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A022099
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Fibonacci sequence beginning 1 9.
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4
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1, 9, 10, 19, 29, 48, 77, 125, 202, 327, 529, 856, 1385, 2241, 3626, 5867, 9493, 15360, 24853, 40213, 65066, 105279, 170345, 275624, 445969, 721593, 1167562, 1889155, 3056717, 4945872, 8002589, 12948461, 20951050, 33899511, 54850561, 88750072, 143600633
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| a(n-1)=sum(P(9;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=8. These are the SW-NE diagonals in P(9;n,k), the (9,1) Pascal triangle A093644. Observation by Paul Barry (pbarry(AT)wit.ie, 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
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LINKS
| Tanya Khovanova, Recursive Sequences
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FORMULA
| a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=9. a(-1):=8.
G.f.: (1+8*x)/(1-x-x^2).
a(n)=((1+sqrt5)^n-(1-sqrt5)^n)/(2^n*sqrt5)+ 4*((1+sqrt5)^(n-1)-(1-sqrt5)^(n-1))/(2^(n-2)*sqrt5). Offset 1. a(3)=10. [From Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009]
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MATHEMATICA
| a={}; b=1; c=9; AppendTo[a, b]; AppendTo[a, c]; Do[b=b+c; AppendTo[a, b]; c=b+c; AppendTo[a, c], {n, 1, 9, 1}]; a (Vladimir Orlovsky, Jul 22 2008)
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CROSSREFS
| a(n) = A109754(8, n+1) = A101220(8, 0, n+1).
Sequence in context: A141640 A015898 A050551 * A042113 A041166 A042613
Adjacent sequences: A022096 A022097 A022098 * A022100 A022101 A022102
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KEYWORD
| nonn,easy
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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