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%I #16 Nov 25 2015 02:10:39
%S 0,1,1,5,13,173,3501,1420717,7343549869,24407739551034797,
%T 264579267653248177273154989,
%U 15107659029337673520218077770654501397966253,5900314832748922900613950065282124787723453785544193308390237364661677
%N Define binary strings S(0)=0, S(1)=1, S(n) = S(n-2)S(n-1); a(n) = S(n) converted to decimal.
%C Note that S(n) in general has leading zeros.
%F S(0) = 0, S(1) = 1, so S(2) = 01, a(2) = 1.
%F Use the substitution system 0->1 and 1->01. The values generated from a(0)=0 are 1, 01, 101, 01101, which in base 10 give the sequence. - _Jon Perry_, Feb 06 2011
%e S(3) = 01 (base 2) = 1 (base 10) so a(3) = 1.
%e S(4) = 101 (base 2) = 5 (base 10) so a(4) = 5.
%e S(5) = 01.101 = 01101 (base 2) = 13 (base 10) so a(5) = 13.
%e S(6) = 101.01101 = 10101101 (base 2) = 173 (base 10) so a(6) = 173.
%e S(7) = 01101.10101101 = 0110110101101 (base 2) = 3501 (base 10).
%t a[1] = 0; a[2] = 1; a[n_] := a[n] = If[ OddQ@n, FromDigits[ Join[ IntegerDigits[ a[n - 2], 2], IntegerDigits[ a[n - 1], 2]], 2], FromDigits[ Join[ IntegerDigits[ a[n - 2], 2], {0}, IntegerDigits[ a[n - 1], 2]], 2]]; Array[a, 13] (* _Robert G. Wilson v_, Apr 20 2006 *)
%Y Cf. A063896.
%K base,nonn
%O 0,4
%A Jordan Goldstein (jboymicro20X6(AT)aim.com), Apr 18 2006
%E More terms from _Robert G. Wilson v_, Apr 20 2006
%E Edited by _N. J. A. Sloane_, Apr 23 2006
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