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A022097 Fibonacci sequence beginning 1, 7. 17
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 (list; graph; refs; listen; history; text; internal format)
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
LINKS
Tanya Khovanova, Recursive Sequences
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).
Sequence in context: A231390 A231458 A070424 * A041100 A129658 A041693
KEYWORD
nonn,easy,changed
AUTHOR
STATUS
approved

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Last modified February 20 20:53 EST 2024. Contains 370217 sequences. (Running on oeis4.)