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(n+1)-st term of the n-th binomial transform of this sequence equals 1 for all n>=0.
0

%I #15 Jun 20 2022 04:42:15

%S 1,0,-3,1,17,-49,-71,1079,-3263,-10529,156401,-624601,-1858319,

%T 42109055,-249633047,-187690049,17346988289,-159631097281,

%U 375894263521,8905140874583,-137481700309199,830604602372959,3549053339380121,-139582700722475473,1526848956867592513

%N (n+1)-st term of the n-th binomial transform of this sequence equals 1 for all n>=0.

%C Equals row sums of triangle A089962, which is the matrix inverse of triangle A071207.

%F a(n) = Sum_{k=0..n} (-1)^(n-k)*C(n, k)*n*k^(n-k-1).

%F Sum_{k=0..n} C(n, k)*n^(n-k)*a(k) = 1.

%F O.g.f.: A(x) = (1-x)*sum_{n>=0} x^n/(1+n*x)^(n+2).

%F E.g.f.: (1-x)*exp(x*exp(-x)).

%e Note the diagonal of 1's in the array of iterated binomial transforms:

%e [1, 0, -3, 1, 17, -49, -71, 1079,..]

%e [1, 1, -2, -7, 4, 17, -134, 121,..]

%e [1, 2, 1, -9, -31, -47, -135, -599,..]

%e [1, 3, 6, 1, -52, -271, -1034, -3895,..]

%e [1, 4, 13, 29, 1, -445, -3311, -18461,..]

%e [1, 5, 22, 81, 212, 1, -5166, -49631,..]

%e [1, 6, 33, 163, 689, 2117, 1, -75019,..]

%e [1, 7, 46, 281, 1564, 7553, 26830, 1,..]

%t Join[{1, 0}, Table[Sum[(-1)^(n - k)*Binomial[n, k]*n*k^(n - k - 1), {k, 0, n} ], {n, 2, 25}]] (* _Vaclav Kotesovec_, Jun 20 2022 *)

%o (PARI) a(n)=if(n<0,0,if(n==0,1,sum(k=0,n,(-1)^(n-k)*binomial(n,k)*n*k^(n-k-1) )))

%Y Cf. A089962, A071207.

%K sign

%O 0,3

%A _Paul D. Hanna_, Nov 18 2003

%E a(19) and a(20) concatenated by _Georg Fischer_, Jun 20 2022