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Expand 1/(1 - (3/2)*x + (2/3)*x^4 - x^5) in powers of x, then multiply coefficient of x^n by 3^floor(n/4)*2^n to get integers.
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%I #59 Aug 01 2020 14:21:27

%S 1,3,9,27,211,633,1899,5697,52297,156891,470673,1412019,12675403,

%T 38026209,114078627,342235881,3081171505,9243514515,27730543545,

%U 83191630635,748691121283,2246073363849,6738220091547

%N Expand 1/(1 - (3/2)*x + (2/3)*x^4 - x^5) in powers of x, then multiply coefficient of x^n by 3^floor(n/4)*2^n to get integers.

%H Vincenzo Librandi, <a href="/A202907/b202907.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,211,0,0,0,7776).

%F From _Chai Wah Wu_, Aug 01 2020: (Start)

%F a(n) = 211*a(n-4) + 7776*a(n-8) for n > 7.

%F G.f.: -(3*x + 1)*(9*x^2 + 1)/(7776*x^8 + 211*x^4 - 1). (End)

%t Table[3^Floor[k/4]*2^k*SeriesCoefficient[ Series[1/(1 - (3/2)* x + (2/3) x^4 - x^5), {x, 0, 30}], k], {k, 0, 30}] (* Bagula *)

%t a[ n_] := 2^n 3^Quotient[ n, 4] SeriesCoefficient[ 1 / (1 - 3/2 x + 2/3 x^4 - x^5), {x, 0, n}] (* _Michael Somos_, Jan 27 2012 *)

%Y Cf. A167602, A167602, A117791, A107293, A204631, A185357.

%K nonn,easy

%O 0,2

%A _Roger L. Bagula_, Jan 27 2012