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

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.

Links

Table of n, a(n) for n=0..5.

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

Adjacent sequences: A131259 A131260 A131261 * A131263 A131264 A131265

Keyword

hard,more,nonn

Author

Alexander Adamchuk, Jun 24 2007

Extensions

a(5) = 1008 from Alexander Adamchuk, May 02 2010

Status

approved