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A022102
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Fibonacci sequence beginning 1 12.
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3
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1, 12, 13, 25, 38, 63, 101, 164, 265, 429, 694, 1123, 1817, 2940, 4757, 7697, 12454, 20151, 32605, 52756, 85361, 138117, 223478, 361595, 585073, 946668, 1531741, 2478409, 4010150, 6488559, 10498709
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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(12;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=11. These are the SW-NE diagonals in P(12;n,k), the (12,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
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
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (is this A001175?) - R. J. Mathar, Aug 10 2012
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LINKS
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Table of n, a(n) for n=0..30.
Tanya Khovanova, Recursive Sequences
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FORMULA
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a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=12. a(-1):=11.
G.f.: (1+11*x)/(1-x-x^2).
a(n)=((1+sqrt5)^n-(1-sqrt5)^n)/(2^n*sqrt5)+ 5.5*((1+sqrt5)^(n-1)-(1-sqrt5)^(n-1))/(2^(n-2)*sqrt5). Offset 1. a(3)=13. [From Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009]
a(n) = 11*A000045(n)+A000045(n+1). - R. J. Mathar, Aug 10 2012
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MATHEMATICA
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a={}; b=1; c=12; AppendTo[a, b]; AppendTo[a, c]; Do[b=b+c; AppendTo[a, b]; c=b+c; AppendTo[a, c], {n, 1, 12, 1}]; a (Vladimir Orlovsky, Jul 23 2008)
LinearRecurrence[{1, 1}, {1, 12}, 40] (* From Harvey P. Dale, Jan 23 2012 *)
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PROG
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(MAGMA) a0:=1; a1:=12; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
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CROSSREFS
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a(n) = A109754(11, n+1) = A101220(11, 0, n+1).
Sequence in context: A108710 A108709 A138821 * A041292 A041679 A041294
Adjacent sequences: A022099 A022100 A022101 * A022103 A022104 A022105
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KEYWORD
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nonn,easy
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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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