1,14

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 ref. Carlitz).

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.

L. Carlitz, Some polynomials related to Fibonacci and Eulerian numbers, Fib. Quart., 16 (1978), 216-226.

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

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.

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} ]

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).

Sequence in context: A073484 A081396 A100544 this_sequence A053259 A143842 A092876

Adjacent sequences: A130651 A130652 A130653 this_sequence A130655 A130656 A130657

nonn

Alexander Adamchuk (alex(AT)kolmogorov.com), Jun 20 2007, Jun 23 2007