A131049 Triangle read by rows: (1/4) * (A007318^3 - A007318^(-1)) as infinite lower triangular matrices.

1, 2, 2, 7, 6, 3, 20, 28, 12, 4, 61, 100, 70, 20, 5, 182, 366, 300, 140, 30, 6, 547, 1274, 1281, 700, 245, 42, 7, 1640, 4376, 5096, 3416, 1400, 392, 56, 8, 4921, 14760, 19692, 15288, 7686, 2520, 588, 72, 9, 14762, 49210, 73800, 65640, 38220, 15372, 4200, 840, 90, 10

Offset

1,2

Comments

Let P = Pascal's triangle, A007318. Then form (1/4) * (P^3 - 1/P). Exclude the initial row of zeros. The resulting triangle T(n,k) is for n >= 1, 0 <= k < n.

Row sums are powers of 4.

Links

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)

Formula

From Andrew Howroyd, Sep 23 2025: (Start)

T(n,k) = binomial(n,k)*(3^(n-k) - (-1)^(n-k)) / 4.

G.f.: x/((1 + (1 - y)*x)*(1 - (3 + y)*x)). (End)

Example

First few rows of the triangle:
    1;
    2,    2;
    7,    6,    3;
   20,   28,   12,   4;
   61,  100,   70,  20,   5;
  182,  366,  300, 140,  30,  6;
  547, 1274, 1281, 700, 245, 42, 7;
  ...

Prog

(PARI) M(n)={my(t=matrix(n+1, n+1, i, j, binomial(i-1, j-1))); (t^3 - t^(-1))[2..n+1, 1..n]/4}
{ my(A=M(10)); for(i=1, #A, print(A[i, 1..i])) } \\ Andrew Howroyd, Sep 23 2025
(PARI) T(n, k) = binomial(n, k)*(3^(n-k) - (-1)^(n-k)) / 4
{ for(n=1, 10, print(vector(n, k, T(n, k-1)))) } \\ Andrew Howroyd, Sep 23 2025

Crossrefs

Column 0 is A015518.

Row sums are A000302(n-1).

Cf. A007318, A131047, A131048, A131050.

Sequence in context: A006748 A193548 A396928 * A142070 A152825 A347073

Adjacent sequences: A131046 A131047 A131048 * A131050 A131051 A131052

Keyword

nonn,tabl

Author

Gary W. Adamson, Jun 12 2007

Extensions

a(46) onwards from Andrew Howroyd, Sep 23 2025

Status

approved