|
%I #16 Sep 14 2023 03:40:24
%S 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,5,5,1,
%T 1,1,1,1,1,11,17,11,1,1,1,1,1,1,21,41,41,21,1,1,1,1,1,1,36,81,101,81,
%U 36,1,1,1,1,1,1,57,141,201,201,141,57,1,1,1
%N Number triangle T(n,k) = Sum_{j=0..n-k} C(k,3j)*C(n-k,3j).
%C Row sums are A119336. Product of Pascal's triangle and A119337.
%H Seiichi Manyama, <a href="/A119335/b119335.txt">Rows n = 0..139, flattened</a>
%F Column k has g.f. (x^k/(1-x)) * Sum_{j=0..k} C(k,3j)(x/(1-x))^(3j).
%e Triangle begins
%e 1;
%e 1, 1;
%e 1, 1, 1;
%e 1, 1, 1, 1;
%e 1, 1, 1, 1, 1;
%e 1, 1, 1, 1, 1, 1;
%e 1, 1, 1, 2, 1, 1, 1;
%e 1, 1, 1, 5, 5, 1, 1, 1;
%e 1, 1, 1, 11, 17, 11, 1, 1, 1;
%e 1, 1, 1, 21, 41, 41, 21, 1, 1, 1;
%e 1, 1, 1, 36, 81, 101, 81, 36, 1, 1, 1;
%t T[n_, k_] := Sum[Binomial[k, 3j] Binomial[n-k, 3j], {j, 0, n-k}];
%t Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Sep 14 2023 *)
%Y T(2n,n) gives A119363.
%Y Cf. A119326.
%K easy,nonn,tabl
%O 0,25
%A _Paul Barry_, May 14 2006
%E More terms from _Seiichi Manyama_, Mar 12 2019
|
