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Decimal expansion of zeta(5).
+10
125
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
COMMENTS
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
REFERENCES
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.
LINKS
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]. Yong-Cheol Kim, zeta(5) is irrational, arXiv:1105.0730 [math.CA], 2011. [Jonathan Vos Post, May 4, 2011].
FORMULA
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) = (-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).
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
EXAMPLE
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...
Expansion of 1 / Product_{k>=1} (1-x^k)^(k+1).
(Formerly M1601)
+10
34
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
COMMENTS
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
REFERENCES
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.
LINKS
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.
FORMULA
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
EXAMPLE
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 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 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)
MAPLE
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
MATHEMATICA
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 *)
PROG
(PARI) a(n)=polcoeff(prod(i=1, n, (1-x^i+x*O(x^n))^-(i+1)), n)
Decimal expansion of zeta'(-3) (the derivative of Riemann's zeta function at -3).
+10
25
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
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 2.15.1 Generalized Glaisher constants, p. 136-137.
FORMULA
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
EXAMPLE
0.0053785763577743011444169742104138428956644397422955070594470232233245...
MATHEMATICA
Join[{0, 0}, RealDigits[Zeta'[-3], 10, 105] // First]
CROSSREFS
Cf. A000335, A000391, A000417, A000428, A023872, A057527, A057528, A255050, A255052, A258350, A258351, A258352, A260404.
Table T(n,k) by antidiagonals: T(n,k) = number of partitions of n balls of k colors.
+10
12
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
COMMENTS
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
EXAMPLE
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, ...
MAPLE
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:
MATHEMATICA
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 *)
Number of partitions of n objects of 3 colors.
+10
8
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
COMMENTS
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
FORMULA
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
EXAMPLE
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.
MAPLE
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:
MATHEMATICA
nn=30; p=Product[1/(1- x^i)^Binomial[i+2, 2], {i, 1, nn}]; CoefficientList[Series[p, {x, 0, nn}], x]
G.f.: 1 / Product_{n>=0} (1 - x^(n+4))^((n+1)*(n+2)*(n+3)/3!).
+10
4
1, 0, 0, 0, 1, 4, 10, 20, 36, 60, 104, 180, 336, 620, 1174, 2160, 3961, 7100, 12690, 22424, 39651, 69820, 122970, 215904, 378470, 660872, 1150740, 1996200, 3452685, 5952916, 10237576, 17559460, 30049285, 51301020, 87390872, 148534232, 251916041, 426329040, 720003646, 1213481344, 2041155052, 3426721080
COMMENTS
Number of partitions of n objects of 4 colors, where each part must contain at least one of each color. [Conjecture - see comment by Franklin T. Adams-Watters in A052847].
FORMULA
G.f.: exp( Sum_{n>=1} ( x^n/(1-x^n) )^4 /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)/3!.
a(n) ~ Zeta(5)^(109/3600) / (2^(791/1800) * n^(1909/3600) * sqrt(5*Pi)) * exp(11*Zeta'(-1)/6 + log(2*Pi)/2 + Zeta(3)/(4*Pi^2) - Pi^16/(194400000 * Zeta(5)^3) + 11*Pi^8 * Zeta(3)/(108000 * Zeta(5)^2) - Pi^6/(1800*Zeta(5)) - 121*Zeta(3)^2/(360*Zeta(5)) + Zeta'(-3)/6 + (-Pi^12/(1350000 * 2^(2/5) * Zeta(5)^(11/5)) + 11*Pi^4 * Zeta(3)/(900 * 2^(2/5) * Zeta(5)^(6/5)) - Pi^2/(3*2^(7/5) * Zeta(5)^(1/5))) * n^(1/5) + (-Pi^8/(9000 * 2^(4/5) * Zeta(5)^(7/5)) + 11*Zeta(3)/(3*2^(9/5) * Zeta(5)^(2/5))) * n^(2/5) - Pi^4/(90 * 2^(1/5) * Zeta(5)^(3/5)) * n^(3/5) + 5*Zeta(5)^(1/5) / 2^(8/5) * n^(4/5)). - Vaclav Kotesovec, Dec 09 2015
EXAMPLE
G.f.: A(x) = 1 + x^4 + 4*x^5 + 10*x^6 + 20*x^7 + 36*x^8 + 60*x^9 + 104*x^10 + 180*x^11 +...
where
1/A(x) = (1-x^4) * (1-x^5)^4 * (1-x^6)^10 * (1-x^7)^20 * (1-x^8)^35 * (1-x^9)^56 * (1-x^10)^84 * (1-x^11)^120 * (1-x^12)^165 *...
Also,
log(A(x)) = (x/(1-x))^4 + (x^2/(1-x^2))^4/2 + (x^3/(1-x^3))^4/3 + (x^4/(1-x^4))^4/4 + (x^5/(1-x^5))^4/5 + (x^6/(1-x^6))^4/6 +...
MATHEMATICA
nmax = 50; CoefficientList[Series[Product[1/(1-x^k)^((k-3)*(k-2)*(k-1)/6), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 09 2015 *)
PROG
(PARI) {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), ", "))
(PARI) {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), ", "))
(PARI) {L(n) = sumdiv(n, d, d*(d-1)*(d-2)*(d-3)/3! )}
{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), ", "))
Expansion of Product_{k>=1} (1 + x^k)^binomial(k+3,3).
+10
4
1, 4, 16, 64, 221, 736, 2338, 7132, 21093, 60652, 170172, 467140, 1257571, 3325824, 8654576, 22189340, 56116043, 140122760, 345769094, 843827436, 2038017983, 4874329024, 11550814704, 27134195608, 63215468883, 146120097736, 335227455982, 763592477104, 1727482413548
FORMULA
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} ( Sum_{d|k} (-1)^(k/d+1) * A033488(d) ) * a(n-k). a(n) ~ (3*zeta(5))^(1/10) / (2^(7/10) * 5^(2/5) * sqrt(Pi) * n^(3/5)) * exp(-469*log(2)/720 - 2401*Pi^16 / (656100000000*zeta(5)^3) + 539*Pi^8*zeta(3) / (8100000*zeta(5)^2) - 7*Pi^6 / (27000*zeta(5)) - 121*zeta(3)^2 / (600*zeta(5)) + (343*Pi^12 / (303750000 * 2^(3/5) * 15^(1/5) * zeta(5)^(11/5)) - 77*Pi^4*zeta(3) / (4500 * 2^(3/5) * 15^(1/5) * zeta(5)^(6/5)) + Pi^2 / (6*2^(3/5) * (15*zeta(5))^(1/5))) * n^(1/5) + (-49*Pi^8 / (270000 * 2^(1/5) * 15^(2/5) * zeta(5)^(7/5)) + 11*zeta(3) / (4*2^(1/5) * (15*zeta(5))^(2/5))) * n^(2/5) + (7*Pi^4 / (90*2^(4/5) * (15*zeta(5))^(3/5))) * n^(3/5) + (5*(15*zeta(5))^(1/5) / (4*2^(2/5))) * n^(4/5)). - Vaclav Kotesovec, May 12 2021
MATHEMATICA
nmax = 28; CoefficientList[Series[Product[(1 + x^k)^Binomial[k + 3, 3], {k, 1, nmax}], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d Binomial[d + 3, 3], {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 28}]
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