|
%I #14 Feb 25 2024 15:37:25
%S 1,-5,1,7,-4,1,-3,3,-3,1,0,0,0,-2,1,0,0,0,-2,-1,1,0,0,0,-2,-3,0,1,0,0,
%T 0,-2,-5,-3,1,1,0,0,0,-2,-7,-8,-2,2,1,0,0,0,-2,-9,-15,-10,0,3,1,0,0,0,
%U -2,-11,-24,-25,-10,3,4
%N Triangle of numbers read by rows: T(n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)*Stirling2(i+3,i+1)*(-1)^i for n >= 0, 0 <= k <= n.
%C Generalized binomial coefficients of the second order.
%H Igor Victorovich Statsenko, <a href="https://aeterna-ufa.ru/sbornik/IN-2024-02-2.pdf#page=15">On the ordinal numbers of triangles of generalized special numbers</a>, Innovation science No 2-2, State Ufa, Aeterna Publishing House, 2024, pp. 15-19. In Russian.
%F T(n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)*Stirling2(i+m+1,i+1)*(-1)^i where m = 2 for n >= 0, 0 <= k <= n.
%e n\k 0 1 2 3 4 5 6
%e 0: 1
%e 1: -5 1
%e 2: 7 -4 1
%e 3: -3 3 -3 1
%e 4: 0 0 0 -2 1
%e 5: 0 0 0 -2 -1 1
%e 6: 0 0 0 -2 -3 0 1
%p C:=(n,k)->n!/(k!*(n-k)!) : T:=(m,n,k)->sum(C(n+1,n-k-r)*Stirling2(r+m+1,r+1)*((-1)^r), r=0..n-k) : m:=2 : seq(seq T(m,n,k), k=0..n), n=0..10);
%Y For m=0 the formula gives the sequence A007318; for m=1 the formula gives the sequence A159854. In this case, we assume that A007318 consists of generalized binomial coefficients of order zero and A159854 consists of generalized binomial coefficients of order one.
%K tabl,sign
%O 0,2
%A _Igor Victorovich Statsenko_, Feb 21 2024
|
