OFFSET
0,2
COMMENTS
a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(9;n-1-k,k) with n>=1, a(-1)=8. These are the SW-NE diagonals in P(9;n,k), the (9,1) Pascal triangle A093644. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have: b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
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
No, it is not the same as in A001175. The Pisano periods are different for moduli 71 and 142, where they are 35 and 105 instead of 70 and 210. Otherwise they coincide with those of the Fibonacci sequence. - Klaus Purath, Jun 26 2022
LINKS
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).
FORMULA
a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=9. a(-1):=8.
G.f.: (1+8*x)/(1-x-x^2).
a(n+1) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5))+ 4*((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) = Lucas(n+3) + Fibonacci(n-4). - Greg Dresden and Mary Beth Pittman, Mar 12 2022
MATHEMATICA
LinearRecurrence[{1, 1}, {1, 9}, 36] (* Robert G. Wilson v, Apr 11 2014 *)
PROG
(Magma) a0:=1; a1:=9; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Jun 14 1998
STATUS
approved