A120249 Numerator of cfenc[n] (see definition in comments).

1, 2, 3, 3, 5, 5, 8, 4, 4, 8, 13, 7, 21, 13, 7, 5, 34, 7, 55, 11, 11, 21, 89, 9, 7, 34, 5, 18, 144, 12, 233, 6, 18, 55, 12, 10, 377, 89, 29, 14, 610, 19, 987, 29, 9, 144, 1597, 11, 11, 11, 47, 47, 2584, 9, 19, 23, 76, 233, 4181, 17, 6765, 377, 14, 7, 31, 31, 10946, 76, 123, 19

Offset: 1

Author

Joseph Biberstine (jrbibers(AT)indiana.edu), Jun 12 2006, Jun 25 2006

Keywords

• frac
• nonn

Comments

a[n] := numerator of cfenc[n]. cfenc[n] := number given by interpreting as a continued fraction expansion (indexed from 1) the sequence whose i-th entry is one plus the exponent on the i-th prime factor of n (fix cfenc[1]=1). a[2^k] = cfenc[2^k] = k+1.

Examples

			a(2646) = numerator[cfenc[2646]]= numerator[cfenc[2^1 * 3^3 * 7^2]] = numerator[FromContinuedFraction[{2; 4, 1, 3}]] = numerator[2 + 1/(4 + 1/(1 + 1/3))] = numerator[42/19] = 42.
From _Antti Karttunen_, Oct 29 2019: (Start)
a(6) = 3 because 6 = 2^1 * 3^1, and the numerator of the continued fraction 1+1 + 1/(1+1) = 5/2 is 5.
a(12) = 7 because 12 = 2^2 * 3^1, and the numerator of the continued fraction 2+1 + 1/(1+1) = 7/2 is 7.
a(15) = 7 because 15 = 2^0 * 3^1 * 5^1, and the numerator of the continued fraction 0+1 + 1/(1+1 + 1/(1+1)) = 1 + 1/(2 + 1/2) = 1 + 2/5 = 7/5 is 7.
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Links

• Hans Havermann, Table of n, a(n) for n = 1..10000
• Wikipedia, Continued fraction

Crossrefs

Corresponding denominators in A120250. Numerators modulo respective denominators in A120251.

Cf. A000040, A000045, A001511, A003961, A061395, A286561, A323080.

Cf. also A007305, A075158, A273671, A273672.

Programs

Mathematica

Table[If[n == 1, 1, (fl = FactorInteger[n]; pq = Table[1, {i, 1, PrimePi[Last[fl][[1]]]}]; While[Length[fl] > 0, pp = First[fl]; fl = Drop[fl, 1]; pq[[PrimePi[pp[[1]]]]] = pp[[2]] + 1;]; Numerator[FromContinuedFraction[pq]])],{n,1,80}]

PARI

A120249(n) = if(1==n,n, my(pi=primepi(vecmax(factor(n)[, 1])), cf=1+valuation(n,prime(pi))); pi--; while(pi, cf = (1+valuation(n,prime(pi)))+(1/cf); pi--); numerator(cf)); \\ Antti Karttunen, Oct 26 2019

Formula

a(2^k) = k+1.

a(A000040(n)) = A000045(n+2).

From Antti Karttunen, Oct 29 2019: (Start)

The following formula employs Gauss's notation for continued fractions (see the section "Notations" in the Wikipedia-article), for example, K_{i=1..3} u(i) stands for 1/(u(1) + 1/(u(2) + 1/u(3))):

Let c(n) = A001511(n) + K_{i=2..A061395(n)} 1/(1+A286561(n,A000040(i))). Then a(n) is the numerator of c(n), and A120250(n) is the denominator of c(n).

For all n >= 2, a(2n) = a(A003961(n)), thus a(n) = f(A323080(n)) for some function f.

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