%I #2 Mar 30 2012 18:37:10
%S 1,2,6,33,241,2391,30903,499000,9804344,230270387,6364661087,
%T 204142831017,7508991442137,313657014920304,14753881974826196,
%U 775751424297987671,45294591976852153687,2919681469388967044215
%N a(n) = (n+1)-th term of the (n+1)-th inverse binomial transform of this sequence for n>=0.
%F O.g.f. satisfies: a(n-1) = [x^n] A( x/(1+n*x) )/(1+n*x) for n>=1 with a(0)=1.
%e If the successive inverse binomial transforms are placed in a table,
%e then we see that the diagonal consists of this sequence shift right:
%e n=0:[(1),2,6,33,241,2391,30903,499000,9804344,230270387,...];
%e n=1:[1, (1),3,20,138,1465,19591,325497,6558907,157672912,...];
%e n=2:[1,0, (2),13,73,949,12511,214938,4430056,108883779,...];
%e n=3:[1,-1,3, (6),34,693,7683,145147,3012155,75811514,...];
%e n=4:[1,-2,6,-7, (33),547,3967,104868,2029432,53365459,...];
%e n=5:[1,-3,11,-32,106, (241),1423,87045,1273819,38606532,...];
%e n=6:[1,-4,18,-75,313,-735, (2391),77062,613352,30170147,...];
%e n=7:[1,-5,27,-142,738,-3251,13291, (30903),131611,27084334,...];
%e n=8:[1,-6,38,-239,1489,-8657,47143,-161808, (499000),25380339,...];
%e n=9:[1,-7,51,-372,2698,-18903,126807,-734927,3716987, (9804344),...].
%o (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]}
%K nonn
%O 0,2
%A _Paul D. Hanna_, Apr 05 2008