A014153 Expansion of 1/((1-x)^2*Product_{k>=1} (1-x^k)).

1, 3, 7, 14, 26, 45, 75, 120, 187, 284, 423, 618, 890, 1263, 1771, 2455, 3370, 4582, 6179, 8266, 10980, 14486, 18994, 24757, 32095, 41391, 53123, 67865, 86325, 109350, 137979, 173450, 217270, 271233, 337506, 418662, 517795, 638565, 785350, 963320, 1178628

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Comments

Number of partitions of n with three kinds of 1. E.g., a(2)=7 because we have 2, 1+1, 1+1', 1+1", 1'+1', 1'+1", 1"+1". - Emeric Deutsch, Mar 22 2005

Partial sums of the partial sums of the partition numbers A000041. Partial sums of A000070. Euler transform of 3,1,1,1,...

Also sum of parts, counted without multiplicity, in all partitions of n, offset 1. Also Sum phi(p), where the sum is taken over all parts p of all partitions of n, offset 1. - Vladeta Jovovic, Mar 26 2005

Equals row sums of triangle A141157. - Gary W. Adamson, Jun 12 2008

A014153 convolved with A010815 = (1, 2, 3, ...). n-th partial sum sequence of A000041 convolved with A010815 = (n-1)-th column of Pascal's triangle, starting (1, n, ...). - Gary W. Adamson, Nov 09 2008

From Omar E. Pol, May 25 2012: (Start)

a(n) is also the sum of all parts of the (n+1)st column of a version of the section model of partitions in which every section has its parts aligned to the right margin (cf. A210953, A210970, A135010).

Rows of triangle A210952 converge to this sequence. (End)

Using the above result (see Jovovic's comment) of Jovovic and Mertens's theorem on the average order of the phi function, we can obtain the estimate a(n-1) = (6/Pi^2)*n*p(n) + O(log(n)*A006128(n)), where p(n) is the partition function A000041(n). It can be shown that A006128(n) = O(sqrt(n)*log(n)*p(n)), so we have the asymptotic result a(n) ~ (6/Pi^2)*n*p(n). - Peter Bala, Dec 23 2013

a(n-2) is the number of partitions of 2n or 2n-1 with palindromicity 2; that is, partitions that can be listed in palindromic order except for a central sequence of two distinct parts. - Gregory L. Simay, Nov 01 2015

Convolution of A000041 and A000027. - Omar E. Pol, Jun 17 2021

Convolution of A002865 and the positive terms of A000217. Partial sums give A014160. - Omar E. Pol, Mar 01 2023

Links

• Alois P. Heinz and Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Alois P. Heinz)
• Mircea Merca and Maxie D. Schmidt, The partition function p(n) in terms of the classical Möbius function, arXiv:2310.13658 [math.CO], 2023.

Crossrefs

Cf. A000041, A000070, A116930, A141157.

Cf. A010815. - Gary W. Adamson, Nov 09 2008

Column k=3 of A292508.

Cf. A000217, A002865, A014160.

Programs

Magma

m:=45; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( 1/((1-x)^2*(&*[1-x^k: k in [1..50]])) )); // G. C. Greubel, Oct 15 2018

Maple

with(numtheory):
a:= proc(n) option remember;
      `if`(n=0, 1, add((2+sigma(j)) *a(n-j), j=1..n)/n)
    end:
seq(a(n), n=0..40);  # Alois P. Heinz, Feb 13 2012

Mathematica

a[n_] := a[n] = If[n == 0, 1, Sum[(2+DivisorSigma[1, j])*a[n-j], {j, 1, n}]/n]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 03 2014, after Alois P. Heinz *)
Table[Sum[(n-k)*PartitionsP[k],{k,0,n}],{n,1,50}] (* Vaclav Kotesovec, Jun 23 2015 *)
t[n_, k_] := Sum[StirlingS1[n, j]* Binomial[i + j - 1, i]* PartitionsP[k - n - i], {j, 0, n}, {i, 0, k - n}]; Print@ Table[t[n, k], {k, 10}, {n, 0, k - 1}]; Table[t[2, k], {k, 3, 43}] (* George Beck, May 25 2016 *)

PARI

x='x+O('x^45); Vec(1/((1-x)^2*prod(k=1,50, 1-x^k))) \\ G. C. Greubel, Oct 15 2018

Formula

Let t(n_, k_) = Sum_{i = 0..k} Sum_{j = 0..n} s(n, j)*C(i, j)*p(k - n - i), where s(n, j) are Stirling numbers of the first kind, C(i, j) are the number of compositions of i distinct objects into j parts, and p is the integer partition function. Then a(k) = t(2, k+2) (conjectured). The formula for t(n, k) is the same as at A126442 except that there the Stirling numbers are of the second kind. - George Beck, May 21 2016

a(n) = (n+1)*A000070(n+1) - A182738(n+1). - Vaclav Kotesovec, Nov 04 2016

a(n) ~ exp(sqrt(2*n/3)*Pi)*sqrt(3)/(2*Pi^2) * (1 + 23*Pi/(24*sqrt(6*n))). - Vaclav Kotesovec, Nov 04 2016