A380290 -id:A380290 - cp's OEIS frontend

A023871 Expansion of Product_{k>=1} (1 - x^k)^(-k^2).

1, 1, 5, 14, 40, 101, 266, 649, 1593, 3765, 8813, 20168, 45649, 101591, 223654, 486046, 1045541, 2225167, 4692421, 9804734, 20318249, 41766843, 85218989, 172628766, 347338117, 694330731, 1379437080, 2724353422, 5350185097, 10449901555, 20304465729, 39254599832

Offset: 0

Olivier Gérard

In general, if g.f. = Product_{k>=1} 1/(1 - x^k)^(c2*k^2 + c1*k + c0) and c2 > 0, then a(n) ~ exp(4*Pi * c2^(1/4) * n^(3/4) / (3*15^(1/4)) + c1*Zeta(3) / Pi^2 * sqrt(15*n/c2) + (Pi * 5^(1/4) * c0 / (2*3^(3/4) * c2^(1/4)) - 15^(5/4) * c1^2 * Zeta(3)^2 / (2*c2^(5/4) * Pi^5)) * n^(1/4) + c1/12 + 75 * c1^3 * Zeta(3)^3 / (c2^2 * Pi^8) - 5*c0 * c1 * Zeta(3) / (4*c2 * Pi^2) - c2*Zeta(3) / (4*Pi^2)) * Pi^(c1/12) * (c2/15)^(1/8 + c0/8 + c1/48) / (A^c1 * 2^((c0 + 3)/2) * n^(5/8 + c0/8 + c1/48)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Nov 08 2017
Let A(x) = Product_{k >= 1} (1 - x^k)^(-k^2). The sequence defined by u(n) := [x^n] A(x)^n is conjectured to satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 7 and all positive integers n and r. See A380290. - Peter Bala, Feb 02 2025
a(n) is the number of partitions of n where there are k^2 sorts of part k. - Joerg Arndt, Feb 02 2025

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (first 1001 terms from Alois P. Heinz)
G. Almkvist, Asymptotic formulas and generalized Dedekind sums, Exper. Math., 7 (No. 4, 1998), pp. 343-359.
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.

Euler transform of squares (A000290).
Cf. A000219, A023872-A023878, A294530, A380290.
Column k=2 of A144048. - Alois P. Heinz, Nov 02 2012

m:=40; R:=PowerSeriesRing(Rationals(), m); Coefficients(R! ( (&*[1/(1-x^k)^k^2: k in [1..m]]) )); // G. C. Greubel, Oct 29 2018

with(numtheory):
a:= proc(n) option remember; `if`(n=0, 1,
      add(add(d*d^2, d=divisors(j)) *a(n-j), j=1..n)/n)
    end:
seq(a(n), n=0..35); # Alois P. Heinz, Nov 02 2012

max = 31; Series[ Product[ 1/(1-x^k)^k^2, {k, 1, max}], {x, 0, max}] // CoefficientList[#, x]& (* Jean-François Alcover, Mar 05 2013 *)

m=40; x='x+O('x^m); Vec(prod(k=1, m, 1/(1-x^k)^k^2)) \\ G. C. Greubel, Oct 29 2018

# uses[EulerTransform from A166861]
b = EulerTransform(lambda n: n^2)
print([b(n) for n in range(32)]) # Peter Luschny, Nov 11 2020

a(n) = 1/n * Sum_{k=1..n} a(n-k)*sigma_3(k), n > 0, a(0)=1, where sigma_3(n) = A001158(n) = sum of cubes of divisors of n. - Vladeta Jovovic, Jan 20 2002
G.f.: Prod_{n>=1} exp(sigma_3(n)*x^n/n), where sigma_3(n) is the sum of cubes of divisors of n (=A001158(n)). - N-E. Fahssi, Mar 28 2010
G.f. (conjectured): 1/Product_{n>=1} E(x^n)^J2(n) where E(x) = Product_{n>=1} 1-x^n and J2(n) = A007434(n) [follows from the identity Sum_{d|n} J2(d) = n^2 - Peter Bala, Feb 02 2025]. - Joerg Arndt, Jan 25 2011
a(n) ~ exp(4 * Pi * n^(3/4) / (3^(5/4) * 5^(1/4)) - Zeta(3) / (4*Pi^2)) / (2^(3/2) * 15^(1/8) * n^(5/8)), where Zeta(3) = A002117 = 1.2020569031595942853997... . - Vaclav Kotesovec, Feb 27 2015

Definition corrected by Franklin T. Adams-Watters and R. J. Mathar, Dec 04 2006

A255672 Coefficient of x^n in Product_{k>=1} 1/(1-x^k)^(k*n).

Original entry on oeis.org

1, 1, 7, 37, 215, 1251, 7459, 44885, 272727, 1668313, 10263057, 63423482, 393440867, 2448542136, 15280435191, 95588065737, 599213418327, 3763242239317, 23673166664695, 149138199543613, 940796936557265, 5941862248557566, 37568309060087582, 237767215209245583

Offset: 0

Vaclav Kotesovec, Mar 01 2015

nonn

Number of partitions of n when parts i are of n*i kinds. - Alois P. Heinz, Nov 23 2018
From Peter Bala, Apr 18 2023: (Start)
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and all positive integers n and k.
Conjecture: the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all primes p >= 3 and all positive integers n and k. (End)

Alois P. Heinz, Table of n, a(n) for n = 0..1000 (first 501 terms from Vaclav Kotesovec)

Cf. A008485, A252782, A270913, A270922, A380290.

Main diagonal of A255961.

b:= proc(n, k) option remember; `if`(n=0, 1, k*add(
      b(n-j, k)*numtheory[sigma][2](j), j=1..n)/n)
    end:
a:= n-> b(n$2):
seq(a(n), n=0..30);  # Alois P. Heinz, Mar 11 2015

Table[SeriesCoefficient[Product[1/(1-x^k)^(k*n),{k,1,n}],{x,0,n}], {n,0,20}] (* Vaclav Kotesovec, Mar 01 2015 *)

a(n) ~ c * d^n / sqrt(n), where d = 6.468409145117839606941857350154192468889057616577..., c = 0.25864792865819067933968646380369970564... . - Vaclav Kotesovec, Mar 01 2015

a(n) = [x^n] exp(n*Sum_{k>=1} x^k/(k*(1 - x^k)^2)). - Ilya Gutkovskiy, May 30 2018

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.

Original entry on oeis.org

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).

A380291 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} (1 + x^k)^(k^2) is the g.f. of A027998.

Original entry on oeis.org

1, 1, 9, 64, 425, 3026, 21672, 157095, 1149289, 8464240, 62683134, 466307865, 3482008904, 26083955002, 195932407939, 1475267031164, 11131100990825, 84140066313620, 637054366975740, 4830417047590165, 36674477204674750, 278779034863684377, 2121418004609211361, 16159262748227985561

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 supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 7 and positive integers n and r. Some examples are given below.

			Examples of supercongruences:
a(7) - a(1) = 157095 - 1 = 2*(7^3)*229 == 0 (mod 7^3)
a(11) - a(1) = 466307865 - 1 = (2^3)*(11^3)*43793 == 0 (mod 11^3)
a(3*7) - a(3) = 278779034863684377 - 64 = (7^4)*43*26891*100413601 == 0 (mod 7^4)

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

Cf. A027998, A078307, A380290.

Cf. A270913, A270922.

with(numtheory):
s_3 := n-> add((-1)^(n/d+1)*d^3, d in divisors(n)):
G(x) := series(exp(add(s_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 + 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[Sum[(-1)^(k/d + 1)*d^3, {d, Divisors[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} s_3(k)*x^k/k), where s_3(n) = Sum_{d divides n} (-1)^(n/d+1)*d^3 = A078307(n).

a(n) ~ c * d^n / sqrt(n), where d = 7.7846790125019502578773343468308844201627754275100035492213697757399421948... and c = 0.2484592487737716543953469621097743519172686743284742545545347906986158... - Vaclav Kotesovec, Jul 30 2025

A380582 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^k)/(1 - x^k))^(k^2) is the g.f. of A206622.

Original entry on oeis.org

1, 2, 24, 236, 2432, 25752, 277152, 3019088, 33186816, 367378814, 4089875024, 45741207228, 513537853952, 5784253405192, 65332622356032, 739706089046736, 8392732289277952, 95401363286044260, 1086232605119042424, 12386037358495697292, 141422619808922418432, 1616691574828234720352

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, B be integers and let C be a positive integer. Define u(n) = [x^(C*n)] Product_{k >= 1} ((1 + x^k)^A * (1 - x^k)^B)^(k^2). The present sequence is the case A = 1, B = -1 and C = 1. We conjecture that the above supercongruences also hold for the sequence {u(n)} for all primes p >= 7.

			Examples of supercongruences:
a(7) - a(1) = 3019088 - 2 = 2*(3^3)*(7^3)*163 == 0 (mod 7^3)
a(13) - a(1) = 5784253405192 - 2 = 2*5*(13^4)*20252279 == 0 (mod 13^4)
a(2*11) - a(2) = 18501616629347623668448 - 24 = (2^3)*(11^3)*17*1951*4243*9817*1257719 == 0 (mod 11^3)
a(5^2) - a(5) = 1884578634304981694792832319004 - 256504 = (2^2)*(5^6)*193381* 155926684363405438573 == 0 (mod 5^6)

Cf. A001158, A015128, A206622, A380290, A380581, A380583.

with(numtheory):
G(x) := series(exp(add( (1/4)*(sigma[3](2*k) - sigma[3](k))*x^k/k, k = 1..23 )),x,24):
seq(coeftayl(G(x)^n, x = 0, n), n = 0..23);

a(n) = [x^n] exp( n*Sum_{k >= 1} (sigma_3(2*k) - sigma_3(k))/4 * x^k/k ).

A380583 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^(2*k))/(1 - x^k))^(k^2).

Original entry on oeis.org

1, 1, 13, 82, 665, 5026, 40180, 319677, 2583401, 20965150, 171276238, 1405008925, 11571476120, 95601033542, 792038546739, 6577523807332, 54737967873385, 456368114019558, 3811136362823056, 31873576059000827, 266919720010452190, 2237944814420991135, 18784073017650350445

Offset: 0

Peter Bala, Jan 27 2025

The sequence satisfies the Gauss congruences: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and for all positive integers n and r. In fact, stronger congruences appear to hold for this sequence.
We conjecture that a(p) == 1 (mod p^3) for all primes p >= 7.
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.
Let A and B be integers. Let C be a positive integer. Define u(n) = [x^(C*n)] Product_{k >= 1} ((1 + - x^(2*k))^A * (1 + - x^k)^B)^(k^2). The present sequence is the case A = 1, B = -1 and C = 1, with the appropriate choice of signs. We conjecture that the above supercongruences also hold for the sequence {u(n)} for all primes p >= 7.

			Examples of supercongruences:
a(7) - a(1) = 319677 - 1 = (2^2)*(7^3)*233 == 0 (mod 7^3)
a(11) - a(1) = 1405008925 - 1 = (2^2)*3*(11^5)*727 == 0 (mod 11^5)
a(22) - a(2) = 18784073017650350445 - 13 = (2^5)*(11^3)*222773*1979699077 == 0 (mod 11^3)

Cf. A380290, A380581, A380582.

G(x) := series(mul( ( (1 + x^(2*k))/(1 - x^k) )^(k^2), k = 1..22), x, 23):
a:= n-> coeftayl(G(x)^n, x = 0, n):
seq(a(n), n = 0..22);

A386720 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^3) is the g.f. of A023872.

Original entry on oeis.org

1, 1, 19, 163, 1571, 15276, 152029, 1525420, 15460771, 157716235, 1617959044, 16672687769, 172459185341, 1789587777849, 18621317408384, 194222638392213, 2029985619026851, 21256104343844595, 222937740908641405, 2341629730618924374, 24627719497316157396, 259326672761381979574

Offset: 0

Vaclav Kotesovec, Jul 31 2025

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..960

Cf. A023872.

Cf. A008485, A255672, A380290.

with(numtheory):
G(x) := series(exp(add(sigma[4](k)*x^k/k, k = 1..25)), x, 26):
seq(coeftayl(G(x)^n, x = 0, n), n = 0..25);

Table[SeriesCoefficient[Product[1/(1-x^k)^(n*k^3), {k, 1, n}], {x, 0, n}], {n, 0, 25}]
Table[SeriesCoefficient[Exp[n*Sum[DivisorSigma[4, k]*x^k/k, {k, 1, n}]], {x, 0, n}], {n, 0, 25}]

a(n) = [x^n] exp(n*Sum_{k >= 1} sigma_4(k)*x^k/k).

a(n) ~ c * d^n / sqrt(n), where d = 10.783710654896500462544161711323081108292517438268962307143535279238... and c = 0.2464683956609371456774144752559018514863700235623819263696832303304...

cp's OEIS Frontend

A023871 Expansion of Product_{k>=1} (1 - x^k)^(-k^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

SageMath

Formula

Extensions

A255672 Coefficient of x^n in Product_{k>=1} 1/(1-x^k)^(k*n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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.

Views

Author

Keywords

Comments

Examples

References

Crossrefs

Programs

Maple

Formula

A380291 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} (1 + x^k)^(k^2) is the g.f. of A027998.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A380582 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^k)/(1 - x^k))^(k^2) is the g.f. of A206622.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Formula

A380583 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^(2*k))/(1 - x^k))^(k^2).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

A386720 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^3) is the g.f. of A023872.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Formula