A022098 Fibonacci sequence beginning 1, 8.

1, 8, 9, 17, 26, 43, 69, 112, 181, 293, 474, 767, 1241, 2008, 3249, 5257, 8506, 13763, 22269, 36032, 58301, 94333, 152634, 246967, 399601, 646568, 1046169, 1692737, 2738906, 4431643, 7170549, 11602192, 18772741, 30374933, 49147674, 79522607, 128670281

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(8; n-1-k, k) with n >= 1, a(-1) = 7. These are the SW-NE diagonals in P(8; n, k), the (8, 1) Pascal triangle A093565. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

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 the same as A106291?). - R. J. Mathar, Aug 10 2012

Also the sum of five consecutive Lucas numbers starting with L(-3). - Alonso del Arte, Sep 26 2013

The Pisano period lengths of this sequence are exactly the same as those of the Lucas sequence (A000032), given in A106291. - Klaus Purath, Apr 20 2019

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Tanya Khovanova, Recursive Sequences
• Index entries for linear recurrences with constant coefficients, signature (1,1).

Crossrefs

Cf. A000032, A000045.

a(n) = A109754(7, n+1) = A101220(7, 0, n+1).

Programs

Magma

a0:=1; a1:=8; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

Mathematica

LinearRecurrence[{1, 1}, {1, 8}, 40] (* Alonso del Arte, Sep 26 2013 *)
CoefficientList[Series[(1 + 7 x)/(1 - x - x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 27 2013 *)
Table[LucasL[n + 3] + LucasL[n - 3] - 3 LucasL[n], {n, 2, 40}] (* Bruno Berselli, Dec 30 2016 *)

PARI

a(n)=([0,1; 1,1]^n*[1;8])[1,1] \\ Charles R Greathouse IV, Oct 07 2016

Formula

a(n) = a(n-1) + a(n-2) for n>1, a(0)=1, a(1)=8, and a(-1):=7.

G.f.: (1 + 7*x)/(1 - x - x^2).

a(n) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5)) + 3.5*((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) = 2^(-1-n)*((1 - sqrt(5))^n*(-15 + sqrt(5)) + (1 + sqrt(5))^n*(15 + sqrt(5)))/sqrt(5). - Herbert Kociemba, Dec 18 2011

a(n) = 7*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012

a(n) = 9*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017

a(n) = Lucas(n+5) + Lucas(n-1) - 3*Lucas(n+2). - Bruno Berselli, Dec 29 2016, corrected by Greg Dresden, Mar 02 2022

a(n) = Lucas(n+3) - Lucas(n-2). - Greg Dresden and Michael Nyc, Mar 02 2022