OFFSET
0,2
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).
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) = 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
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 *)
PROG
(Magma) a0:=1; a1:=8; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013
(PARI) a(n)=([0, 1; 1, 1]^n*[1; 8])[1, 1] \\ Charles R Greathouse IV, Oct 07 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved