k-1),{k,1,nmax}],{x,0,nmax}],x]

a(n) ~ Zeta(3)^(1/6) * exp(-Pi^4 / (2592*Zeta(3)) - Pi^2 * n^(1/3) / (12*(3*Zeta(3))^(1/3)) + 3^(4/3)/2 * Zeta(3)^(1/3) * n^(2/3)) / (2^(1/6) * 3^(1/3) * sqrt(Pi) * n^(2/3)), where Zeta(3) = A002117.

G.f.: exp(Sum_{k>=1} (-1)^(k+1)*x^k*(1 + x^k)/(k*(1 - x^k)^2)). - Ilya Gutkovskiy, Jun 07 2018

A002760 Squares and cubes.

Original entry on oeis.org

0, 1, 4, 8, 9, 16, 25, 27, 36, 49, 64, 81, 100, 121, 125, 144, 169, 196, 216, 225, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024, 1089, 1156, 1225, 1296, 1331, 1369, 1444, 1521, 1600, 1681, 1728, 1764, 1849

Offset: 1

N. J. A. Sloane

nonn

Catalan's Conjecture states that 8 and 9 are the only pair of consecutive numbers in this sequence. The conjecture was established in 2003 by Mihilescu.

Subsequence of A022549. - Reinhard Zumkeller, Jul 17 2010

Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 68.
Clifford A. Pickover, The Math Book, Sterling, NY, 2009; see p. 236.

Michael De Vlieger, Table of n, a(n) for n = 1..10443 (first 1000 terms from Zak Seidov)
Yuri F. Bilu, Catalan's Conjecture (After Mihilescu), Astérisque, No. 294, 1-26, 2004.
Yuri F. Bilu, Catalan Without Logarithmic Forms (after Bugeaud, Hanrot and Mihailescu), J. Théor. Nombres Bordeaux 17, 69-85, 2005.
David Masser, Alan Baker, arXiv:2010.10256 [math.HO], 2020. See p. 4.
Tauno Metsänkylä, Catalan's conjecture: another old Diophantine problem solved, Bull. Amer. Math. Soc. (NS), Vol. 41, No. 1 (2004), pp. 43-57.
Preda Mihǎilescu, A Class Number Free Criterion for Catalan's Conjecture, J. Number Th. 99 225-231, 2003.
Preda Mihǎilescu, Primary Cyclotomic Units and a Proof of Catalan's Conjecture, J. Reine angew. Math. 572 (2004): 167-195. MR 2076124.
Paulo Ribenboim, Catalan's Conjecture, Séminaire de Philosophie et Mathématiques, 6 (1994), p. 1-11.
Paulo Ribenboim, Catalan's Conjecture, Amer. Math. Monthly, Vol. 103(7) Aug-Sept 1996, pp. 529-538.

Cf. A131799; union of A000290 and A000578.
First differences in A075052. [From Zak Seidov, May 10 2010]
Cf. A002117, A013661, A013664.

[n: n in [0..1600] | IsIntegral(n^(1/3)) or IsIntegral(n^(1/2))]; // Bruno Berselli, Feb 09 2016

nMax=2000;Union[Range[0,nMax^(1/2)]^2,Range[0,nMax^(1/3)]^3] (* Vladimir Joseph Stephan Orlovsky, Apr 11 2011 *)
nxt[n_] := Min[ Floor[1 + Sqrt[n]]^2, Floor[1 + n^(1/3)]^3]; NestList[ nxt, 0, 55] (* Robert G. Wilson v, Aug 16 2014 *)

isok(n) = issquare(n) || ispower(n, 3); \\ Michel Marcus, Mar 29 2016

from math import isqrt
from sympy import integer_nthroot
def A002760(n):
    def f(x): return n-1+x+integer_nthroot(x,6)[0]-integer_nthroot(x,3)[0]-isqrt(x)
    m, k = n-1, f(n-1)
    while m != k:
        m, k = k, f(k)
    return m # Chai Wah Wu, Aug 09 2024

Sum_{n>=2} 1/a(n) = zeta(2) + zeta(3) - zeta(6). - Amiram Eldar, Dec 19 2020

A006501 Expansion of (1+x^2) / ( (1-x)^2 * (1-x^3)^2 ).

Original entry on oeis.org

1, 2, 4, 8, 12, 18, 27, 36, 48, 64, 80, 100, 125, 150, 180, 216, 252, 294, 343, 392, 448, 512, 576, 648, 729, 810, 900, 1000, 1100, 1210, 1331, 1452, 1584, 1728, 1872, 2028, 2197, 2366, 2548, 2744, 2940, 3150, 3375, 3600, 3840, 4096, 4352, 4624, 4913, 5202

Offset: 0

N. J. A. Sloane

a(n+3) = maximal product of three numbers with sum n: a(n) = max(r*s*t), n = r+s+t. - Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jul 10 2003
It appears that k is a term of the sequence if and only if k is a positive integer such that floor(v) * ceiling(v) * round(v) = k, where v = k^(1/3). - John W. Layman, Mar 21 2012
The sequence floor(n/3)*floor((n+1)/3)*floor((n+2)/3) is essentially the same: 0, 0, 0, 1, 2, 4, 8, 12, 18, 27, 36, 48, 64, 80, 100, 125, 150, 180, 216, 252, ... - N. J. A. Sloane, Dec 27 2013

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
G. E. Bergum and V. E. Hoggatt, Jr., A combinatorial problem involving recursive sequences and tridiagonal matrices, Fib. Quart., 16 (1978), 113-118.
Dhruv Mubayi, Counting substructures II: Hypergraphs, Combinatorica 33 (2013), no. 5, 591--612. MR3132928.
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.
Index entries for linear recurrences with constant coefficients, signature (2,-1,2,-4,2,-1,2,-1).

Cf. A000578, A002117, A144677, A210433.

Maximal product of k positive integers with sum n, for k = 2..10: A002620 (k=2), this sequence (k=3), A008233 (k=4), A008382 (k=5), A008881 (k=6), A009641 (k=7), A009694 (k=8), A009714 (k=9), A354600 (k=10).

A006501:=(1+z**2)/(z**2+z+1)**2/(z-1)**4; # Simon Plouffe in his 1992 dissertation

CoefficientList[Series[(1+x^2)/(1-x)^2 /(1-x^3)^2,{x,0,50}],x] (* Vincenzo Librandi, Jun 16 2012 *)

a(n) = [(n+3)/3] * [(n+4)/3] * [(n+5)/3]. - Reinhard Zumkeller, May 18 2004
a(n-3) = Sum_{k=0..n} [k/3]*[(k+1)/3]. - Mitch Harris, Dec 02 2004
Conjecture: a(n) = A144677(n) + A144677(n-2). - R. J. Mathar, Mar 15 2011
Sum_{n>=0} 1/a(n) = 1 + zeta(3). - Amiram Eldar, Jan 10 2023
a(3*m) = (m+1)^3 (A000578). - Bernard Schott, Feb 22 2023

More terms from Reinhard Zumkeller, May 18 2004

A091360 Partial sums of A000219.

Original entry on oeis.org

1, 2, 5, 11, 24, 48, 96, 182, 342, 624, 1124, 1983, 3462, 5947, 10114, 16993, 28290, 46624, 76225, 123555, 198833, 317627, 504102, 794885, 1246079, 1942112, 3010857, 4643515, 7126749, 10886361, 16555324, 25067633, 37801062, 56776035, 84951990, 126643036, 188127997, 278507781, 410949776, 604437277, 886284200, 1295668181

Offset: 0

Christian G. Bower, Jan 02 2004

nonn

Convergent of columns of A091355.

Joerg Arndt and Vaclav Kotesovec, Table of n, a(n) for n = 0..5000 (first 500 terms from Joerg Arndt)
N. J. A. Sloane, Transforms

Cf. A000219, A002117, A074962.

CoefficientList[Series[1/(1-x)*Product[1/(1-x^k)^k,{k,1,50}],{x,0,50}],x] (* Vaclav Kotesovec, Aug 16 2015 *)

N=66; x='x+O('x^N); Vec( 1/((1-x)*prod(n=1,N, (1-x^n)^n )) ) \\ Joerg Arndt, Mar 15 2014

Euler transform of 2, 2, 3, 4, 5, 6, 7, 8, 9, ...
G.f.: 1/( (1-x) * prod(n>=1, (1-x^n)^n ) ). [Joerg Arndt, Mar 15 2014]
From Vaclav Kotesovec, Aug 16 2015: (Start)
a(n) = Sum_{k=0..n} A000219(k).
a(n) ~ (n/(2*Zeta(3)))^(1/3) * A000219(n).
a(n) ~ exp(1/12 + 3 * Zeta(3)^(1/3) * n^(2/3) / 2^(2/3)) / (A * 2^(23/36) * sqrt(3*Pi) * Zeta(3)^(5/36) * n^(13/36)), where Zeta(3) = A002117 and A = A074962 is the Glaisher-Kinkelin constant.
(End)
G.f.: exp(Sum_{k>=1} (sigma_2(k) + 1)*x^k/k). - Ilya Gutkovskiy, Aug 21 2018

A206622 G.f.: Product_{n>0} ( (1+x^n)/(1-x^n) )^(n^2).

Original entry on oeis.org

1, 2, 10, 36, 118, 376, 1148, 3376, 9654, 26894, 73192, 195188, 510948, 1315048, 3332720, 8326448, 20529526, 49998884, 120379574, 286726340, 676057144, 1578880480, 3654180236, 8385122192, 19085029540, 43103203626, 96630606968, 215105226728, 475608824400

Offset: 0

Paul D. Hanna, Feb 10 2012

nonn

Compare g.f. to: Product_{n>0} (1+x^n)/(1-x^n) = exp( Sum_{n>=1} (sigma(2*n) - sigma(n))*x^n/n ) which equals 1/theta_4(x) = 1/(1 + 2*Sum_{n>=1} (-x)^(n^2)).
Convolution of A023871 and A027998. - Vaclav Kotesovec, Aug 19 2015
In general, if g.f. = Product_{k>=1} ((1 + x^k)/(1 - x^k))^(c2*k^2 + c1*k + c0) and c2>0, then a(n) ~ exp(Pi * 2^(5/4) * c2^(1/4) * n^(3/4) / 3 + 7*c1 * Zeta(3) * sqrt(n) / (Pi^2 * sqrt(2*c2)) + (c0*Pi / (2^(5/4) * c2^(1/4)) - 49*c1^2 * Zeta(3)^2 / (2^(5/4) * c2^(5/4) * Pi^5)) * n^(1/4) + 22411 * c1^3 * Zeta(3)^3 / (196 * c2^2 * Pi^8) - 7*c0*c1 * Zeta(3) / (4*c2 * Pi^2) - c2 * Zeta(3) / (4*Pi^2) + c1/12) * Pi^(c1/12) * c2^(1/8 + c0/8 + c1/48) / (A^c1 * 2^(15/8 + 11*c0/8 + 7*c1/48) * n^(5/8 + c0/8 + c1/48)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Nov 08 2017
Let A(x) denote the g.f. and let m be an integer. Define a sequence by u(n) = [x^n] A(x)^(m*n). We conjecture that the supercongruence u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) holds for all positive integers n and r and all primes p >= 5. Cf. A380582. - Peter Bala, Jan 21 2025

			G.f.: A(x) = 1 + 2*x + 10*x^2 + 36*x^3 + 118*x^4 + 376*x^5 + 1148*x^6 +...
where A(x) = (1+x)/(1-x) * (1+x^2)^4/(1-x^2)^4 * (1+x^3)^9/(1-x^3)^9 *...
Also, A(x) = Euler transform of [2,7,18,28,50,63,98,112,162,175,...]:
A(x) = 1/((1-x)^2*(1-x^2)^7*(1-x^3)^18*(1-x^4)^28*(1-x^5)^50*(1-x^6)^63*...).

Seiichi Manyama, Table of n, a(n) for n = 0..9042 (terms 0..1000 from Vaclav Kotesovec)
Vaclav Kotesovec, A method of finding the asymptotics of q-series based on the convolution of generating functions, arXiv:1509.08708 [math.CO], Sep 30 2015, p. 23.

Cf. A156616, A206623, A206624, A001158 (sigma_3), A380582.

nmax = 40; CoefficientList[Series[Product[((1+x^k)/(1-x^k))^(k^2), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 19 2015 *)

{a(n)=polcoeff(prod(m=1,n+1,((1+x^m)/(1-x^m+x*O(x^n)))^(m^2)),n)}

{a(n)=polcoeff(exp(sum(m=1, n, (sigma(2*m, 3)-sigma(m, 3))/4*x^m/m)+x*O(x^n)), n)}

{a(n)=local(InvEulerGF=x*(2+7*x+12*x^2+7*x^3+2*x^4)/(1-x^2+x*O(x^n))^3);polcoeff(1/prod(k=1,n,(1-x^k+x*O(x^n))^polcoeff(InvEulerGF,k)),n)}
for(n=0,35,print1(a(n),", "))

G.f.: exp( Sum_{n>=1} (sigma_3(2*n) - sigma_3(n))/4 * x^n/n ), where sigma_3(n) is the sum of cubes of divisors of n (A001158).
The inverse Euler transform has g.f.: x*(2 + 7*x + 12*x^2 + 7*x^3 + 2*x^4)/(1-x^2)^3.
a(n) ~ exp(2^(5/4)*Pi*n^(3/4)/3 - Zeta(3)/(4*Pi^2)) / (2^(15/8) * n^(5/8)), where Zeta(3) = A002117. - Vaclav Kotesovec, Aug 19 2015
a(0) = 1, a(n) = (2/n)*Sum_{k=1..n} A007331(k)*a(n-k) for n > 0. - Seiichi Manyama, Apr 30 2017

cp's OEIS Frontend

A253905 Decimal expansion of zeta(3)/zeta(2).

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A255269 a(n) = Product_{k=1..n} k!^k.

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A062775 Number of Pythagorean triples mod n: total number of solutions to x^2 + y^2 = z^2 mod n.

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A067558 Sum of squares of proper divisors of n.

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A152651 Decimal expansion of 3*Zeta(5) - Zeta(3)*Pi^2/6.

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A255835 G.f.: Product_{k>=1} (1+x^k)^(2*k-1).

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A002760 Squares and cubes.

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A006501 Expansion of (1+x^2) / ( (1-x)^2 * (1-x^3)^2 ).

A152651 Decimal expansion of 3Zeta(5) - Zeta(3)Pi^2/6.