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A176818 a(n) = (3^(2*n+1) + 2^(n+2))/7. 0

%I #8 Feb 18 2016 14:34:15

%S 1,5,37,317,2821,25325,227797,2049917,18448741,166037645,1494336757,

%T 13449026717,121041232261,1089371073965,9804339632917,88239056630717,

%U 794151509545381,7147363585646285,64326272270292277,578936450431581917

%N a(n) = (3^(2*n+1) + 2^(n+2))/7.

%C 3^(2*n+1) + 2^(n+2) = 3*(3^2)^n + 4*2^n == 3*2^n + 4*2^n = 7*2^n == 0 (mod 7).

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-18).

%F G.f.: ( 1-6*x ) / ( (9*x-1)*(2*x-1) ). - _R. J. Mathar_, Feb 18 2016

%e a(2) = (3^5 + 2^4)/7 = 259/7 = 37.

%p a:= n-> (3^(2*n+1) + 2^(n+2))/7:

%p seq (a(n), n=0..30);

%K nonn,easy

%O 0,2

%A _Michel Lagneau_, Apr 26 2010

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Last modified March 28 20:05 EDT 2024. Contains 371254 sequences. (Running on oeis4.)