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A192240
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Constant term in the reduction of the polynomial (x+3)^n by x^2 -> x+1.
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3
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1, 3, 10, 37, 149, 636, 2813, 12695, 57922, 265809, 1223521, 5640748, 26026505, 120137307, 554669594, 2561176781, 11826871933, 54615158940, 252210521317, 1164706900879, 5378632571666, 24838652091993, 114705606355625, 529714071477452
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OFFSET
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0,2
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COMMENTS
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LINKS
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FORMULA
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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
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
a(n) = Sum_{i=0..n} (-1)^i*Fibonacci(i+1)*binomial(n,i)*4^(n-i) (conjecture). - Rigoberto Florez, Mar 25 2020
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MAPLE
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seq(eval(rem((x+3)^n, x^2-x-1, x), x=0), n=0..50); # Robert Israel, Mar 14 2023
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MATHEMATICA
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q[x_] := x + 1;
p[n_, x_] := (x + 3)^n;
reductionRules = {x^y_?EvenQ -> q[x]^(y/2),
x^y_?OddQ -> x q[x]^((y - 1)/2)};
t = Table[
Last[Most[
FixedPointList[Expand[#1 /. reductionRules] &, p[n, x]]]], {n, 0,
30}];
Table[Coefficient[Part[t, n], x, 0], {n, 30}] (* A192240 *)
Table[Coefficient[Part[t, n], x, 1], {n, 30}] (* A099453 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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