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Pascal-like triangle with A000012 as the left border and A080956 as the right border.
4

%I #15 Jun 24 2018 11:45:26

%S 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 7,20,28,14,-14,-28,-20,1,8,27,48,42,0,-42,-48,-27,1,9,35,75,90,42,

%U -42,-90,-75,-35,1,10,44,110,165,132,0,-132,-165,-110,-44

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

%C 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 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 Triangle begins:

%e 1;

%e 1, 1;

%e 1, 2, 0;

%e 1, 3, 2, -2;

%e 1, 4, 5, 0, -5;

%e 1, 5, 9, 5, -5, -9;

%e 1, 6, 14, 14, 0, -14, -14;

%e 1, 7, 20, 28, 14, -14, -28, -20;

%e ...

%o (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 tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ _Michel Marcus_, Apr 21 2018

%Y Cf. A000012, A007318, A024537, A080956, A195350, A301417, A301420, A301421, A301424.

%K tabl,sign

%O 1,5

%A _Gregory Gerard Wojnar_, Apr 12 2018