A092505 -id:A092505 - cp's OEIS frontend

A130654 Exponent m such that 2^m = A092505(n) = A002430(n) / A046990(n).

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, 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, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1

Offset: 1

Alexander Adamchuk, Jun 20 2007, Jun 23 2007

nonn

Conjecture: A092505(n) is always a power of 2. a(n) = Log[ 2, A092505(n) ]. Note that a(n) = 0 iff n is a power of 2; or A002430(2^n) = A046990(2^n) and A092505(2^n) = 1. It appears that a(2k+1) = 1 for k>0. Note that least index k such that a(k) = n is {1, 3, 14, 60, ...} which apparently coincides with A006502(n) = {1, 3, 14, 60, 279, 1251, ...} Related to Fibonacci numbers (see Carlitz reference).
Least index k such that a(k) = n is listed in A131262(n) = {1, 3, 14, 60, 248, ...}. Conjecture: A131262(n) = Sigma(2^n)*EulerPhi(2^n) = 2^(2n) - Floor(2^n/2) = A062354(2^n). If this conjecture is true then a(1008) = 5 and a(n)<5 for all n<1008.
Positions of records indeed continue as 1, 3, 14, 60, 248, 1008, 4064, 16320, ..., strongly suggesting union of {1} and A171499. - Antti Karttunen, Jan 13 2019

			A092505(n) begins {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, ...}.
Thus a(1) = Log[2,1] = 0, a(2) = Log[2,1] = 0, a(3) = Log[2,2] = 1.

Antti Karttunen, Table of n, a(n) for n = 1..16385
L. Carlitz, Some polynomials related to Fibonacci and Eulerian numbers, Fib. Quart., 16 (1978), 216-226.

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. A006502 = Related to Fibonacci numbers.
Cf. A131262 = Least index k such that A130654(k) = n. Cf. A062354 = Sigma(n)*EulerPhi(n).
Cf. also A171499.

a=Series[ Tan[x], {x,0,256} ]; b=Series[ Log[ 1/Cos[x] ], {x,0,256}]; Table[ Log[ 2, Numerator[ SeriesCoefficient[ a, 2n-1 ] ] / Numerator[ SeriesCoefficient[ b, 2n ] ] ], {n,1,128} ]

a(n) = Log[ 2, A092505(n) ]. a(n) = Log[ 2, A002430(n) / A046990(n) ] = A007814(A092505(n)).

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

Original entry on oeis.org

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

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A130654 Exponent m such that 2^m = A092505(n) = A002430(n) / A046990(n).

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