A022095 Fibonacci sequence beginning 1, 5.

1, 5, 6, 11, 17, 28, 45, 73, 118, 191, 309, 500, 809, 1309, 2118, 3427, 5545, 8972, 14517, 23489, 38006, 61495, 99501, 160996, 260497, 421493, 681990, 1103483, 1785473, 2888956, 4674429, 7563385, 12237814, 19801199, 32039013, 51840212, 83879225, 135719437

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• changed

Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(5; n-1-k, k), n >= 1, with a(-1)=4. These are the sums of the SW-NE diagonals in P(5; n, k), the (5,1) Pascal triangle A093562. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also sums of the SW-NE diagonals in the (1,4)-Pascal triangle A095666.

Row sums of triangle A131776 starting (1, 5, 6, 11, 17, 28, ...). - Gary W. Adamson, Jul 14 2007

In general, for a Fibonacci sequence beginning with 1,b we have:

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

Subsequence of primes: 5, 11, 17, 73, 191, 809, 421493, 1103483, ... . - R. J. Mathar, Aug 09 2012

Pisano periods: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 9, 60, ... (differs from A001175). - R. J. Mathar, Aug 10 2012

Links

• Muniru A Asiru, Table of n, a(n) for n = 0..960
• Élis Gardel da Costa Mesquita, Eudes Antonio Costa, Paula M. M. C. Catarino, and Francisco R. V. Alves, Jacobsthal-Mulatu Numbers, Latin Amer. J. Math. (2025) Vol. 4, No. 1, 23-45. See p. 24.
• Tanya Khovanova, Recursive Sequences
• José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368 [cs.DM], 2012.
• José L. Ramírez and Gustavo N. Rubiano, Properties and Generalizations of the Fibonacci Word Fractal, The Mathematica Journal, Vol. 16 (2014).
• Matty van Son, Uniqueness conjectures for extended Markov numbers, arXiv:1911.00746 [math.NT], 2019.
• Index entries for linear recurrences with constant coefficients, signature (1,1).

Crossrefs

Cf. A000032, A000045, A001175, A093562, A095666, A101220, A109754, A131776.

Row n=4 of A109754 (shifted).

Programs

GAP

List([0..40],n->4*Fibonacci(n)+Fibonacci(n+1)); # Muniru A Asiru, Mar 04 2018

Magma

a0:=1; a1:=5; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

Maple

with(combinat): a:= n-> 4*fibonacci(n)+fibonacci(n+1): seq(a(n), n=0..32); # Zerinvary Lajos, Oct 05 2007

Mathematica

f[n_] := (LucasL[n - 2] + 8*LucasL[n - 1] + 4*LucasL[n] + 2*LucasL[n + 1])/5; Array[f, 38, 0] (* or *)
LinearRecurrence[{1, 1}, {1, 5}, 38] (* Robert G. Wilson v, Oct 22 2012 *)

PARI

a(n)=fibonacci(n-1)+5*fibonacci(n) \\ Charles R Greathouse IV, Jun 05 2011

SageMath

A022095=BinaryRecurrenceSequence(1,1,1,5)
[A022095(n) for n in range(41)] # G. C. Greubel, Jun 02 2025

Formula

a(n) = a(n-1) + a(n-2), n >= 2, a(0)=1, a(1)=5.

G.f.: (1+4*x)/(1-x-x^2).

a(n) = 4*Fibonacci(n) + Fibonacci(n+1), n >= 1. - Zerinvary Lajos, Oct 05 2007, corrected by R. J. Mathar, Apr 07 2011

a(n-1) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5)) + 2*((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) = 4*Fibonacci(n+2) - 3*Fibonacci(n+1). - Gary Detlefs, Dec 21 2010

a(n) = (L(n-2) + 8*L(n-1) + 4*L(n) + 2*L(n+1))/5 for the Lucas numbers L(n). - J. M. Bergot, Oct 22 2012

a(n) = ((2*sqrt(5) - 1)*(((1 + sqrt(5))/2)^(n+1)) + (2*sqrt(5) + 1)*(((1 - sqrt(5))/2)^(n+1)))/(sqrt(5)). - Bogart B. Strauss, Jul 19 2013

a(n) = Lucas(n-1) + Fibonacci(n+3) = Lucas(n+2) - Fibonacci(n-3). - Greg Dresden and Griffin Donaldson, Mar 03 2022