|
%I #9 Jul 26 2016 20:49:14
%S 0,2,4,20,64,248,880,3248,11776,43040,156736,571712,2083840,7597952,
%T 27698944,100985600,368164864,1342243328,4893475840,17840411648,
%U 65041678336,237125826560,864501723136,3151758405632,11490527150080,41891604733952,152726372368384
%N a(n) = ((1 + sqrt(7))^n - (1 - sqrt(7))^n)/sqrt(7).
%C Number of zeros in substitution system {0 -> 111, 1 -> 1001} at step n from initial string "1" (see example).
%H Ilya Gutkovskiy, <a href="/A274520/a274520.pdf">Illustration (substitution system {0 -> 111, 1 -> 1001})</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SubstitutionSystem.html">Substitution System</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,6)
%F O.g.f.: 2*x/(1 - 2*x - 6*x^2).
%F E.g.f.: 2*exp(x)*sinh(sqrt(7)*x)/sqrt(7).
%F Dirichlet g.f.: (PolyLog(s,1+sqrt(7)) - PolyLog(s,1-sqrt(7)))/sqrt(7), where PolyLog(s,x) is the polylogarithm function.
%F a(n) = 2*a(n-1) + 6*a(n-2).
%F a(n) = 2*A083099(n).
%F Lim_{n->infinity} a(n+1)/a(n) = 1 + sqrt(7) = 1 + A010465.
%e Evolution from initial string "1": 1 -> 1001 -> 10011111111001 -> 1001111111100110011001100110011001100110011111111001 -> ...
%e Therefore, number of zeros at step n:
%e a(0) = 0;
%e a(1) = 2;
%e a(2) = 4;
%e a(3) = 20, etc.
%t LinearRecurrence[{2, 6}, {0, 2}, 27]
%o (PARI) a(n)=([0,1; 6,2]^n*[0;2])[1,1] \\ _Charles R Greathouse IV_, Jul 26 2016
%Y Cf. A010465, A083099.
%K nonn,easy
%O 0,2
%A _Ilya Gutkovskiy_, Jun 26 2016
|
