A130654 -id:A130654 - cp's OEIS frontend

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

1, 3, 14, 60, 248, 1008

Offset: 0

Alexander Adamchuk, Jun 24 2007

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.

			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.

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).

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

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

A092505 a(n) = A002430(n) / A046990(n).

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 4, 2, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 2, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 4, 2, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4

Offset: 1

Ralf Stephan, Apr 05 2004

nonn

Antti Karttunen, Table of n, a(n) for n = 1..16385

Cf. A002430, A046990, A130654.

[Numerator((-1)^(n - 1)*2^(2*n)*(2^(2*n) - 1)*Bernoulli(2*n) / Factorial(2*n)) / (Numerator(((-4)^n-(-16)^n) * Bernoulli(2*n) / 2 / n / Factorial(2*n))): n in [1..100]]; // Vincenzo Librandi, Jan 13 2019

a(n)=if(n<1,0,numerator(polcoeff(Ser(tan(x)),2*n-1))/numerator(polcoeff(Ser(log(1/cos(x))),2*n)))

\\ Quite wasteful, especially as there is the same bernfrac(2*n) in both. Should reduce to a much simpler form?
A002430(n) = numerator(((-1)^(n-1)) * 2^(2*n) * (2^(2*n)-1)*bernfrac(2*n)/((2*n)!)); \\ After Johannes W. Meijer's May 24 2009 formula in A002430.
A046990(n) = numerator(((-4)^n-(-16)^n)*bernfrac(2*n)/2/n/(2*n)!); \\ From A046990
A092505(n) = (A002430(n) / A046990(n)); \\ Antti Karttunen, Jan 12 2019

A007814(a(n)) = A130654(n). - Antti Karttunen, Jan 12 2019

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

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A092505 a(n) = A002430(n) / A046990(n).

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