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.

A257448 a(n) = 13*(2^n - 1) - 3*n^2 - 9*n.

Original entry on oeis.org

1, 9, 37, 111, 283, 657, 1441, 3051, 6319, 12909, 26149, 52695, 105859, 212265, 425161, 851043, 1702903, 3406725, 6814477, 13630095, 27261451, 54524289, 109050097, 218101851, 436205503, 872412957, 1744828021, 3489658311, 6979319059, 13958640729
Offset: 1

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Author

Luciano Ancora, Apr 23 2015

Keywords

Comments

These numbers belong to a family of sequences obtained as follows:
. A000225: 1*(2^n-1);
. A050488: 3*(2^n-1) - 2*n;
. a(n): 13*(2^n-1) - 3*n^2 - 9*n;
. A257449: 75*(2^n-1) - 4*n^3 - 18*n^2 - 52*n;
. A257450: 541*(2^n-1) - 5*n^4 - 30*n^3 - 130*n^2 - 375*n,
where the sequence 1, 3, 13, 75, 541, ... is A000670 (after the first term), and A208744 gives the triangle of coefficients:
2;
3, 9;
4, 18, 52;
5, 30, 130, 375;
6, 45, 260, 1125, 3246;
7, 63, 455, 2625, 11361, 32781, etc.
Also, the antidiagonal sums in the array are given by the formula (6*n^2 + 6*k*n + (k-1)*k)*(k+n)!/((k+3)!*(n-1)!) for k = 0, 1, 2, 3, 4, ... (see Example field).

Examples

			By the second comment, the array begins (antidiagonals in A046902):
k=0: 1,  8, 27,  64,  125,  216, ...  A000578
k=1: 1,  9, 36, 100,  225,  441, ...  A000537
k=2: 1, 10, 46, 146,  371,  812, ...  A024166
k=3: 1, 11, 57, 203,  574, 1386, ...  A101094
k=4: 1, 12, 69, 272,  846, 2232, ...  A101097
k=5: 1, 13, 82, 354, 1200, 3432, ...  A101102
k=6: 1, 14, 96, 450, 1650, 5082, ...  A254469
...
See also A254469 (Example field).
		

Crossrefs

Programs

  • Magma
    [13*(2^n-1)-3*n^2-9*n: n in [1..30]]; // Vincenzo Librandi, Apr 24 2015
  • Mathematica
    Table[13 (2^n - 1) - 3 n^2 - 9n, {n, 30}]
    CoefficientList[Series[x (1 + 4 x + x^2)/((1 - x)^3*(1 - 2 x)), {x, 0, 30}], x] (* Michael De Vlieger, Nov 14 2016 *)

Formula

G.f.: x*(1+4*x+x^2)/((1-x)^3*(1-2*x)).
a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4) for n>4. - Ray Chandler, Jul 25 2015

Extensions

Edited by Bruno Berselli, Apr 28 2015