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.

A131047 (1/2) * (A007318 - A007318^(-1)).

Original entry on oeis.org

1, 0, 2, 1, 0, 3, 0, 4, 0, 4, 1, 0, 10, 0, 5, 0, 6, 0, 20, 0, 6, 1, 0, 21, 0, 35, 0, 7, 0, 8, 0, 56, 0, 56, 0, 8, 1, 0, 36, 0, 126, 0, 84, 0, 9
Offset: 1

Views

Author

Gary W. Adamson, Jun 12 2007

Keywords

Comments

Row sums = (1, 2, 4, 8, ...). A131047 * (1,2,3, ...) = A087447 starting (1, 4, 10, 24, 56, ...). A generalized set of analogous triangles: (1/(Q+1)) * (P^Q - 1/P), Q an integer, generates triangles with row sums = powers of (Q+1). Cf. A131048, A131049, A131050, A131051 for triangles having Q = 2,3,4 and 5, respectively.
A007318, Pascal's triangle, = this triangle + A119467, since one triangle = the zeros or masks of the other. - Gary W. Adamson, Jun 12 2007

Examples

			First few rows of the triangle:
  1;
  0, 2;
  1, 0,  3;
  0, 4,  0,  4;
  1, 0, 10,  0,  5;
  0, 6,  0, 20,  0, 6;
  1, 0, 21,  0, 35, 0, 7;
  ...

Crossrefs

Cf. A131048, A131049, A131050, A131051.
Cf. A119467.

Formula

Let A007318 (Pascal's triangle) = P, then A131047 = (1/2) * (P - 1/P); deleting the right border of zeros.

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

Original entry on oeis.org

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

Views

Author

Gary W. Adamson, Jun 12 2007

Keywords

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.

Examples

			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.

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)

A131049 (1/4) * (A007318^3 - A007318^(-1)).

Original entry on oeis.org

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
Offset: 1

Views

Author

Gary W. Adamson, Jun 12 2007

Keywords

Comments

Row sums = powers of 4: (1, 4, 16, 64, ...).
Left border = A015518: (1, 2, 7, 20, 61, 182, ...).

Examples

			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;
  ...

Crossrefs

Cf. A015518, A007318, A131047, A131048, A131050, A131051.

Formula

(1/4) * (P^3 - 1/P), where P = Pascal's triangle, A007318. Delete right border of zeros.
Showing 1-3 of 3 results.

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