OFFSET
0,4
COMMENTS
Inverse binomial transform A000254.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..450
N. J. A. Sloane, Transforms
FORMULA
a(n) = Sum_{k=1..n} (-1)^(n-k)*binomial(n,k)*k!*H(k), where H(k) is the k-th harmonic number.
a(n) ~ n! * (log(n) + gamma) / exp(1), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jun 23 2018
Recurrence: a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 5, a(n+4) = (2*n+5)*a(n+3) - (n^2+2*n-1)*a(n+2) - (2*n+3)*(n+2)*a(n+1) - (n+2)*(n+1)*a(n). - Vladimir Reshetnikov, Mar 30 2022
EXAMPLE
-exp(-x)*log(1 - x)/(1 - x) = x/1! + x^2/2! + 5*x^3/3! + 20*x^4/4! + 109*x^5/5! + 689*x^6/6! + 5053*x^7/7! + ...
MAPLE
b:= proc(n) option remember; `if`(n<2, n, n*b(n-1)+(n-1)!) end:
a:= proc(n) add(b(k)*(-1)^(n-k)*binomial(n, k), k=0..n) end:
seq(a(n), n=0..25); # Alois P. Heinz, Mar 07 2018
MATHEMATICA
nmax = 22; CoefficientList[Series[-Exp[-x] Log[1 - x]/(1 - x), {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[(-1)^(n - k) Binomial[n, k] k! HarmonicNumber[k], {k, 1, n}], {n, 0, 22}]
a[0] = 0; a[1] = a[2] = 1; a[3] = 5; a[n_Integer] := a[n] = (2 n - 3) a[n - 1] - (n^2 - 6 n + 7) a[n - 2] - (n - 2) (2 n - 5) a[n - 3] - (n - 3) (n - 2) a[n - 4]; Table[a[n], {n, 0, 22}] (* Vladimir Reshetnikov, Mar 30 2022 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Mar 07 2018
STATUS
approved