cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A122176 Triangle, read by rows, where T(n,k) = C( k*(k+1)/2 + n-k + 1, n-k) for n>=k>=0.

Original entry on oeis.org

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, 8, 28, 126, 330, 364, 153, 23, 1, 9, 36, 210, 792, 1365, 969, 276, 30, 1, 10, 45, 330, 1716, 4368, 4845, 2300, 465, 38, 1, 11, 55, 495, 3432, 12376, 20349, 14950, 4960, 741, 47, 1
Offset: 0

Views

Author

Paul D. Hanna, Aug 25 2006

Keywords

Comments

Remarkably, column k of the matrix inverse (A121436) equals signed column k of matrix power: A107876^(k*(k+1)/2 + 2).

Examples

			Triangle begins:
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;
8, 28, 126, 330, 364, 153, 23, 1;
9, 36, 210, 792, 1365, 969, 276, 30, 1; ...

Links

Crossrefs

Cf. A121436 (inverse); variants: A098568, A122175, A122177.

Programs

  • Mathematica
    Flatten[Table[Binomial[(k(k+1))/2+n-k+1,n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Mar 18 2013 *)
  • PARI
    T(n,k)=if(n

A121440 Matrix inverse of triangle A121335, where A121335(n,k) = C( n*(n+1)/2 + n-k + 1, n-k) for n>=k>=0.

Original entry on oeis.org

1, -3, 1, 0, -5, 1, -12, 4, -8, 1, -129, -22, 18, -12, 1, -1785, -238, -51, 51, -17, 1, -30291, -3634, -345, -161, 115, -23, 1, -608565, -66750, -6111, -285, -505, 225, -30, 1, -14112744, -1432296, -122227, -9177, 665, -1387, 399, -38, 1, -370746528, -35129022, -2818543, -196037, -14335, 4841, -3337, 658
Offset: 0

Views

Author

Paul D. Hanna, Aug 29 2006

Keywords

Comments

A triangle having similar properties and complementary construction is the dual triangle A121436.

Examples

			Triangle, A121335^-1, begins:
1;
-3, 1;
0, -5, 1;
-12, 4, -8, 1;
-129, -22, 18, -12, 1;
-1785, -238, -51, 51, -17, 1;
-30291, -3634, -345, -161, 115, -23, 1;
-608565, -66750, -6111, -285, -505, 225, -30, 1;
-14112744, -1432296, -122227, -9177, 665, -1387, 399, -38, 1; ...
Triangle A121412 begins:
1;
1, 1;
3, 1, 1;
18, 4, 1, 1;
170, 30, 5, 1, 1; ...
Row 3 of A121335^-1 equals row 3 of A121412^(-8), which begins:
1;
-8, 1;
12, -8, 1;
-12, 4, -8, 1; ...
Row 4 of A121335^-1 equals row 4 of A121412^(-12), which begins:
1;
-12, 1;
42, -12, 1;
-34, 30, -12, 1;
-129, -22, 18, -12, 1; ...

Crossrefs

Cf. A121335 (matrix inverse); A121412; variants: A121438, A121439, A121441; A121436 (dual).

Programs

  • PARI
    /* Matrix Inverse of A121335 */ {T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial(r*(r-1)/2+r-c+1,r-c)))); return((M^-1)[n+1,k+1])}

Formula

T(n,k) = [A121412^(-n*(n+1)/2 - 2)](n,k) for n>=k>=0; i.e., row n of A121335^-1 equals row n of matrix power A121412^(-n*(n+1)/2 - 2).

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

Original entry on oeis.org

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

Views

Author

Paul D. Hanna, Aug 27 2006

Keywords

Examples

			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; ...

Crossrefs

Cf. A098568, A107876, A122177, A107885.

Programs

  • PARI
    /* 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])
  • PARI
    /* 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)

Formula

(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)
Showing 1-3 of 3 results.

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