login
a(n+1) = the n-th term of the n-th binomial transform.
0

%I #5 Mar 30 2012 18:36:39

%S 1,1,2,10,82,946,14246,267974,6117202,166015698,5273053710,

%T 193534712510,8119820921626,385777848702394,20583872009571798,

%U 1224407374239009622,80669343513439179922,5852864801437926734482,465237079520383362585598

%N a(n+1) = the n-th term of the n-th binomial transform.

%C Form a square array where the n-th row is the n-th binomial transform of this sequence, starting with this sequence in the zeroth row; then the diagonal of the square array so formed is this sequence shifted 1 place left.

%F a(n+1) = sum(k=0, n, a(k)*binomial(n, k)*n^(n-k))

%e Note the diagonal in the array of iterated binomial transforms:

%e [_1,1,2,10,82,946,14246,267974,..]

%e [1,_2,5,20,139,1482,21389,390832,..]

%e [1,3,_10,42,258,2438,32854,577362,..]

%e [1,4,17,_82,499,4264,52361,869270,..]

%e [1,5,26,146,_946,7770,87350,1346062,..]

%e [1,6,37,240,1707,_14246,151501,2159484,..]

%e [1,7,50,370,2914,25582,_267974,3588122,..]

%e [1,8,65,542,4723,44388,473369,_6117202,..]

%o (PARI) {L=20; a=[1]; for(i=1,L,b=a; for(n=0,length(a)-1, b[n+1]=sum(k=0,n,a[k+1]*binomial(n,k)*n^(n-k)); ); a=concat(1,b); ); for(j=1,L,print1(a[j],","))}

%Y Cf. A071207, A088956.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Nov 08 2003