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A022100
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Fibonacci sequence beginning 1 10.
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6
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1, 10, 11, 21, 32, 53, 85, 138, 223, 361, 584, 945, 1529, 2474, 4003, 6477, 10480, 16957, 27437, 44394, 71831, 116225, 188056, 304281, 492337, 796618, 1288955, 2085573, 3374528, 5460101, 8834629, 14294730, 23129359, 37424089, 60553448, 97977537, 158530985
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| a(n-1)=sum(P(10;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=9. These are the SW-NE diagonals in P(10;n,k), the (10,1) Pascal triangle A093645. 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)=10. a(-1):=9.
G.f.: (1+9*x)/(1-x-x^2).
a(n)=sum{k=0..n, Fib(n-k+1)(9*binomial(1, k)-8*binomial(0, k))} - Paul Barry (pbarry(AT)wit.ie), May 05 2005
a(n)=((1+sqrt5)^n-(1-sqrt5)^n)/(2^n*sqrt5)+ 4.5*((1+sqrt5)^(n-1)-(1-sqrt5)^(n-1))/(2^(n-2)*sqrt5). Offset 1. a(3)=11. [From Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009]
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MATHEMATICA
| a={}; b=1; c=10; 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(9, n+1) = A101220(9, 0, n+1).
Sequence in context: A014418 A089591 A064039 * A041475 A041204 A041202
Adjacent sequences: A022097 A022098 A022099 * A022101 A022102 A022103
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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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