A131048 (1/3) * (A007318^2 - A007318^(-1)).

1, 1, 2, 3, 3, 3, 5, 12, 6, 4, 11, 25, 30, 10, 5, 21, 66, 75, 60, 15, 6, 43, 147, 231, 175, 105, 21, 7, 85, 344, 588, 616, 350, 168, 28, 8, 171, 765, 1548, 1764, 1386, 630, 252, 36, 9

Offset

1,3

Comments

Left border = A001045: (1, 1, 3, 5, 11, 21, 43, 85, ...).

Row sums = (1, 3, 9, 27, ...).

Analogous triangles for other powers of P are: A131047, A131049, A131050 and A131051.

Links

Table of n, a(n) for n=1..45.

Formula

Let A007318 (Pascal's triangle) = P. then A131048 = (1/3) * (P^2 - 1/P). Delete right border of zeros.

From Peter Bala, Oct 24 2007: (Start)

O.g.f.: 1/(1 - (2*x + 1)*t + (x^2 + x - 2)*t^2) = 1 + (1 + 2*x)*t + (3 + 3*x + 3*x^2)*t^2 + ....

T(n,n-k) = (1/3)*C(n,k)*(2^k - (-1)^k) = C(n,k)*A001045(k).

The row polynomials R(n,x) := Sum_{k = 0..n} T(n,n-k)*x^(n-k) = (1/3)*((x + 2)^n - (x - 1)^n) and have the divisibility property R(n,x) divides R(m,x) in the polynomial ring Z[x] if n divides m.

The polynomials R(n,-x), n >= 2, satisfy a Riemann hypothesis: their zeros lie on the vertical line Re x = 1/2 in the complex plane. Compare with A094440. (End)

Example

First few rows of the triangle:
   1;
   1,   2;
   3,   3,   3;
   5,  12,   6,   4;
  11,  25,  30,  10,   5;
  21,  66,  75,  60,  15,  6;
  43, 147, 231, 175, 105, 21, 7;
  ...

Crossrefs

Cf. A131047, A131049, A131050, A131051, A001045, A007318.

Cf. A001045, A094440, A132148.

Sequence in context: A343885 A264870 A346048 * A119688 A126868 A373093

Adjacent sequences: A131045 A131046 A131047 * A131049 A131050 A131051

Keyword

nonn,tabl

Author

Gary W. Adamson, Jun 12 2007

Status

approved