A022102 Fibonacci sequence beginning 1, 12.

1, 12, 13, 25, 38, 63, 101, 164, 265, 429, 694, 1123, 1817, 2940, 4757, 7697, 12454, 20151, 32605, 52756, 85361, 138117, 223478, 361595, 585073, 946668, 1531741, 2478409, 4010150, 6488559, 10498709

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(12;n-1-k,k) with n>=1, a(-1)=11. These are the SW-NE diagonals in P(12;n,k), the (12,1) 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

Links

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

Crossrefs

Cf. A000045, A001175, A093645, A101220, A109754.

Programs

Magma

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

Mathematica

LinearRecurrence[{1,1},{1,12},40] (* Harvey P. Dale, Jan 23 2012 *)

PARI

a(n) = if(n==0, 1, if(n==1, 12, a(n-1)+a(n-2))) \\ Felix Fröhlich, Jun 09 2022

PARI

Vec((1+11*x)/(1-x-x^2) + O(x^20)) \\ Felix Fröhlich, Jun 09 2022

Formula

a(n) = a(n-1) + a(n-2), n >= 2, a(0)=1, a(1)=12. a(-1):=11.

G.f.: (1+11*x)/(1-x-x^2).

a(n) = A109754(11, n+1) = A101220(11, 0, n+1).

a(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + (11/2)*((1+sqrt(5))^(n-1)-(1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=13. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009

a(n) = 11*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012

a(n) = 13*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017

a(n) = F(n+5) + F(n-1) - F(n-5) for F(n) the Fibonacci number A000045(n). - Greg Dresden and Aamen Muharram, Jun 09 2022