0,2

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.

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),...].

(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) )}

Cf. A138912 (inverse binomial transform).

Sequence in context: A150026 A150027 A058131 this_sequence A107377 A058132 A002851

Adjacent sequences: A138908 A138909 A138910 this_sequence A138912 A138913 A138914

nonn

Paul D. Hanna (pauldhanna(AT)juno.com), Apr 05 2008, Apr 06 2008