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A022100
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Fibonacci sequence beginning 1, 10.
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8
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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
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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(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, 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
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LINKS
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FORMULA
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a(n)= a(n-1) + a(n-2) for 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, 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. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
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MATHEMATICA
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LinearRecurrence[{1, 1}, {1, 10}, 40] (* Harvey P. Dale, May 17 2017 *)
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PROG
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(Magma) a0:=1; a1:=10; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // 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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