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%I #28 Mar 14 2023 08:38:37
%S 1,3,10,37,149,636,2813,12695,57922,265809,1223521,5640748,26026505,
%T 120137307,554669594,2561176781,11826871933,54615158940,252210521317,
%U 1164706900879,5378632571666,24838652091993,114705606355625,529714071477452
%N Constant term in the reduction of the polynomial (x+3)^n by x^2 -> x+1.
%C See A192232.
%H Robert Israel, <a href="/A192240/b192240.txt">Table of n, a(n) for n = 0..1498</a>
%F Empirical g.f. and recurrence: (1-4*x)/(1-7*x+11*x^2). a(n) = 7*a(n-1) - 11*a(n-2). - _Colin Barker_, Feb 09 2012
%F Proof of recurrence: if r(n) == (x+3)^n mod (x^2-x-1), then r(n+j) == (x+1)^(n+j) mod (x^2 - x - 1). Now r(n+2) - 7*r(n+1) + 11*r(n) == ((x+3)^2 - 7*(x+3) + 11)*r(n) == 0 mod (x^2-x-1) since ((x+3)^2 - 7*(x+3) + 11 = x^2 - x - 1. - _Robert Israel_, Mar 14 2023
%F a(n) = Sum_{i=0..n} (-1)^i*Fibonacci(i+1)*binomial(n,i)*4^(n-i) (conjecture). - _Rigoberto Florez_, Mar 25 2020
%p seq(eval(rem((x+3)^n,x^2-x-1,x),x=0),n=0..50); # _Robert Israel_, Mar 14 2023
%t q[x_] := x + 1;
%t p[n_, x_] := (x + 3)^n;
%t reductionRules = {x^y_?EvenQ -> q[x]^(y/2),
%t x^y_?OddQ -> x q[x]^((y - 1)/2)};
%t t = Table[
%t Last[Most[
%t FixedPointList[Expand[#1 /. reductionRules] &, p[n, x]]]], {n, 0,
%t 30}];
%t Table[Coefficient[Part[t, n], x, 0], {n, 30}] (* A192240 *)
%t Table[Coefficient[Part[t, n], x, 1], {n, 30}] (* A099453 *)
%t (* _Peter J. C. Moses_, Jun 26 2011 *)
%Y Cf. A192232.
%K nonn
%O 0,2
%A _Clark Kimberling_, Jun 26 2011
%E Offset corrected by _Robert Israel_, Mar 14 2023
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