A022101 Fibonacci sequence beginning 1, 11.

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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

Links

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

Crossrefs

Cf. A000032, A000045.

a(n) = A109754(10, n+1) = A101220(10, 0, n+1).

Programs

Magma

a0:=1; a1:=11; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

Mathematica

LinearRecurrence[{1,1},{1,11},40] (* Harvey P. Dale, Aug 16 2015 *)

PARI

a(n) = 10*fibonacci(n)+fibonacci(n+1) \\ Charles R Greathouse IV, Jun 11 2015

Formula

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+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + 5*((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009

a(n) = 10*A000045(n) + A000045(n+1). - R. J. Mathar, Apr 07 2011

a(n) = 12*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017

a(n) = A000045(n+4) + A000032(n-4) for n > 0. - Bruno Berselli, Sep 27 2017