A022097 Fibonacci sequence beginning 1, 7.

1, 7, 8, 15, 23, 38, 61, 99, 160, 259, 419, 678, 1097, 1775, 2872, 4647, 7519, 12166, 19685, 31851, 51536, 83387, 134923, 218310, 353233, 571543, 924776, 1496319, 2421095, 3917414, 6338509, 10255923, 16594432, 26850355, 43444787, 70295142, 113739929

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(7;n-1-k,k) with n>=1, a(-1)=6. These are the SW-NE diagonals in P(7;n,k), the (7,1) Pascal triangle A093564. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (perhaps the same as A001175). - R. J. Mathar, Aug 10 2012

For n >= 1, a(n) is the number of edge covers of the tadpole graph T_{4,n-1} with T_{4,0} interpreted as just the cycle graph C_4. Example: If n=2, we have C_4 and path P_1 joined by a bridge. This is the cycle with a pendant and has 7 edge covers. - Feryal Alayont, Sep 22 2024

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Tanya Khovanova, Recursive Sequences
• Eric Weisstein's World of Mathematics, Tadpole Graph.
• Index entries for linear recurrences with constant coefficients, signature (1,1).

Crossrefs

Cf. A000045, A001175, A093564, A101220, A109754, A118654, A131778.

Programs

Magma

a0:=1; a1:=7; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

Mathematica

First /@ NestList[{Last@ #, Total@ #} &, {1, 7}, 36] (* or *)
CoefficientList[Series[(1 + 6 x)/(1 - x - x^2), {x, 0, 36}], x] (* Michael De Vlieger, Feb 20 2017 *)
LinearRecurrence[{1,1},{1,7},40] (* Harvey P. Dale, May 17 2018 *)

PARI

a(n)=([0,1; 1,1]^n*[1;7])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

SageMath

A022097=BinaryRecurrenceSequence(1,1,1,7)
print([A022097(n) for n in range(41)]) # G. C. Greubel, Jun 03 2025

Formula

a(n) = a(n-1) + a(n-2) for n>=2, a(0)=1, a(1)=7, a(-1):=6.

G.f.: (1+6*x)/(1-x-x^2).

a(n) = A109754(6, n+1).

a(n) = A118654(3, n).

a(n) = (2^(-1-n)*((1 - sqrt(5))^n*(-13 + sqrt(5)) + (1 + sqrt(5))^n*(13 + sqrt(5))))/sqrt(5). - Herbert Kociemba

a(n) = 6*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012

a(n) = 8*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017

From Aamen Muharram, Aug 05 2022: (Start)

a(n) = F(n-4) + F(n-1) + F(n+4),

a(n) = F(n) + F(n+4) - F(n-3),

where F(n) = A000045(n) is the Fibonacci numbers. (End)