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.

A096886 Expansion of (1+3*x)/(1-8*x^2).

Original entry on oeis.org

1, 3, 8, 24, 64, 192, 512, 1536, 4096, 12288, 32768, 98304, 262144, 786432, 2097152, 6291456, 16777216, 50331648, 134217728, 402653184, 1073741824, 3221225472, 8589934592, 25769803776, 68719476736, 206158430208, 549755813888
Offset: 0

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Author

Paul Barry, Jul 14 2004

Keywords

Comments

From R. J. Mathar, Oct 12 2010: (Start)
Apparently the row n=3 of an array T(n,k) counting walks with k steps on an n X n board starting at an edge position next to a corner, each step to one of the <= 4 adjacent squares:
1, 3, 8, 24, 64, 192, 512, 1536, 4096, 12288, 32768, 98304, 262144, ...
1, 3, 9, 29, 93, 301, 973, 3149, 10189, 32973, 106701, 345293, 1117389, ...
1, 3, 9, 30, 99, 342, 1161, 4050, 13851, 48438, 165969, 580770, 1990899, ...
1, 3, 9, 30, 100, 349, 1216, 4329, 15381, 55187, 197714, 711458, 2557699, ...
1, 3, 9, 30, 100, 350, 1224, 4400, 15776, 57552, 209088, 768768, 2812160, ...
1, 3, 9, 30, 100, 350, 1225, 4409, 15865, 58091, 212586, 786708, 2909166, ...
1, 3, 9, 30, 100, 350, 1225, 4410, 15875, 58200, 213300, 791700, 2936375, ...
1, 3, 9, 30, 100, 350, 1225, 4410, 15876, 58211, 213431, 792623, 2943291, ...
1, 3, 9, 30, 100, 350, 1225, 4410, 15876, 58212, 213443, 792778, 2944460, ...
1, 3, 9, 30, 100, 350, 1225, 4410, 15876, 58212, 213444, 792791, 2944641, ...
...
(End)

Crossrefs

Cf. A038754.

Programs

  • Mathematica
    CoefficientList[Series[(1+3x)/(1-8x^2),{x,0,30}],x] (* or *) LinearRecurrence[{0,8},{1,3},30] (* Harvey P. Dale, Apr 25 2023 *)

Formula

G.f.: (1+3*x)/(1-8*x^2).
a(n) = (1 + (-1)^n)*8^floor((n+1)/2)/2 + 3*(1-(-1)^n)*8^floor(n/2)/2.
a(n) = 2^(3*n/2)*(3*sqrt(2)/8 + 1/2 - (3*sqrt(2)/8 - 1/2)*(-1)^n).
a(2n+1) = 2*a(2n) + 2*a(2n-1) + 2*a(2n-2).
a(2n) = 2*a(2n-1) + 2*a(2n-2).
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011