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A022095
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Fibonacci sequence beginning 1 5.
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19
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1, 5, 6, 11, 17, 28, 45, 73, 118, 191, 309, 500, 809, 1309, 2118, 3427, 5545, 8972, 14517, 23489, 38006, 61495, 99501, 160996, 260497, 421493, 681990, 1103483, 1785473, 2888956, 4674429, 7563385, 12237814, 19801199, 32039013, 51840212, 83879225, 135719437
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OFFSET
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0,2
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COMMENTS
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a(n-1) = sum(P(5;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=4. These are the sums of the SW-NE diagonals in P(5;n,k), the (5,1) Pascal triangle A093562. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also sums of the SW-NE diagonals in the (1,4)-Pascal triangle A095666.
Row sums of triangle A131776 starting (1, 5, 6, 11, 17, 28,...). - Gary W. Adamson, Jul 14 2007
In general, for a Fibonacci sequence beginning with 1,b we have:
a(n) = (2^(-1-n)((1-sqrt(5))^n*(1+sqrt(5)-2b)+(1+sqrt(5))^n*(-1+sqrt(5)+2b)))/sqrt(5). - Herbert Kociemba(kociemba(AT)t-online.de), Dec 18 2011
Subsequence of primes: 5, 11, 17, 73, 191, 809, 421493, 1103483,... - R. J. Mathar, Aug 09 2012
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 9, 60,.. (differs from A001175). - R. J. Mathar, Aug 10 2012
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LINKS
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Table of n, a(n) for n=0..37.
Tanya Khovanova, Recursive Sequences
José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368, 2012
Index entries for sequences related to linear recurrences with constant coefficients, signature (1,1)
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FORMULA
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a(n) = a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=5, a(-1):=4.
G.f.: (1+4*x)/(1-x-x^2).
a(n) = 4*fibonacci(n)+fibonacci(n+1), n>=1 - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 05 2007, corrected by R. J. Mathar, Apr 07 2011
a(n-1) = ((1+sqrt5)^n-(1-sqrt5)^n)/(2^n*sqrt5)+ 2*((1+sqrt5)^(n-1)-(1-sqrt5)^(n-1))/(2^(n-2)*sqrt5). [Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009]
a(n) = 4*Fibonacci(n+2) - 3*Fibonacci(n+1). [Gary Detlefs, Dec 21 2010]
a(n) = (L(n-2) + 8*L(n-1) + 4*L(n) + 2*L(n+1))/5 for the Lucas numbers L(n). - J. M. Bergot, Oct 22 2012
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MAPLE
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a:=n->4*fibonacci(n)+fibonacci(n+1): seq(a(n), n=0..32); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 05 2007
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MATHEMATICA
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lst = {1, 5}; b = 1; c = 5; Do[b = b + c; AppendTo[lst, b]; c = b + c; AppendTo[lst, c], {n, 18}]; lst (* _Vladimir Orlovsky_, Jul 22 2008 and modified by Robert G. Wilson v, Oct 22 2012 *)
f[n_] := (LucasL[n - 2] + 8*LucasL[n - 1] + 4*LucasL[n] + 2*LucasL[n + 1])/5; Array[f, 38, 0] (* or *)
LinearRecurrence[{1, 1}, {1, 5}, 38] (* Robert G. Wilson v, Oct 22 2012 *)
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PROG
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(PARI) a(n)=fibonacci(n-1)+5*fibonacci(n) \\ Charles R Greathouse IV, Jun 05, 2011
(MAGMA) a0:=1; a1:=5; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013
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CROSSREFS
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a(n) = A101220(4, 0, n+1).
a(n) = A109754(4, n+1).
Cf. A131776.
Sequence in context: A136974 A101187 A070373 * A042531 A042839 A041373
Adjacent sequences: A022092 A022093 A022094 * A022096 A022097 A022098
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KEYWORD
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nonn,easy,changed
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AUTHOR
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N. J. A. Sloane.
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STATUS
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approved
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