|
| |
|
|
A302764
|
|
Pascal-like triangle with A000012 as the left border and A080956 as the right border.
|
|
4
|
|
|
|
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, 7, 20, 28, 14, -14, -28, -20, 1, 8, 27, 48, 42, 0, -42, -48, -27, 1, 9, 35, 75, 90, 42, -42, -90, -75, -35, 1, 10, 44, 110, 165, 132, 0, -132, -165, -110, -44
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,5
|
|
|
COMMENTS
|
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).
|
|
|
LINKS
|
|
|
|
FORMULA
|
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.
|
|
|
EXAMPLE
|
Triangle begins:
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, 7, 20, 28, 14, -14, -28, -20;
...
|
|
|
PROG
|
(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))));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Apr 21 2018
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
