OFFSET
0,2
COMMENTS
Number of zeros in substitution system {0 -> 11111, 1 -> 1001} at step n from initial string "1" (see example).
LINKS
Ilya Gutkovskiy, Illustration (substitution system {0 -> 11111, 1 -> 1001})
Eric Weisstein's World of Mathematics, Substitution System
Index entries for linear recurrences with constant coefficients, signature (2,10)
FORMULA
O.g.f.: 2*x/(1 - 2*x - 10*x^2).
E.g.f.: 2*exp(x)*sinh(sqrt(11)*x)/sqrt(11).
Dirichlet g.f.: (PolyLog(s,1+sqrt(11)) - PolyLog(s,1-sqrt(11)))/sqrt(11), where PolyLog(s,x) is the polylogarithm function.
a(n) = 2*a(n-1) + 10*a(n-2).
a(n) = 2*A083102(n-1), n>0.
Lim_{n->infinity} a(n+1)/a(n) = 1 + sqrt(11) = 1 + A010468.
EXAMPLE
Evolution from initial string "1": 1 -> 1001 -> 100111111111111001 -> 1001111111111110011001100110011001100110011001100110011001100111111111111001 -> ...
Therefore, number of zeros at step n:
a(0) = 0;
a(1) = 2;
a(2) = 4;
a(3) = 28, etc.
MATHEMATICA
LinearRecurrence[{2, 10}, {0, 2}, 25]
PROG
(PARI) concat(0, Vec(2*x/(1-2*x-10*x^2) + O(x^99))) \\ Altug Alkan, Jun 27 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ilya Gutkovskiy, Jun 27 2016
STATUS
approved