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a(n) = a(n-1) + a(n-2) + 1, with a(0) = 0 and a(1) = 12.
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%I #11 Aug 25 2017 23:35:29

%S 0,12,13,26,40,67,108,176,285,462,748,1211,1960,3172,5133,8306,13440,

%T 21747,35188,56936,92125,149062,241188,390251,631440,1021692,1653133,

%U 2674826,4327960,7002787,11330748

%N a(n) = a(n-1) + a(n-2) + 1, with a(0) = 0 and a(1) = 12.

%H G. C. Greubel, <a href="/A022317/b022317.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-1).

%F From _R. J. Mathar_, Apr 07 2011: (Start)

%F G.f.: x*(12-11*x)/( (1-x)*(1-x-x^2) ).

%F a(n) = A022103(n) - 1. (End)

%F a(n) = F(n+2) + 11*F(n) - 1, where F = A000045. - _G. C. Greubel_, Aug 25 2017

%t LinearRecurrence[{2,0,-1}, {0,12,13}, 50] (* _G. C. Greubel_, Aug 25 2017 *)

%o (PARI) x='x+O('x^50); concat([0], Vec(x*(12-11*x)/( (1-x)*(1-x-x^2) ))) \\ _G. C. Greubel_, Aug 25 2017

%K nonn

%O 0,2

%A _N. J. A. Sloane_