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%I #18 Jul 19 2015 08:59:30
%S 0,2,6,10,14,18,22,30,34,42,46,54,58,62,70,78,82,90,94,102,106,114,
%T 118,122,130,138,142,146,154,158,162,166,174,182,190,194,210,214,222,
%U 230,238,242,250,254,270,274,278,286,294,298,302,310,314,330,334,342,346,354,358,366,374,390,394,402,410,418,426,434,442
%N Infinite sequence starting with a(0)=0 such that A(a(k)) = a(k-1) for all k>=1, where A(n) = n - A037445(n).
%C The first infinitary analog (see also A260124) of A259934 (see comment there). Using Falcao's method (2015) one can prove that such an infinite sequence exists.
%C All the first differences are powers of 2 (A260085). The infinitary case is interesting because here we have at least two analogs of sequences A259934, A259935 (respectively A260084, A260124 and A260085, A260123).
%C It is a corollary of the fact that all terms of A037445, except for n=1, are even (powers of 2). Therefore, in the analogs of A259934 we can begin with not only 0,2 (as in this sequence), but also with 0,1 (as in A260124). Then this sequence contains only the even terms, while A260124 - only the odd ones.
%C A generalization. For an even m, the multiplication of A260124 by 2^m and 2^(m+1) gives two infinite solutions of the system of equations for integer x_n, n>=1: A037445(x_1 + ... + x_n) = x_n/2^A005187(m), n>=1. In particular, for m=0, we obtain A260124 and A260084.
%F a(n) = 2 * A260124(n).
%Y Cf. A037445, A259934, A259935, A260085, A260124.
%K nonn
%O 0,2
%A _Vladimir Shevelev_, Jul 15 2015
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