A103924 Number of partitions of n into parts but with two kinds of parts of sizes 1,2,3,4 and 5.

1, 2, 5, 10, 20, 36, 64, 107, 177, 282, 443, 678, 1026, 1522, 2234, 3231, 4628, 6550, 9193, 12774, 17619, 24098, 32740, 44161, 59213, 78894, 104553, 137787, 180702, 235806, 306354, 396226, 510392, 654787, 836911, 1065734, 1352475, 1710535, 2156536, 2710318

Offset: 0

Author

Wolfdieter Lang, Mar 24 2005

Keywords

• nonn
• easy

Comments

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

Convolution of A001401 and A000041. - Vaclav Kotesovec, Aug 28 2015

Also the sum of binomial (D(p), 5) over partitions p of n+15, 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, Table of n, a(n) for n = 0..10000

Crossrefs

Sixth column (m=5) of Fine-Riordan triangle A008951 and 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) a(n):=`if`(n=0, 1, add(add(d*`if`(d<6, 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

a[n_] := a[n] = If[n==0, 1, Sum[Sum[d*If[d<6, 2, 1], {d, Divisors[j]}] * a[n-j], {j, 1, n}]/n]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Aug 28 2015, after Alois P. Heinz *)
nmax=60; CoefficientList[Series[Product[1/(1-x^k), {k, 1, 5}] * 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@5],  {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+15, 5];
Table[a[n], {n, 0, 60}] (* Jean-François Alcover, May 30 2021 *)

Formula

G.f.: (product(1/(1-x^k), k=1..5)^2)*product(1/(1-x^j), j=6..infty).

a(n) = sum(A000710(n-5*j), j=0..floor(n/5)), n>=0.

a(n) ~ 3*n^(3/2) * exp(Pi*sqrt(2*n/3)) / (20*sqrt(2)*Pi^5). - Vaclav Kotesovec, Aug 28 2015