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A022103
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Fibonacci sequence beginning 1, 13.
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6
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1, 13, 14, 27, 41, 68, 109, 177, 286, 463, 749, 1212, 1961, 3173, 5134, 8307, 13441, 21748, 35189, 56937, 92126, 149063, 241189, 390252, 631441, 1021693, 1653134, 2674827, 4327961, 7002788, 11330749, 18333537, 29664286, 47997823, 77662109, 125659932, 203322041, 328981973
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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(13;n-1-k,k) for n>=1, a(-1)=12. These are the SW-NE diagonals in P(13;n,k), the (13,1) 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, 4, 24, 16, 12, 24, 12, 10, 24, 28, 48, 8, 24, 36, 24, 18, 12, ... (is this A106291?). - 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) for n>=2, a(0)=1, a(1)=13, and a(-1):=12.
G.f.: (1 + 12*x)/(1 - x - x^2).
a(n) = ((1 + sqrt(5))^n-(1 - sqrt(5))^n)/(2^n*sqrt(5))+ 6*((1 + sqrt(5))^(n-1)-(1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)) for n>0. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
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MATHEMATICA
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LinearRecurrence[{1, 1}, {1, 13}, 40] (* or *) Table[LucasL[n + 5] - 5 LucasL[n], {n, 0, 40}] (* Bruno Berselli, Dec 30 2016 *)
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PROG
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(Magma) a0:=1; a1:=13; [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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