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If n mod 2 = 0 then 2^n*3^(n-1)+2^(n+1)*3^(n/2-1) otherwise 2^n*3^(n-1)+2^n*3^((n-1)/2).
1

%I #15 Sep 08 2022 08:46:06

%S 1,4,20,96,528,2880,16704,96768,573696,3400704,20321280,121430016,

%T 727584768,4359536640,26145275904,156799991808,940656623616,

%U 5643079778304,33856758743040,203130232897536,1218760758263808,7312440714854400,43874396619669504,263244893701275648,1579466390174171136

%N If n mod 2 = 0 then 2^n*3^(n-1)+2^(n+1)*3^(n/2-1) otherwise 2^n*3^(n-1)+2^n*3^((n-1)/2).

%H Vincenzo Librandi, <a href="/A232493/b232493.txt">Table of n, a(n) for n = 0..1000</a>

%H G. Wu, M. G. Parker, <a href="http://arxiv.org/abs/1309.0157">A complementary construction using mutually unbiased bases</a>, arXiv preprint arXiv:1309.0157 [cs.IT], 2013 [See Th. 1].

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,12,-72).

%F G.f.: ( 1-2*x-16*x^2 ) / ( (6*x-1)*(12*x^2-1) ). - _R. J. Mathar_, Dec 04 2013

%p f:=proc(n)

%p if (n mod 2) = 0 then 2^n*3^(n-1)+2^(n+1)*3^(n/2-1) else

%p 2^n*3^(n-1)+2^n*3^((n-1)/2) fi; end;

%p [seq(f(n),n=0..40)];

%t LinearRecurrence[{6,12,-72},{1,4,20},40] (* _Harvey P. Dale_, May 03 2017 *)

%t CoefficientList[Series[(1 - 2 x - 16 x^2)/((6 x - 1) (12 x^2 - 1)), {x, 0, 33}], x] (* _Vincenzo Librandi_, May 07 2017 *)

%o (Magma) I:=[1,4,20]; [n le 3 select I[n] else 6*Self(n-1)+12*Self(n-2)-72*Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, May 07 2017

%K nonn

%O 0,2

%A _N. J. A. Sloane_, Dec 02 2013

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Last modified September 24 00:34 EDT 2024. Contains 376185 sequences. (Running on oeis4.)