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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, 15640999, 25307668, 40948667, 66256335, 107205002, 173461337, 280666339
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OFFSET
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0,2
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COMMENTS
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a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(11;n-1-k,k) with n>=1, 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, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have:
b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, 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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FORMULA
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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). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
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MATHEMATICA
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LinearRecurrence[{1, 1}, {1, 11}, 40] (* Harvey P. Dale, Aug 16 2015 *)
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PROG
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(Magma) a0:=1; a1:=11; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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