A022103 Fibonacci sequence beginning 1, 13.

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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

Links

• Tanya Khovanova, Recursive Sequences
• Index entries for linear recurrences with constant coefficients, signature (1,1).

Crossrefs

a(n) = A109754(12, n+1) = A101220(12, 0, n+1).

Cf. A000032, A000045.

Programs

Magma

a0:=1; a1:=13; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

Mathematica

LinearRecurrence[{1, 1}, {1, 13}, 40] (* or *) Table[LucasL[n + 5] - 5 LucasL[n], {n, 0, 40}] (* Bruno Berselli, Dec 30 2016 *)

Formula

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

a(n) = 12*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012

a(n) = 14*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017

a(n) = Lucas(n+5) - 5*Lucas(n). - Bruno Berselli, Dec 30 2016