A266232 Binomial transform of the number of partitions into distinct parts (A000009).

1, 2, 4, 9, 21, 49, 114, 265, 615, 1422, 3272, 7493, 17090, 38850, 88065, 199097, 448953, 1009788, 2265642, 5071611, 11328395, 25254093, 56195143, 124829822, 276839061, 612991848, 1355268779, 2992016128, 6596222234, 14522634554, 31933047707, 70130243427

Offset: 0

Author

Vaclav Kotesovec, Dec 25 2015

Keywords

• nonn

Comments

Let 0 < p < 1, r > 0, v > 0, f(n) = v*exp(r*n^p)/n^b, then

Sum_{k=0..n} binomial(n,k) * f(k) ~ f(n/2) * 2^n * exp(g(n)), where

g(n) = p^2 * r^2 * n^p / (2^(1+2*p)*n^(1-p) + p*r*(1-p)*2^(1+p)).

Special cases:

p < 1/2, g(n) = 0

p = 1/2, g(n) = r^2/16

p = 2/3, g(n) = r^2 * n^(1/3) / (9 * 2^(1/3)) - r^3/81

p = 3/4, g(n) = 9*r^2*sqrt(n)/(64*sqrt(2)) - 27*r^3*n^(1/4)/(2048*2^(1/4)) + 81*r^4/65536

p = 3/5, g(n) = 9*r^2*n^(1/5)/(100*2^(1/5))

p = 4/5, g(n) = 2^(7/5)*r^2*n^(3/5)/25 - 4*2^(3/5)*r^3*n^(2/5)/625 + 8*2^(4/5)*r^4*n^(1/5)/15625 - 32*r^5/390625

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..3200

Crossrefs

Cf. A000009, A294467, A293467, A294468.

Cf. A218481, A294466, A281425, A095051, A294500.

Programs

Mathematica

Table[Sum[Binomial[n, k]*PartitionsQ[k], {k, 0, n}], {n, 0, 50}]
nmax = 30; CoefficientList[Series[Sum[PartitionsQ[k] * x^k / (1-x)^(k+1), {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 31 2022 *)

Formula

a(n) ~ 2^(n-5/4) * exp(Pi*sqrt(n/6) + Pi^2/48) / (3^(1/4)*n^(3/4)).

G.f.: (1/(1 - x))*Product_{k>=1} (1 + x^k/(1 - x)^k). - Ilya Gutkovskiy, Aug 19 2018