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A330795 Evaluation of the polynomials given by the Riordan square of the Fibonacci sequence with a(0) = 1 (A193737) at 1/2 and normalized with 2^n.

%I A330795 #14 Sep 15 2023 01:39:47
%S A330795 1,3,9,39,153,615,2457,9831,39321,157287,629145,2516583,10066329,
%T A330795 40265319,161061273,644245095,2576980377,10307921511,41231686041,
%U A330795 164926744167,659706976665,2638827906663,10555311626649,42221246506599,168884986026393,675539944105575
%N A330795 Evaluation of the polynomials given by the Riordan square of the Fibonacci sequence with a(0) = 1 (A193737) at 1/2 and normalized with 2^n.
%H A330795 G. C. Greubel, <a href="/A330795/b330795.txt">Table of n, a(n) for n = 0..1000</a>
%H A330795 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,4).
%F A330795 a(n) = 2^n*Sum_{k=0..n} A193737(n,k)/2^k.
%F A330795 a(n) = [x^n] (1 - 4*x^2)/(1 - x*(3 + 4*x)).
%F A330795 a(n) = n! [x^n] (3/5)*exp(-x)*(exp(5*x) - 1) + 1.
%F A330795 a(n) = 4*a(n-2) + 3*a(n-1).
%F A330795 a(n) = 3*A015521(n), n>0. - _R. J. Mathar_, Aug 19 2022
%p A330795 gf := (4*x^2 - 1)/(x*(4*x + 3) - 1): ser := series(gf, x, 32):
%p A330795 seq(coeff(ser, x, n), n=0.. 25);
%p A330795 # Alternative:
%p A330795 gf:= (3/5)*exp(-x)*(exp(5*x) - 1) + 1: ser := series(gf, x, 32):
%p A330795 seq(n!*coeff(ser, x, n), n=0.. 25);
%p A330795 # Or:
%p A330795 a := proc(n) option remember; if n < 3 then return [1, 3, 9][n + 1] fi;
%p A330795 4*a(n-2) + 3*a(n-1) end: seq(a(n), n=0..25);
%t A330795 LinearRecurrence[{3,4}, {1,3,9}, 31] (* _G. C. Greubel_, Sep 14 2023 *)
%o A330795 (Magma) [1] cat [3*(4^n -(-1)^n)/5: n in [1..30]]; // _G. C. Greubel_, Sep 14 2023
%o A330795 (SageMath) [3*(4^n -(-1)^n)//5 + int(n==0) for n in range(31)] # _G. C. Greubel_, Sep 14 2023
%Y A330795 Cf. A006131, A015521, A193737, A321620, A324969 (Fibonacci with a(0)=1).
%K A330795 nonn,easy
%O A330795 0,2
%A A330795 _Peter Luschny_, Jan 10 2020