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 A165443 a(n) = ( 16^(2*n+1) + 81^(2*n+1) )/97. 1

%I

%S 1,5521,35957041,235845988561,1547368082644081,10152277523461827601,

%T 66609091687940958003121,437022250271846649679394641,

%U 2867302983958645970747063186161,18812374877733491600234823630721681

%N a(n) = ( 16^(2*n+1) + 81^(2*n+1) )/97.

%C The general form of the g.f. for (A^(2*n+1)+B^(2*n+1))/(A+B) is (1-A*B*x)/((1-A^2*x)(1-B^2*x)).

%H G. C. Greubel, <a href="/A165443/b165443.txt">Table of n, a(n) for n = 0..260</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6817,-1679616).

%F G.f.: (1 - 16*81*x)/((1 - 16^2*x)*(1 - 81^2*x)).

%F a(n) = (16^2+81^2)*a(n-1) - 16^2*81^2*a(n-2).

%e a(0) = (16^1 + 81^1)/97 = 97/97 = 1.

%e a(1) = (16^3 + 81^3)/97 = 535537/97 = 5521.

%p seq(coeff(series((1-16*81*x)/((1-16^2*x)*(1-81^2*x)),x,n+1), x, n), n = 0 .. 10); # _Muniru A Asiru_, Oct 21 2018

%t f[n_]:=Module[{c=2n+1},(16^c+81^c)/97]; Array[f,20,0] (* _Harvey P. Dale_, Oct 02 2012 *)

%o (PARI) a(n)=(16^(2*n+1)+81^(2*n+1))/97

%o (MAGMA) [(2^(8*n+4) + 3^(8*n+4))/97: n in [0..20]]; // _G. C. Greubel_, Oct 20 2018

%o (GAP) List([0..10],n->(16^(2*n+1)+81^(2*n+1))/97); # _Muniru A Asiru_, Oct 21 2018

%o (Python) for n in range(0, 10): print(int((16**(2*n+1)+81**(2*n+1))/97), end=', ') # _Stefano Spezia_, Oct 21 2018

%Y Cf. A007689, A082101, A096951, A165259.

%K nonn,easy

%O 0,2

%A _Jaume Oliver Lafont_, Sep 19 2009

%E Definition replaced with formula by _R. J. Mathar_, Sep 21 2009

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Last modified February 22 12:42 EST 2020. Contains 332136 sequences. (Running on oeis4.)