A103925 Number of partitions of n into parts but with two kinds of parts of sizes 1,2,3,4,5 and 6.

1, 2, 5, 10, 20, 36, 65, 109, 182, 292, 463, 714, 1091, 1631, 2416, 3523, 5091, 7264, 10284, 14405, 20035, 27621, 37831, 51425, 69497, 93299, 124588, 165408, 218533, 287231, 375851, 489525, 634980, 820195, 1055444, 1352965, 1728326, 2200060, 2791516, 3530513

Offset: 0

Author

Wolfdieter Lang, Mar 24 2005

Keywords

• nonn
• easy

Comments

See A103923 for other combinatorial interpretations of a(n).

Also the sum of binomial (D(p), 6) over partitions p of n+21, where D(p) is the number of different part sizes in p. - Emily Anible, Jun 09 2018

References

• H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958 (reprinted 1962), p. 90.
• J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.

Links

• Alois P. Heinz and Vaclav Kotesovec, Table of n, a(n) for n = 0..5000 (first 1000 terms from Alois P. Heinz)

Crossrefs

Seventh column (m=6) of Fine-Riordan triangle A008951, of triangle A103923, i.e. the p_2(n, m) array of the Gupta et al. reference.

Cf. A000712 (all parts of two kinds).

Programs

Maple

with(numtheory): a:= proc(n) option remember; `if`(n=0, 1, add(add(d*`if`(d<7, 2, 1), d=divisors(j)) *a(n-j), j=1..n)/n) end: seq(a(n), n=0..40);  # Alois P. Heinz, Sep 14 2014

Mathematica

nmax=60; CoefficientList[Series[Product[1/(1-x^k), {k, 1, 6}] * Product[1/(1-x^k), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 28 2015 *)
Table[Length@IntegerPartitions[n, All, Range@n~Join~Range@6],  {n,0,39}] (* Robert Price, Jul 29 2020 *)
T[n_, 0] := PartitionsP[n];
T[n_, m_] /; (n >= m(m+1)/2) := T[n, m] = T[n-m, m-1] + T[n-m, m];
T[, ] = 0;
a[n_] := T[n+21, 6];
Table[a[n], {n, 0, 60}] (* Jean-François Alcover, May 30 2021 *)

Formula

G.f.: (product(1/(1-x^k), k=1..6)^2)*product(1/(1-x^j), j=7..infty).

a(n) = sum(A103924(n-6*j), j=0..floor(n/6)), n>=0.

a(n) ~ exp(Pi*sqrt(2*n/3)) * 6^3 * n^2 / (4*sqrt(3) * 6! * Pi^6) = exp(Pi*sqrt(2*n/3)) * sqrt(3) * n^2 / (40*Pi^6). - Vaclav Kotesovec, Aug 28 2015