A138911 The n-th term of n-th inverse binomial transform of this sequence is 1 for n>=0.

1, 2, 5, 19, 81, 401, 2233, 13721, 91969, 666145, 5174001, 42827225, 375850225, 3481908353, 33923685097, 346468507201, 3698848172289, 41173895716289, 476826776487649, 5733422156706473, 71449460236945201, 921311262658989217, 12274120192571550105, 168717589565046739409

Offset: 0

Author

Paul D. Hanna, Apr 05 2008, Apr 06 2008

Keywords

• nonn

Examples

			If the successive inverse binomial transforms are placed in a table,
then we see that the diagonal consists of all 1's:
n=0:[(1), 2, 5, 19, 81, 401, 2233, 13721, 91969, 666145,...];
n=1:[1, (1), 2, 9, 28, 145, 726, 4249, 27000, 186561,...];
n=2:[1, 0, (1), 5, 1, 79, 121, 1511, 6721, 50975,...];
n=3:[1, -1, 2, (1), -12, 113, -422, 2441, -6584, 44929,...];
n=4:[1, -2, 5, -9, (1), 157, -1263, 8173, -45087, 243261,...];
n=5:[1, -3, 10, -31, 76, (1), -1922, 19841, -153896, 1068065,...];
n=6:[1, -4, 17, -71, 273, -805, (1), 29339, -359135, 3316411,...];
n=7:[1, -5, 26, -135, 676, -3071, 10626, (1), -525144, 7462305,...];
n=8:[1, -6, 37, -229, 1393, -8087, 42313, -167839, (1), 10811609,...];
n=9:[1, -7, 50, -359, 2556, -17743, 116902, -691447, 3091768, (1),...].
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..541

Crossrefs

Cf. A138912 (inverse binomial transform).

Programs

PARI

{a(n)=local(A=[1]);for(k=1,n,A=concat(A,0); A[k+1]=1-polcoeff(subst(Ser(A),x,x/(1+k*x+x*O(x^k)))/(1+k*x),k));A[n+1]}

PARI

{a(n)=1 + sum(k=0,n-1,binomial(n,k)*n*k^(n-k-1) )}

Formula

O.g.f. satisfies: [x^n] A( x/(1+n*x) )/(1+n*x) = 1 for n>=0.

E.g.f. satisfies: [x^n] A(x)*exp(-n*x) = 1/n! for n>=0.

a(n) = 1 + Sum_{k=0..n-1} C(n,k)*n*k^(n-k-1) for n>0 with a(0)=1.