%I #9 Jul 22 2024 23:30:58
%S 1,2,1,3,3,1,4,6,5,1,5,10,15,8,1,6,15,35,36,12,1,7,21,70,120,78,17,1,
%T 8,28,126,330,364,153,23,1,9,36,210,792,1365,969,276,30,1,10,45,330,
%U 1716,4368,4845,2300,465,38,1,11,55,495,3432,12376,20349,14950,4960,741,47,1
%N Triangle, read by rows, where T(n,k) = C( k*(k+1)/2 + n-k + 1, n-k) for n>=k>=0.
%C Remarkably, column k of the matrix inverse (A121436) equals signed column k of matrix power: A107876^(k*(k+1)/2 + 2).
%H Harvey P. Dale, <a href="/A122176/b122176.txt">Table of n, a(n) for n = 0..10000</a>
%e Triangle begins:
%e 1;
%e 2, 1;
%e 3, 3, 1;
%e 4, 6, 5, 1;
%e 5, 10, 15, 8, 1;
%e 6, 15, 35, 36, 12, 1;
%e 7, 21, 70, 120, 78, 17, 1;
%e 8, 28, 126, 330, 364, 153, 23, 1;
%e 9, 36, 210, 792, 1365, 969, 276, 30, 1; ...
%t Flatten[Table[Binomial[(k(k+1))/2+n-k+1,n-k],{n,0,10},{k,0,n}]] (* _Harvey P. Dale_, Mar 18 2013 *)
%o (PARI) T(n,k)=if(n<k||k<0,0,binomial(k*(k+1)/2+n-k+1,n-k))
%Y Cf. A121436 (inverse); variants: A098568, A122175, A122177.
%K nonn,tabl
%O 0,2
%A _Paul D. Hanna_, Aug 25 2006