A361547 Expansion of e.g.f. exp(x^5/(120 * (1-x))).

1, 0, 0, 0, 0, 1, 6, 42, 336, 3024, 30366, 335412, 4041576, 52756704, 741620880, 11169844686, 179448036768, 3063069801792, 55360031126400, 1056123043335360, 21208345049147256, 447183762148547424, 9877939209960101280, 228112734232663600320

Offset: 0

Author

Seiichi Manyama, Mar 15 2023

Keywords

• nonn
• changed

Comments

In general, if m>=1 and e.g.f. = exp(x^m / (m! * (1-x))), then a(n) ~ n! * exp(2*sqrt(n/m!) - (2*m-1)/(2*m!)) / (2*sqrt(Pi) * m!^(1/4) * n^(3/4)). - Vaclav Kotesovec, Aug 28 2025

Crossrefs

Cf. A185369, A361533, A361545.

Programs

Mathematica

RecurrenceTable[{4 (-5 + n) (-4 + n) (-3 + n) (-2 + n) (-1 + n) a[-6 + n] - 5 (-4 + n) (-3 + n) (-2 + n) (-1 + n) a[-5 + n] + 120 (-2 + n) (-1 + n) a[-2 + n] - 240 (-1 + n) a[-1 + n] + 120 a[n] == 0, a[1] == 0, a[2] == 0, a[3] == 0, a[4] == 0, a[5] == 1, a[6] == 6}, a, {n, 0, 25}] (* Vaclav Kotesovec, Aug 28 2025 *)

PARI

my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(x^5/(120*(1-x)))))

Formula

a(n) = 2*(n-1) * a(n-1) - (n-1)*(n-2) * a(n-2) + binomial(n-1,4) * a(n-5) - 4*binomial(n-1,5) * a(n-6) for n > 5.

From Seiichi Manyama, Jun 17 2024: (Start)

a(n) = n! * Sum_{k=0..floor(n/5)} binomial(n-4*k-1,n-5*k)/(120^k * k!).

a(0) = 1; a(n) = ((n-1)!/120) * Sum_{k=5..n} k * a(n-k)/(n-k)!. (End)

a(n) ~ 2^(-5/4) * 15^(-1/4) * exp(-3/80 + sqrt(n/30) - n) * n^(n - 1/4). - Vaclav Kotesovec, Aug 28 2025