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A092900 A Jacobsthal sequence (A078008) to base 4.

%I A092900 #15 Apr 01 2016 06:29:17
%S A092900 1,0,2,2,12,22,112,222,1112,2222,11112,22222,111112,222222,1111112,
%T A092900 2222222,11111112,22222222,111111112,222222222,1111111112,2222222222,
%U A092900 11111111112,22222222222,111111111112,222222222222,1111111111112
%N A092900 A Jacobsthal sequence (A078008) to base 4.
%H A092900 Colin Barker, <a href="/A092900/b092900.txt">Table of n, a(n) for n = 0..1000</a>
%H A092900 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,11,0,-10).
%F A092900 For n > 0, a(2*n+1) is represented as a string of n 2's and a(2*n) as a string of (n-1) 1's followed by a 2.
%F A092900 From _Colin Barker_, Apr 01 2016: (Start)
%F A092900 a(n) = (6+10*(-1)^n+10^(1/2*(-1+n))*(2-2*(-1)^n+sqrt(10)+(-1)^n*sqrt(10)))/18.
%F A092900 a(n) = (10^(n/2)+8)/9 for n even.
%F A092900 a(n) = (2^((n+1)/2)*5^((n-1)/2)-2)/9 for n odd.
%F A092900 a(n) = 11*a(n-2)-10*a(n-4) for n>3.
%F A092900 G.f.: (1-9*x^2+2*x^3) / ((1-x)*(1+x)*(1-10*x^2)).
%F A092900 (End)
%e A092900 a(8)= 1112 because A078008(8) = 86 (in base 10) = 64 + 16 + 4 + 2 = 1*(4^3) + 1*(4^2) + 1*(4^1) + 2.
%o A092900 (PARI) Vec((1-9*x^2+2*x^3)/((1-x)*(1+x)*(1-10*x^2)) + O(x^30)) \\ _Colin Barker_, Apr 01 2016
%Y A092900 Cf. A081857.
%K A092900 easy,base,nonn
%O A092900 0,3
%A A092900 _Paul Barry_, Mar 12 2004