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A131262
a(n) = least index k such that A130654(k) = n.
2
1, 3, 14, 60, 248, 1008
OFFSET
0,2
COMMENTS
Also a(n) = least index k such that A092505(k) = A002430(k) / A046990(k) = 2^n.
Note that
a(0) = 1 = 1 - 0 = 2^0 - 0;
a(1) = 3 = 4 - 1 = 2^2 - 1;
a(2) = 14 = 16 - 2 = 2^4 - 2;
a(3) = 60 = 64 - 4 = 2^6 - 4;
a(4) = 248 = 256 - 8 = 2^8 - 8.
Conjecture: a(n) = Sigma(2^n)*EulerPhi(2^n) = 2^(2n) - Floor(2^n/2) = A062354(2^n).
If this conjecture is true the next term would be a(5) = 1008 = 1024 - 16 = 2^10 - 16.
FORMULA
Conjecture: a(n) = Sigma(2^n)*EulerPhi(2^n) = 2^(2n) - Floor(2^n/2) = A062354(2^n).
EXAMPLE
A130654(n) begins
{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, ...}.
Thus a(0) = 1, a(1) = 3, a(2) = 14, a(3) = 60.
CROSSREFS
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).
Sequence in context: A151236 A006224 A219544 * A171499 A006502 A024037
KEYWORD
hard,more,nonn
AUTHOR
Alexander Adamchuk, Jun 24 2007
EXTENSIONS
a(5) = 1008 from Alexander Adamchuk, May 02 2010
STATUS
approved