A131262 - OEIS

%I #3 Mar 31 2012 13:20:36

%S 1,3,14,60,248,1008

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

%C Also a(n) = least index k such that A092505(k) = A002430(k) / A046990(k) = 2^n.

%C Note that

%C a(0) = 1 = 1 - 0 = 2^0 - 0;

%C a(1) = 3 = 4 - 1 = 2^2 - 1;

%C a(2) = 14 = 16 - 2 = 2^4 - 2;

%C a(3) = 60 = 64 - 4 = 2^6 - 4;

%C a(4) = 248 = 256 - 8 = 2^8 - 8.

%C Conjecture: a(n) = Sigma(2^n)*EulerPhi(2^n) = 2^(2n) - Floor(2^n/2) = A062354(2^n).

%C If this conjecture is true the next term would be a(5) = 1008 = 1024 - 16 = 2^10 - 16.

%F Conjecture: a(n) = Sigma(2^n)*EulerPhi(2^n) = 2^(2n) - Floor(2^n/2) = A062354(2^n).

%e A130654(n) begins

%e {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 Thus a(0) = 1, a(1) = 3, a(2) = 14, a(3) = 60.

%Y 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 hard,more,nonn

%O 0,2

%A _Alexander Adamchuk_, Jun 24 2007

%E a(5) = 1008 from _Alexander Adamchuk_, May 02 2010