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A022101
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Fibonacci sequence beginning 1 11.
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5
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1, 11, 12, 23, 35, 58, 93, 151, 244, 395, 639, 1034, 1673, 2707, 4380, 7087, 11467, 18554, 30021, 48575, 78596, 127171, 205767, 332938, 538705, 871643, 1410348, 2281991, 3692339, 5974330, 9666669
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| a(n-1)=sum(P(11;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=10. These are the SW-NE diagonals in P(11;n,k), the (11,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. 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
Index to sequences with linear recurrences with constant coefficients, signature (1,1)
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FORMULA
| a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=11. a(-1):=10.
G.f.: (1+10*x)/(1-x-x^2).
a(n-1)=((1+sqrt5)^n-(1-sqrt5)^n)/(2^n*sqrt5)+ 5*((1+sqrt5)^(n-1)-(1-sqrt5)^(n-1))/(2^(n-2)*sqrt5). [From Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009]
a(n) = 10*A000045(n)+A000045(n+1). - R. J. Mathar, Apr 07 2011
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MATHEMATICA
| a={}; b=1; c=11; 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(10, n+1) = A101220(10, 0, n+1).
Sequence in context: A015903 A105945 A139114 * A041246 A042633 A197221
Adjacent sequences: A022098 A022099 A022100 * A022102 A022103 A022104
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KEYWORD
| nonn
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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