cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A383160 a(n) is the numerator of the mean of the maximum exponents in the prime factorizations of the unitary divisors of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 5, 1, 3, 3, 2, 1, 5, 1, 5, 3, 3, 1, 7, 1, 3, 3, 5, 1, 7, 1, 5, 3, 3, 3, 3, 1, 3, 3, 7, 1, 7, 1, 5, 5, 3, 1, 9, 1, 5, 3, 5, 1, 7, 3, 7, 3, 3, 1, 11, 1, 3, 5, 3, 3, 7, 1, 5, 3, 7, 1, 2, 1, 3, 5, 5, 3, 7, 1, 9, 2, 3, 1, 11, 3, 3, 3
Offset: 1

Views

Author

Amiram Eldar, Apr 18 2025

Keywords

Comments

First differs from A296082 at n = 27.
a(n) depends only on the prime signature of n (A118914).

Examples

			Fractions begin with 0, 1/2, 1/2, 1, 1/2, 3/4, 1/2, 3/2, 1, 3/4, 1/2, 5/4, ...
4 has 2 unitary divisors: 1 and 4 = 2^2. The maximum exponents in their prime factorizations are 0 and 2, respectively. Therefore, a(4) = numerator((0 + 2)/2) = numerator(1) = 1.
12 has 4 unitary divisors: 1, 3 = 3^1, 4 = 2^2 and 12 = 2^2 * 3. The maximum exponents in their prime factorizations are 0, 1, 2 and 2, respectively. Therefore, a(12) = numerator((0 + 1 + 2 + 2)/4) = numerator(5/4) = 5.
		

Crossrefs

Programs

  • Mathematica
    emax[n_] := If[n == 1, 0, Max[FactorInteger[n][[;; , 2]]]]; a[n_] := Numerator[DivisorSum[n, emax[#] &, CoprimeQ[#, n/#] &] / 2^PrimeNu[n]]; Array[a, 100]
  • PARI
    emax(n) = if(n == 1, 0, vecmax(factor(n)[,2]));
    a(n) = my(f = factor(n)); numerator(sumdiv(n, d, emax(d) * (gcd(d, n/d) == 1)) / (1 << omega(f)));

Formula

a(n) = numerator(Sum_{d|n, gcd(d, n/d) = 1} A051903(d) / A034444(n)) = numerator(A383159(n) / A034444(n)).
a(n)/A383161(n) >= A383157(n)/A383158(n), with equality if and only if n is either squarefree (A005117) or a power of prime (A000961), i.e., n is not in A126706.
a(n)/A383161(n) < 1 if and only if n is squarefree.
a(n)/A383161(n) = 1 if and only if n is a square of a prime (A001248).
Sum_{k=1..n} a(k)/A383161(k) ~ c_1 * n - c_2 * n / sqrt(log(n)), where c = m(2) + Sum_{k>=3} (k-1) * (m(k) - m(k-1)) = 1.36914082067166047512... is the asymptotic mean of the fractions a(k)/A383161(k), m(k) = Product_{p prime} (1 - 1/(2*p^k)) is the asymptotic mean of the ratio between the number of k-free unitary divisors and the number of unitary divisors. e.g., m(2) = A383057 and m(3) = A383058, and c_2 = A345288/sqrt(Pi) = 0.6189064491320478904... .

A383058 Decimal expansion of the asymptotic mean of A365498(k)/A034444(k), the ratio between the number of cubefree unitary divisors and the number of unitary divisors over the positive integers.

Original entry on oeis.org

9, 1, 4, 2, 9, 4, 4, 1, 1, 8, 0, 1, 9, 8, 0, 6, 2, 4, 4, 8, 2, 9, 6, 1, 7, 6, 4, 5, 2, 1, 5, 6, 7, 1, 8, 4, 3, 7, 8, 5, 4, 6, 6, 9, 1, 7, 8, 1, 9, 3, 6, 8, 6, 6, 5, 9, 1, 9, 9, 7, 9, 7, 6, 7, 0, 0, 8, 5, 3, 4, 3, 8, 8, 3, 2, 0, 5, 6, 7, 6, 0, 8, 0, 0, 7, 1, 0, 7, 6, 7, 3, 6, 5, 0, 0, 4, 2, 6, 2, 6, 0, 5, 8, 2, 4
Offset: 0

Views

Author

Amiram Eldar, Apr 15 2025

Keywords

Comments

The asymptotic mean of the inverse ratio A034444(k)/A365498(k) is zeta(3)/zeta(6) (A157289).
In general, the asymptotic mean of the inverse ratio, between the number of unitary divisors and the number of k-free (i.e., not divisible by a k-th power other than 1) unitary divisors over the positive integers, for k >= 2, is zeta(k)/zeta(2*k).

Examples

			0.91429441180198062448296176452156718437854669178193...
		

Crossrefs

The unitary analog of A361062.

Programs

  • Mathematica
    $MaxExtraPrecision = 300; m = 300; f[p_] := 1 - 1/(2*p^3); c = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, m}], x]]; RealDigits[Exp[NSum[Indexed[c, n]*(PrimeZetaP[n]), {n, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 120][[1]]
  • PARI
    prodeulerrat(1 - 1/(2*p^3))

Formula

Equals lim_{m->oo} (1/m) * Sum_{k=1..m} A365498(k)/A034444(k).
Equals Product_{p prime} (1 - 1/(2*p^3)).
In general, the asymptotic mean of the ratio between the number of k-free unitary divisors and the number of unitary divisors over the positive integers, for k >= 2, is Product_{p prime} (1 - 1/(2*p^k)).
Showing 1-2 of 2 results.