%I #10 Jul 23 2024 08:38:07
%S 1,3,1,6,4,1,10,10,6,1,15,20,21,9,1,21,35,56,45,13,1,28,56,126,165,91,
%T 18,1,36,84,252,495,455,171,24,1,45,120,462,1287,1820,1140,300,31,1,
%U 55,165,792,3003,6188,5985,2600,496,39,1,66,220,1287,6435,18564,26334
%N Triangle, read by rows, where T(n,k) = C( k*(k+1)/2 + n-k + 2, n-k) for n>=k>=0.
%C Remarkably, column k of the matrix inverse (A121437) equals signed column k of matrix power: A107876^(k*(k+1)/2 + 3) for k>=0.
%e Triangle begins:
%e 1;
%e 3, 1;
%e 6, 4, 1;
%e 10, 10, 6, 1;
%e 15, 20, 21, 9, 1;
%e 21, 35, 56, 45, 13, 1;
%e 28, 56, 126, 165, 91, 18, 1;
%e 36, 84, 252, 495, 455, 171, 24, 1;
%e 45, 120, 462, 1287, 1820, 1140, 300, 31, 1; ...
%t A122177[n_, k_] := Binomial[k*(k + 1)/2 + n - k + 2, n - k];
%t Table[A122177[n, k], {n, 0, 10}, {k, 0, n}] (* _Paolo Xausa_, Jul 23 2024 *)
%o (PARI) T(n,k)=if(n<k||k<0,0,binomial(k*(k+1)/2+n-k+2,n-k))
%Y Cf. A121437 (inverse); variants: A098568, A122175, A122176.
%K nonn,tabl
%O 0,2
%A _Paul D. Hanna_, Aug 25 2006