A135854 a(n) = (n+1)*(2^n+1) for n > 0 with a(0)=1.
1, 6, 15, 36, 85, 198, 455, 1032, 2313, 5130, 11275, 24588, 53261, 114702, 245775, 524304, 1114129, 2359314, 4980755, 10485780, 22020117, 46137366, 96469015, 201326616, 419430425, 872415258, 1811939355, 3758096412
Offset: 0
Examples
a(3) = 15 = sum of row 3 terms of triangle A135853: (6 + 6 + 3). a(4) = 36 = (1, 3, 3, 1) dot (1, 5, 4, 8) = (1 + 15 + 12 + 8).
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (6,-13,12,-4)
Programs
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Maple
A135854:=n->(n+1)*(2^n+1): 1, seq(A135854(n), n=1..50); # Wesley Ivan Hurt, Dec 07 2016
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Mathematica
Join[{1}, LinearRecurrence[{6,-13,12,-4}, {6,15,36,85}, 25]] (* G. C. Greubel, Dec 07 2016 *) -
PARI
Vec((1-8*x^2+12*x^3-4*x^4)/((1-x)^2*(1-2*x)^2) + O(x^50)) \\ G. C. Greubel, Dec 07 2016
Formula
Binomial transform of [1, 5, 4, 8, 8, 12, 12, 16, 16, 20, 20, ...].
G.f.: 1 - x*(-6 + 21*x - 24*x^2 + 8*x^3) / ( (2*x-1)^2*(x-1)^2 ). - R. J. Mathar, Apr 04 2012
G.f.: (1 - 8*x^2 + 12*x^3 - 4*x^4)/((1-x)^2*(1-2*x)^2). - L. Edson Jeffery, Jan 14 2014
a(0) = 1, a(n) = (n+1)*(2^n+1), n>0. - L. Edson Jeffery, Jan 14 2014
E.g.f.: exp(x)*(1 + x + exp(x)*(1 + 2*x)) - 1. - Stefano Spezia, Dec 13 2021
Extensions
Corrected by R. J. Mathar, Apr 04 2012