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A085449
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Horadam sequence (0,1,4,2).
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17
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0, 1, 2, 8, 24, 80, 256, 832, 2688, 8704, 28160, 91136, 294912, 954368, 3088384, 9994240, 32342016, 104660992, 338690048, 1096024064, 3546808320, 11477712896, 37142659072, 120196169728, 388962975744, 1258710630400
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OFFSET
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0,3
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COMMENTS
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a(n) / a(n-1) converges to sqrt(5) + 1 as n approaches infinity. sqrt(5) + 1 can also be written as Phi^3 - 1, 2 * Phi, Phi^2 + Phi - 1 and (L(n) / F(n)) + 1, where L(n) is the n-th Lucas number and F(n) is the n-th Fibonacci number as n approaches infinity.
Binomial transform is A001076. - Paul Barry, Aug 25 2003
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LINKS
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Karl V. Keller, Jr., Table of n, a(n) for n = 0..1000
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Eric Weisstein, Horadam Sequence.
Eric Weisstein, Fibonacci Number.
Eric Weisstein, Pell Number.
Eric Weisstein, Lucas Number.
Eric Weisstein, Lucas Sequence.
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FORMULA
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a(n) = s*a(n-1) + r*a(n-2); for n > 1, where a(0) = 0, a(1) = 1, s = 2, r = 4.
From Paul Barry, Aug 25 2003: (Start)
G.f.: x/(1-2*x-4*x^2).
a(n) = sqrt(5)*((1+sqrt(5))^n - (1-sqrt(5))^n)/10.
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k+1)5^k . (End)
The signed version 0, 1, -2, ... has a(n)=sqrt(5)((sqrt(5)-1)^n-(-sqrt(5)-1)^n)/10. It is the second inverse binomial transform of A085449. - Paul Barry, Aug 25 2003
a(n) = 2^(n-1)*Fib(n). - Paul Barry, Mar 22 2004
Sum_{n>=1} 1/a(n) = A269991. - Amiram Eldar, Feb 01 2021
a(n) = -(-4)^n*a(-n) for all integer n. - Michael Somos, Mar 07 2021
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EXAMPLE
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a(4) = 24 because a(3) = 8, a(2) = 2, s = 2, r = 4 and (2 * 8) + (4 * 2) = 24.
G.f. = x + 2*x^2 + 8*x^3 + 24*x^4 + 80*x^5 + 256*x^6 + 832*x^7 + ... - Michael Somos, Mar 07 2021
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MAPLE
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a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*(a[n-1]+2*a[n-2]) od: seq(a[n], n=0..26); # Zerinvary Lajos, Mar 17 2008
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MATHEMATICA
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Table[2^(n-1)*Fibonacci[n], {n, 0, 50}] (* G. C. Greubel, Oct 08 2018 *)
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PROG
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(PARI) vector(50, n, n--; 2^(n-1)*fibonacci(n)) \\ G. C. Greubel, Oct 08 2018
(MAGMA) [2^(n-1)*Fibonacci(n): n in [0..50]]; // G. C. Greubel, Oct 08 2018
(GAP) a:=[0, 1];; for n in [3..30] do a[n]:=2*a[n-1]+4*a[n-2]; od; a; # Muniru A Asiru, Oct 09 2018
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CROSSREFS
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Cf. A024318, A000032, A000129, A001076, A085939, A269991.
Essentially the same as A063727.
Sequence in context: A327550 A034741 A063727 * A127362 A133443 A094038
Adjacent sequences: A085446 A085447 A085448 * A085450 A085451 A085452
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KEYWORD
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easy,nonn
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AUTHOR
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Ross La Haye, Aug 18 2003
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STATUS
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approved
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