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A131262 a(n) = least index k such that A130654(k) = n.

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%I A131262 #3 Mar 31 2012 13:20:36
%S A131262 1,3,14,60,248,1008
%N A131262 a(n) = least index k such that A130654(k) = n.
%C A131262 Also a(n) = least index k such that A092505(k) = A002430(k) / A046990(k) = 2^n.
%C A131262 Note that
%C A131262 a(0) = 1 = 1 - 0 = 2^0 - 0;
%C A131262 a(1) = 3 = 4 - 1 = 2^2 - 1;
%C A131262 a(2) = 14 = 16 - 2 = 2^4 - 2;
%C A131262 a(3) = 60 = 64 - 4 = 2^6 - 4;
%C A131262 a(4) = 248 = 256 - 8 = 2^8 - 8.
%C A131262 Conjecture: a(n) = Sigma(2^n)*EulerPhi(2^n) = 2^(2n) - Floor(2^n/2) = A062354(2^n).
%C A131262 If this conjecture is true the next term would be a(5) = 1008 = 1024 - 16 = 2^10 - 16.
%F A131262 Conjecture: a(n) = Sigma(2^n)*EulerPhi(2^n) = 2^(2n) - Floor(2^n/2) = A062354(2^n).
%e A131262 A130654(n) begins
%e A131262 {0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 2, 1, 0, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 0, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 0, ...}.
%e A131262 Thus a(0) = 1, a(1) = 3, a(2) = 14, a(3) = 60.
%Y A131262 Cf. A130654 = Exponent m such that 2^m = A092505(n) = A002430(n) / A046990(n). Cf. A092505 = A002430(n) / A046990(n), n>0. Cf. A002430 = Numerators in Taylor series for tan(x). Cf. A046990 = Numerators of Taylor series for log(1/cos(x)). Cf. A062354 = Sigma(n)*EulerPhi(n).
%K A131262 hard,more,nonn
%O A131262 0,2
%A A131262 _Alexander Adamchuk_, Jun 24 2007
%E A131262 a(5) = 1008 from _Alexander Adamchuk_, May 02 2010