A140660 a(n) = 3*4^n + 1.
4, 13, 49, 193, 769, 3073, 12289, 49153, 196609, 786433, 3145729, 12582913, 50331649, 201326593, 805306369, 3221225473, 12884901889, 51539607553, 206158430209, 824633720833, 3298534883329, 13194139533313, 52776558133249
Offset: 0
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..300
- Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
- Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
- Index entries for linear recurrences with constant coefficients, signature (5,-4).
Programs
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Magma
[3*4^n+1: n in [0..30] ]; // Vincenzo Librandi, May 23 2011
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Mathematica
LinearRecurrence[{5,-4}, {4,13}, 50] (* or *) CoefficientList[Series[ (7*x-4)/((1-x)*(4*x-1)), {x,0,50}], x] (* G. C. Greubel, Sep 15 2017 *)
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PARI
x='x+O('x^50); Vec((7*x-4)/((1-x)*(4*x-1))) \\ G. C. Greubel, Sep 15 2017
Formula
a(n) = A002001(n+1) + 1.
a(n) = 4*a(n-1) - 3.
First differences: a(n+1) - a(n) = A002063(n).
O.g.f.: (7*x - 4)/((1 - x)*(4*x - 1)). - R. J. Mathar, Jul 14 2008
From G. C. Greubel, Sep 15 2017: (Start)
E.g.f.: 3*exp(4*x) + exp(x).
a(n) = 5*a(n-1) - 4*a(n-2). (End)
Extensions
Edited and extended R. J. Mathar, Jul 14 2008
Comments