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

0,2

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

Table of n, a(n) for n=0..36.

Tanya Khovanova, Recursive Sequences

Eric Weisstein's World of Mathematics, Tadpole Graph.

Index entries for linear recurrences with constant coefficients, signature (1,1).

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) = (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)

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 *)

Prog

(Magma) a0:=1; a1:=7; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013
(PARI) a(n)=([0, 1; 1, 1]^n*[1; 7])[1, 1] \\ Charles R Greathouse IV, Oct 03 2016

Crossrefs

a(n) = A101220(6, 0, n+1) = A109754(6, n+1) = A118654(3, n).

Cf. A000045, A131778.

Sequence in context: A231390 A231458 A070424 * A041100 A129658 A041693

Adjacent sequences: A022094 A022095 A022096 * A022098 A022099 A022100

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved