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A302764 Pascal-like triangle with A000012 as the left border and A080956 as the right border.

%I A302764 #15 Jun 24 2018 11:45:26
%S A302764 1,1,1,1,2,0,1,3,2,-2,1,4,5,0,-5,1,5,9,5,-5,-9,1,6,14,14,0,-14,-14,1,
%T A302764 7,20,28,14,-14,-28,-20,1,8,27,48,42,0,-42,-48,-27,1,9,35,75,90,42,
%U A302764 -42,-90,-75,-35,1,10,44,110,165,132,0,-132,-165,-110,-44
%N A302764 Pascal-like triangle with A000012 as the left border and A080956 as the right border.
%C A302764 Number the rows of the triangle beginning with n=0. For each row construct a degree n polynomial with regularly decreasing powers, denoting the polynomial as f_n(x); e.g., for row 2 we have f_2(x)=1x^2+2x+0. Then construct g_n(x)=x^2*f_{n-1}(x)-(n+1)x+1. It obtains that g_n(x)=(1-x)(2-(1+x)^n). These g_n(x) are the denominators of the generating functions for the following sequences: A024537 (n=2); A195350 (n=3); A301417 (n=4); A301420 (n=5); A301421 (n=6); A301424 (n=7). For these sequences the asymptotic term-to-term ratios are 1/(2^(1/n)-1). The numerators of the generating functions are 1-x(x+1)^(n-1).
%F A302764 T(n,k) = T(n-1,k) + T(n-1,k-1) with T(n, 0) = 1 and T(n, n) = (n+1)*(2-n)/2.
%e A302764 Triangle begins:
%e A302764   1;
%e A302764   1, 1;
%e A302764   1, 2,  0;
%e A302764   1, 3,  2, -2;
%e A302764   1, 4,  5,  0, -5;
%e A302764   1, 5,  9,  5, -5,  -9;
%e A302764   1, 6, 14, 14,  0, -14, -14;
%e A302764   1, 7, 20, 28, 14, -14, -28, -20;
%e A302764   ...
%o A302764 (PARI) T(n,k) = if (k==0, 1, if (k==n, (n+1)*(2-n)/2, if (k>n, 0, T(n-1,k) + T(n-1,k-1))));
%o A302764 tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ _Michel Marcus_, Apr 21 2018
%Y A302764 Cf. A000012, A007318, A024537, A080956, A195350, A301417, A301420, A301421, A301424.
%K A302764 tabl,sign
%O A302764 1,5
%A A302764 _Gregory Gerard Wojnar_, Apr 12 2018