A098568 -id:A098568 - cp's OEIS frontend

A121437 Matrix inverse of triangle A122177, where A122177(n,k) = C( k*(k+1)/2 + n-k + 2, n-k) for n>=k>=0.

1, -3, 1, 6, -4, 1, -16, 14, -6, 1, 63, -62, 33, -9, 1, -351, 365, -215, 72, -13, 1, 2609, -2790, 1731, -642, 143, -18, 1, -24636, 26749, -17076, 6696, -1664, 261, -24, 1, 284631, -311769, 202356, -81963, 21684, -3831, 444, -31, 1, -3909926, 4305579, -2822991, 1166310, -320515, 60768, -8012, 713, -39, 1

Offset: 0

Paul D. Hanna, Aug 27 2006

			Triangle begins:
  1;
  -3, 1;
  6, -4, 1;
  -16, 14, -6, 1;
  63, -62, 33, -9, 1;
  -351, 365, -215, 72, -13, 1;
  2609, -2790, 1731, -642, 143, -18, 1;
  -24636, 26749, -17076, 6696, -1664, 261, -24, 1;
  284631, -311769, 202356, -81963, 21684, -3831, 444, -31, 1; ...

Cf. A098568, A107876, A122177, A107885.

/* Matrix Inverse of A122177 */ T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial((c-1)*(c-2)/2+r+1,r-c)))); return((M^-1)[n+1,k+1])

/* Obtain by g.f. */ T(n,k)=polcoeff(1-sum(j=0, n-k-1, T(j+k,k)*x^j/(1-x+x*O(x^n))^(j*(j+1)/2+j*k+k*(k+1)/2+3)), n-k)

(1) T(n,k) = A121436(n-1,k) - A121436(n-1,k+1).
(2) T(n,k) = (-1)^(n-k)*[A107876^(k*(k+1)/2 + 3)](n,k); i.e., column k equals signed column k of A107876^(k*(k+1)/2 + 3).
G.f.s for column k:
(3) 1 = Sum_{j>=0} T(j+k,k)*x^j/(1-x)^( j*(j+1)/2) + j*k + k*(k+1)/2 + 3);
(4) 1 = Sum_{j>=0} T(j+k,k)*x^j*(1+x)^( j*(j-1)/2) + j*k + k*(k+1)/2 + 3).
From Benedict W. J. Irwin, Nov 26 2016: (Start)
Conjecture: The sequence (column 2 of triangle) 14, -62, 365, -2790, 26749, ... is described by a series of nested sums:
14    =  Sum_{i=1..4} (i+1),
-62   = -Sum_{i=1..4} (Sum_{j=1..i+1} (j+2)),
365   =  Sum_{i=1..4} (Sum_{j=1..i+1} (Sum_{k=1..j+2} (k+3))),
-2790 = -Sum_{i=1..4} (Sum_{j=1..i+1} (Sum_{k=1..j+2} (Sum_{l=1..k+3} (l+4)))). (End)

A183202 Triangle, read by rows, where T(n,k) equals the sum of (n-k) terms in row n of triangle A131338 starting at position nk - k(k-1)/2, with the main diagonal formed from the row sums.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 3, 3, 3, 5, 4, 6, 10, 9, 14, 5, 10, 22, 34, 29, 43, 6, 15, 40, 84, 122, 100, 143, 7, 21, 65, 169, 334, 463, 367, 510, 8, 28, 98, 300, 738, 1390, 1851, 1426, 1936, 9, 36, 140, 489, 1426, 3345, 6043, 7767, 5839, 7775, 10, 45, 192, 749, 2510, 6990, 15735

Offset: 0

Paul D. Hanna, Dec 30 2010

			Triangle begins:
1;
1,1;
2,1,2;
3,3,3,5;
4,6,10,9,14;
5,10,22,34,29,43;
6,15,40,84,122,100,143;
7,21,65,169,334,463,367,510;
8,28,98,300,738,1390,1851,1426,1936;
9,36,140,489,1426,3345,6043,7767,5839,7775;
10,45,192,749,2510,6990,15735,27374,34097,25094,32869; ...
The rows are derived from triangle A131338 by summing terms in the following manner:
(1);
(1),(1);
(1+1),(1),(2);
(1+1+1),(1+2),(3),(5);
(1+1+1+1),(1+2+3),(4+6),(9),(14);
(1+1+1+1+1),(1+2+3+4),(5+7+10),(14+20),(29),(43);
(1+1+1+1+1+1),(1+2+3+4+5),(6+8+11+15),(20+27+37),(51+71),(100),(143); ...
where row n of triangle A131338 consists of n '1's followed by the partial sums of the prior row.

Cf. A131338, A098568, A098569 (row sums), A183203 (antidiagonal sums).

{A131338(n, k)=if(k>n*(n+1)/2||k<0,0,if(k<=n,1,sum(i=0, k-n,A131338(n-1,i))))}
{T(n,k)=if(n==k,A131338(n,n*(n+1)/2),sum(j=n*k-k*(k-1)/2,n*k-k*(k-1)/2+n-k-1,A131338(n,j)))}

cp's OEIS Frontend

A121437 Matrix inverse of triangle A122177, where A122177(n,k) = C( k*(k+1)/2 + n-k + 2, n-k) for n>=k>=0.

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A183202 Triangle, read by rows, where T(n,k) equals the sum of (n-k) terms in row n of triangle A131338 starting at position nk - k(k-1)/2, with the main diagonal formed from the row sums.

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PARI