0,3

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.

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

Note the diagonal in the array of iterated binomial transforms:

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

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

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

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

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

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

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

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

(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], ", "))}

Cf. A071207, A088956.

Sequence in context: A088351 A062396 A112487 this_sequence A111265 A003093 A006679

Adjacent sequences: A089466 A089467 A089468 this_sequence A089470 A089471 A089472

nonn

Paul D. Hanna (pauldhanna(AT)juno.com), Nov 08 2003