A023873 -id:A023873 - cp's OEIS frontend

A380581 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^4) is the g.f. of A023873.

1, 1, 35, 397, 5075, 67126, 897911, 12144945, 165880531, 2280262825, 31522512910, 437730330357, 6101414176535, 85317965576325, 1196299277106675, 16813979471920522, 236812229975204563, 3341448338530887015, 47225228515043980715, 668417245247747877735, 9473101371364286661950, 134416752857691389968377, 1909344928242571795580255

Offset: 0

Peter Bala, Jan 27 2025

Given an integer sequence {f(n) : n >= 0} with f(0) = 1, there is a unique power series F(x) with rational coefficients, where F(0) = 1, such that f(n) = [x^n] F(x)^n. F(x) is given by F(x) = series_reversion(x/E(x)), where E(x) = exp(Sum_{n >= 1} f(n)*x^n/n). Furthermore, if the series E(x) has integer coefficients then the series F(x) also has integer coefficients and the sequence {f(n)} satisfies the Gauss congruences: f(n*p^r) == f(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r (by Stanley, Ch. 5, Ex. 5.2(a), p. 72 and the Lagrange inversion formula).
Thus the present sequence satisfies the Gauss congruences. In fact, stronger congruences appear to hold for this sequence.
We conjecture that a(p) == 1 (mod p^3) for all primes p >= 5 (checked up to p = 61).
More generally, we conjecture that the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 5 and positive integers n and r. Some examples are given below.
Let A >= 1 and B be integers. Define a sequence {u(n) : n >= 0} by u(n) = [x^(A*n)] G(x)^(B*n), so the present sequence is the case A = B = 1. We conjecture that the above supercongruences also hold for the sequence {u(n)}.

			Examples of supercongruences:
a(7) - a(1) = 12144945 - 1 = (2^4)*(7^3)*2213 = 0 (mod 7^3)
a(3*7) - a(3) = 134416752857691389968377 - 397 = (2^2)*5*(7^3)*17*223*5168630662682423 == 0 (mod 7^3)
a(2*11) - a(2) = 1909344928242571795580255 - 35 = (2^2)*(3^4)*5*7*(11^4)*17*23* 29411951377843 == 0 (mod 11^4)

R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Cf. A001160, A023873, A380290, A380582, A380583.

with(numtheory):
G(x) := series(exp(add(sigma[5](k)*x^k/k, k = 1..22)), x, 23):
seq(coeftayl(G(x)^n, x = 0, n), n = 0..22);

a(n) = [x^n] exp(n*Sum_{k >= 1} sigma_5(k)*x^k/k).

A013663 Decimal expansion of zeta(5).

Original entry on oeis.org

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

A144048 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is Euler transform of (j->j^k).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 6, 5, 1, 1, 9, 14, 13, 7, 1, 1, 17, 36, 40, 24, 11, 1, 1, 33, 98, 136, 101, 48, 15, 1, 1, 65, 276, 490, 477, 266, 86, 22, 1, 1, 129, 794, 1828, 2411, 1703, 649, 160, 30, 1, 1, 257, 2316, 6970, 12729, 11940, 5746, 1593, 282, 42, 1, 1, 513

Offset: 0

Alois P. Heinz, Sep 08 2008

In general, column k > 0 is asymptotic to (Gamma(k+2)*Zeta(k+2))^((1-2*Zeta(-k)) /(2*k+4)) * exp((k+2)/(k+1) * (Gamma(k+2)*Zeta(k+2))^(1/(k+2)) * n^((k+1)/(k+2)) + Zeta'(-k)) / (sqrt(2*Pi*(k+2)) * n^((k+3-2*Zeta(-k))/(2*k+4))). - Vaclav Kotesovec, Mar 01 2015

			Square array begins:
  1,  1,   1,   1,    1,     1, ...
  1,  1,   1,   1,    1,     1, ...
  2,  3,   5,   9,   17,    33, ...
  3,  6,  14,  36,   98,   276, ...
  5, 13,  40, 136,  490,  1828, ...
  7, 24, 101, 477, 2411, 12729, ...

Alois P. Heinz, Antidiagonals = 0..99, flattened
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. 21.
N. J. A. Sloane, Transforms

Columns k=0-9 give: A000041, A000219, A023871, A023872, A023873, A023874, A023875, A023876, A023877, A023878.
Rows give: 0-1: A000012, 2: A000051, A094373, 3: A001550, 4: A283456, 5: A283457.
Main diagonal gives A252782.
Cf. A283272.

with(numtheory): etr:= proc(p) local b; b:= proc(n) option remember; `if`(n=0,1, add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n)/n) end end: A:= (n,k)-> etr(j->j^k)(n); seq(seq(A(n,d-n), n=0..d), d=0..13);

etr[p_] := Module[{ b}, b[n_] := b[n] = If[n == 0, 1, Sum[Sum[d*p[d], {d, Divisors[j]}]*b[n - j], {j, 1, n}]/n]; b]; A[n_, k_] := etr[Function[j, j^k]][n]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 13}] // Flatten (* Jean-François Alcover, Dec 27 2013, translated from Maple *)

G.f. of column k: Product_{j>=1} 1/(1-x^j)^(j^k).

A283271 Expansion of exp( Sum_{n>=1} -sigma_5(n)*x^n/n ) in powers of x.

Original entry on oeis.org

1, -1, -16, -65, -55, 807, 4809, 13135, 550, -169070, -862710, -2281174, -1221309, 20194565, 114391575, 346400092, 486546751, -1239516671, -11089537215, -41702958960, -93143227027, -45337210750, 674845109986, 3682196642725, 11405949184465, 20796945542222

Offset: 0

Seiichi Manyama, Mar 04 2017

sign

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 >= 7. Cf. A380581. - Peter Bala, Jan 21 2025

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Column k=4 of A283272.
Cf. A023873 (exp( Sum_{n>=1} sigma_5(n)*x^n/n )).
Cf. exp( Sum_{n>=1} -sigma_k(n)*x^n/n ): A010815 (k=1), A073592 (k=2), A283263 (k=3), A283264 (k=4), this sequence (k=5).

G.f.: Product_{n>=1} (1 - x^n)^(n^4).

a(n) = -(1/n)*Sum_{k=1..n} sigma_5(k)*a(n-k).

A206624 G.f.: Product_{n>0} ( (1+x^n)/(1-x^n) )^(n^4).

Original entry on oeis.org

1, 2, 34, 228, 1414, 8872, 52876, 301136, 1662614, 8929406, 46738920, 239036116, 1197187780, 5882369976, 28397283056, 134864166352, 630819797174, 2908948327780, 13236421303742, 59477002686404, 264104800719672, 1159649708139680, 5037895127964316

Offset: 0

Paul D. Hanna, Feb 12 2012

nonn

Convolution of A023873 and A248883. - Vaclav Kotesovec, Aug 19 2015
In general, for m >= 0, if g.f. = Product_{k>=1} ((1+x^k)/(1-x^k))^(k^m), then a(n) ~ ((2^(m+2)-1) * Gamma(m+2) * Zeta(m+2) / (2^(2*m+3) * n))^((1-2*Zeta(-m))/(2*m+4)) * exp((m+2)/(m+1) * ((2^(m+2)-1) * n^(m+1) * Gamma(m+2) * Zeta(m+2) / 2^(m+1))^(1/(m+2)) + Zeta'(-m)) / sqrt((m+2)*Pi*n). - Vaclav Kotesovec, Aug 19 2015
If m is even and m >= 2, then can be simplified as: a(n) ~ ((2^(m+2)-1) * Gamma(m+2) * Zeta(m+2) / (2^(2*m+3) * n))^(1/(2*m+4)) * exp((m+2)/(m+1) * ((2^(m+2)-1) * n^(m+1) * Gamma(m+2) * Zeta(m+2) / 2^(m+1))^(1/(m+2)) + (-1)^(m/2) * Gamma(m+1) * Zeta(m+1) / (2^(m+1) * Pi^m)) / sqrt((m+2)*Pi*n). - Vaclav Kotesovec, Aug 19 2015

			G.f.: A(x) = 1 + 2*x + 18*x^2 + 88*x^3 + 398*x^4 + 1768*x^5 + 7508*x^6 +...
where A(x) = (1+x)/(1-x) * (1+x^2)^16/(1-x^2)^16 * (1+x^3)^81/(1-x^3)^81 *...
Also, A(x) = Euler transform of [2,31,162,496,1250,2511,4802,7936,...]:
A(x) = 1/((1-x)^2*(1-x^2)^31*(1-x^3)^162*(1-x^4)^496*(1-x^5)^1250*(1-x^6)^2511*...).

Seiichi Manyama, Table of n, a(n) for n = 0..2916 (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. A015128 (m=0), A156616 (m=1), A206622 (m=2), A206623 (m=3), A001160 (sigma_5).

nmax = 40; CoefficientList[Series[Product[((1+x^k)/(1-x^k))^(k^4), {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^4)),n)}

{a(n)=polcoeff(exp(sum(m=1, n, (sigma(2*m, 5)-sigma(m, 5))/16*x^m/m)+x*O(x^n)), n)}

{a(n)=local(InvEulerGF=x*(2+31*x+152*x^2+341*x^3+460*x^4+341*x^5+152*x^6+31*x^7+2*x^8)/(1-x^2+x*O(x^n))^5); polcoeff(1/prod(k=1,n,(1-x^k+x*O(x^n))^polcoeff(InvEulerGF,k)),n)}
for(n=0,30,print1(a(n),", "))

G.f.: exp( Sum_{n>=1} (sigma_5(2*n) - sigma_5(n))/16 * x^n/n ), where sigma_5(n) is the sum of 5th powers of divisors of n (A001160).
Inverse Euler transform has g.f.: x*(2 + 31*x + 152*x^2 + 341*x^3 + 460*x^4 + 341*x^5 + 152*x^6 + 31*x^7 + 2*x^8)/(1-x^2)^5.
a(n) ~ exp(3*2^(2/3)*Pi*n^(5/6)/5 + 3*Zeta(5)/(4*Pi^4)) / (2^(7/6) * 3^(1/2) * n^(7/12)), where Zeta(5) = A013663. - Vaclav Kotesovec, Aug 19 2015
a(0) = 1, a(n) = (2/n)*Sum_{k=1..n} A096960(k)*a(n-k) for n > 0. - Seiichi Manyama, Apr 30 2017

A380290 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^2) is the g.f. of A023871.

Original entry on oeis.org

1, 1, 11, 73, 539, 3976, 30107, 229811, 1771803, 13749742, 107305836, 841211966, 6619647419, 52258136399, 413682035393, 3282569032273, 26101575743771, 207930807629248, 1659134361686186, 13258065574274885, 106084302933126364, 849845499077000534, 6815530442695480418, 54712839001004065090

Offset: 0

Peter Bala, Jan 19 2025

Given an integer sequence {f(n) : n >= 0} with f(0) = 1, there is a unique power series F(x) with rational coefficients, where F(0) = 1, such that f(n) = [x^n] F(x)^n. F(x) is given by F(x) = series_reversion(x/E(x)), where E(x) = exp(Sum_{n >= 1} f(n)*x^n/n). Furthermore, if the series E(x) has integer coefficients then the series F(x) also has integer coefficients and the sequence {f(n)} satisfies the Gauss congruences: f(n*p^r) == f(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r (by Stanley, Ch. 5, Ex. 5.2(a), p. 72 and the Lagrange inversion formula).
Thus the present sequence satisfies the Gauss congruences. In fact, stronger congruences appear to hold for the present sequence.
We conjecture that a(p) == 1 (mod p^3) for all primes p >= 7 (checked up to p = 61).
More generally, we conjecture that the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 7 and positive integers n and r. Some examples are given below.

			Examples of supercongruences:
a(7) - a(1) = 229811 - 1 = 2*5*(7^3)*67 == 0 (mod 7^3)
a(3*7) - a(3) = 849845499077000534 - 73 = (7^3)*29243*84727410689 == 0 (mod 7^3)
a(19) - a(1) = 13258065574274885 - 1 = (2^2)*11*(19^3)*29*26723*56687 == 0 (mod 19^3)

R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
Peter Bala, Notes on A380290 and A380291

Cf. A023871, A001158, A023873, A283263, A380291, A380581, A380582, A380583.

Cf. A008485, A255672, A386720.

with(numtheory):
G(x) := series(exp(add(sigma[3](k)*x^k/k, k = 1..23)),x,24):
seq(coeftayl(G(x)^n, x = 0, n), n = 0..23);

Table[SeriesCoefficient[Product[1/(1 - x^k)^(n*k^2), {k, 1, n}], {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jul 30 2025 *)
(* or *)
Table[SeriesCoefficient[Exp[n*Sum[DivisorSigma[3, k]*x^k/k, {k, 1, n}]], {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jul 30 2025 *)

a(n) = [x^n] exp(n*Sum_{k >= 1} sigma_3(k)*x^k/k).

a(n) ~ c * d^n / sqrt(n), where d = 8.20432131153340331179513077696629277558952852444670658917204305357709... and c = 0.2513708881073263860977360125648021910598660424705749139651716452651... - Vaclav Kotesovec, Jul 30 2025

cp's OEIS Frontend

A380581 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^4) is the g.f. of A023873.

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A013663 Decimal expansion of zeta(5).

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A144048 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is Euler transform of (j->j^k).

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A283271 Expansion of exp( Sum_{n>=1} -sigma_5(n)*x^n/n ) in powers of x.

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A206624 G.f.: Product_{n>0} ( (1+x^n)/(1-x^n) )^(n^4).

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A380290 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^2) is the g.f. of A023871.

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A343284 Dirichlet g.f.: Product_{k>=2} 1 / (1 - k^(-s))^(k^4).

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