0,2

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

If the successive inverse binomial transforms are placed in a table,

then we see that the diagonal consists of this sequence shift right:

n=0:[(1),2,6,33,241,2391,30903,499000,9804344,230270387,...];

n=1:[1, (1),3,20,138,1465,19591,325497,6558907,157672912,...];

n=2:[1,0, (2),13,73,949,12511,214938,4430056,108883779,...];

n=3:[1,-1,3, (6),34,693,7683,145147,3012155,75811514,...];

n=4:[1,-2,6,-7, (33),547,3967,104868,2029432,53365459,...];

n=5:[1,-3,11,-32,106, (241),1423,87045,1273819,38606532,...];

n=6:[1,-4,18,-75,313,-735, (2391),77062,613352,30170147,...];

n=7:[1,-5,27,-142,738,-3251,13291, (30903),131611,27084334,...];

n=8:[1,-6,38,-239,1489,-8657,47143,-161808, (499000),25380339,...];

n=9:[1,-7,51,-372,2698,-18903,126807,-734927,3716987, (9804344),...].

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

Sequence in context: A019028 A127114 A138909 this_sequence A121774 A053042 A012874

Adjacent sequences: A138980 A138981 A138982 this_sequence A138984 A138985 A138986

nonn

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