|
|
A202193
|
|
Triangle T(n,m) = coefficient of x^n in expansion of (x/(1 - x - x^2 - x^3 - x^4))^m = Sum_{n>=m} T(n,m) x^n.
|
|
1
|
|
|
1, 1, 1, 2, 2, 1, 4, 5, 3, 1, 8, 12, 9, 4, 1, 15, 28, 25, 14, 5, 1, 29, 62, 66, 44, 20, 6, 1, 56, 136, 165, 129, 70, 27, 7, 1, 108, 294, 401, 356, 225, 104, 35, 8, 1, 208, 628, 951, 944, 676, 363, 147, 44, 9, 1, 401, 1328, 2211, 2424, 1935, 1176, 553, 200
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
COMMENTS
|
As a Riordan array, this is (1/(1 - x - x^2 - x^3 - x^4), x/(1 - x - x^2 - x^3 - x^4)).
Diagonal sums are A077926(n)*(-1)^n.
Tetranacci convolution triangle. (End)
|
|
LINKS
|
|
|
FORMULA
|
T(n,m) = Sum_{k=1..n-m} Sum_{i=0..floor((n-m-k)/4)} (-1)^i*binomial(k,k-i)*binomial(n-m-4*i-1,k-1))*binomial(k+m-1,m-1)), n > m, T(n,n)=1.
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-3,k) + T(n-4,k), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Feb 16 2014
|
|
EXAMPLE
|
Triangle begins:
1;
1, 1;
2, 2, 1;
4, 5, 3, 1;
8, 12, 9, 4, 1;
15, 28, 25, 14, 5, 1;
29, 62, 66, 44, 20, 6, 1;
|
|
PROG
|
(Maxima)
T(n, m):=if n=m then 1 else sum(sum((-1)^i*binomial(k, k-i)*binomial(n-m-4*i-1, k-1), i, 0, (n-m-k)/4)*binomial(k+m-1, m-1), k, 1, n-m);
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|