OFFSET
0,2
COMMENTS
[Empirical] a(base,n)=a(base-1,n)+11^(n-1) for base>=5n-4; a(base,n)=a(base-1,n)+11^(n-1)-2 when base=5n-5.
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,6} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,5).
FORMULA
From Philippe Deléham, Mar 24 2012: (Start)
G.f.: (1+x)/(1-6*x-5*x^2).
a(n) = 6*a(n-1) + 5*a(n-2), a(0) = 1, a(1) = 7 .
a(n) = Sum_{k=0..=n} A054458(n,k)*4^k.
(End)
a(n) = A091928(n+1)/5. - Philippe Deléham, Mar 27 2012
a(n) = (((3-sqrt(14))^n * (-4+sqrt(14)) + (3+sqrt(14))^n * (4+sqrt(14)))) / (2*sqrt(14)). - Colin Barker, Sep 08 2016
MATHEMATICA
LinearRecurrence[{6, 5}, {1, 7}, 25] (* Paolo Xausa, Aug 08 2024 *)
PROG
(S/R) stvar $[N]:(0..M-1) init $[]:=0 asgn $[]->{*} kill +[i in 0..N-2](($[i]`-$[i+1]`>5)+($[i+1]`-$[i]`>5))
(PARI) Vec((1+x)/(1-6*x-5*x^2) + O(x^30)) \\ Colin Barker, Sep 08 2016
CROSSREFS
KEYWORD
nonn,base
AUTHOR
R. H. Hardin, Dec 28 2006
STATUS
approved