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.

Showing 1-6 of 6 results.

A361061 Decimal expansion of the asymptotic mean of A000005(k)/A073184(k), the ratio between the number of divisors and the number of cubefree divisors.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Mar 01 2023

Keywords

Examples

			1.109049677998737336345288587781671766009752629677303...
		

Crossrefs

Cf. A000005, A073184, A361062 (mean of the inverse ratio).
Cf. A307869 (squarefree analog), A308043.

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; m = 1000; f[p_] := 1 + 1/(3*(p - 1)*p^2); 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, 100][[1]]
  • PARI
    prodeulerrat(1 + 1/(3*(p-1)*p^2))

Formula

Equals lim_{m->oo} (1/m) * Sum_{k=1..m} A000005(k)/A073184(k).
Equals Product_{p prime} (1 + 1/(3*(p-1)*p^2)).

A383157 a(n) is the numerator of the mean of the maximum exponents in the prime factorizations of the divisors of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 7, 1, 3, 3, 2, 1, 7, 1, 7, 3, 3, 1, 13, 1, 3, 3, 7, 1, 7, 1, 5, 3, 3, 3, 13, 1, 3, 3, 13, 1, 7, 1, 7, 7, 3, 1, 21, 1, 7, 3, 7, 1, 13, 3, 13, 3, 3, 1, 5, 1, 3, 7, 3, 3, 7, 1, 7, 3, 7, 1, 11, 1, 3, 7, 7, 3, 7, 1, 21, 2, 3, 1, 5, 3
Offset: 1

Views

Author

Amiram Eldar, Apr 18 2025

Keywords

Comments

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, 7/6, ...
4 has 3 divisors: 1, 2 = 2^1 and 4 = 2^2. The maximum exponents in their prime factorizations are 0, 1 and 2, respectively. Therefore, a(4) = numerator((0 + 1 + 2)/3) = numerator(1) = 1.
12 has 6 divisors: 1, 2 = 2^1, 3 = 3^1, 4 = 2^2, 6 = 2 * 3 and 12 = 2^2 * 3. The maximum exponents in their prime factorizations are 0, 1, 1, 2, 1 and 2, respectively. Therefore, a(12) = numerator((0 + 1 + 1 + 2 + 1 + 2)/6) = numerator(7/6) = 7.
		

Crossrefs

Programs

  • Mathematica
    emax[n_] := If[n == 1, 0, Max[FactorInteger[n][[;; , 2]]]]; a[n_] := Numerator[DivisorSum[n, emax[#] &] / DivisorSigma[0, 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)) / numdiv(f));

Formula

a(n) = numerator(Sum_{d|n} A051903(d) / A000005(n)) = numerator(A383156(n) / A000005(n)).
a(n)/A383158(n) = 1 if and only if n is a square of a prime (A001248).
Sum_{k=1..n} a(k)/A383158(k) ~ c_1 * n - c_2 * n /sqrt(log(n)), where c_1 = m(2) + Sum_{k>=3} (k-1) * (m(k) - m(k-1)) = 1.27968644485944694957... is the asymptotic mean of the fractions a(k)/A383158(k), m(k) = Product_{p prime} (1 + (1-1/p) * Sum_{i>=k} (k/(i+1) - 1)/p^i is the asymptotic mean of the ratio between the number of k-free divisors and the number of divisors, e.g., m(2) = A308043 and m(3) = A361062, and c_2 = A345231 = 0.54685595528047446684... .

A366586 Decimal expansion of the asymptotic mean of the ratio between the number of cubefree divisors and the number of squarefree divisors.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Oct 14 2023

Keywords

Comments

For a positive integer k the ratio between the number of cubefree divisors and the number of squarefree divisors is r(k) = A073184(k)/A034444(k).
r(k) >= 1 with equality if and only if k is squarefree (A005117).
The indices of records of this ratio are the squares of primorial numbers (A061742), and the corresponding record values are r(A061742(k)) = (3/2)^k. Therefore, this ratio is unbounded.
The asymptotic second raw moment is = Product_{p prime} (1 + 5/(4*p^2)) = 1.67242666864454336962... and the asymptotic standard deviation is 0.35851843008068965078... .

Examples

			1.24253418622467728695963000629433770800015253305890...
		

Crossrefs

Similar constants: A307869, A308042, A308043, A358659, A361059, A361060, A361061, A361062, A366587 (mean of the inverse ratio).

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; m = 1000; c = LinearRecurrence[{0, -(1/2)}, {0, 1}, m]; RealDigits[Exp[NSum[Indexed[c, n] * PrimeZetaP[n]/n, {n, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 105][[1]]
  • PARI
    prodeulerrat(1 + 1/(2*p^2))

Formula

Equals lim_{m->oo} (1/m) * Sum_{k=1..m} A073184(k)/A034444(k).
Equals Product_{p prime} (1 + 1/(2*p^2)).
In general, the asymptotic mean of the ratio between the number of (k+1)-free divisors and the number of k-free divisors, for k >= 2, is Product_{p prime} (1 + 1/(k*p^2)).

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)).

A366587 Decimal expansion of the asymptotic mean of the ratio between the number of squarefree divisors and the number of cubefree divisors.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Oct 14 2023

Keywords

Comments

For a positive integer k the ratio between the number of squarefree divisors and the number of cubefree divisors is r(k) = A034444(k)/A073184(k).
r(k) <= 1 with equality if and only if k is squarefree (A005117).
The asymptotic second raw moment is = Product_{p prime} (1 - 5/(9*p^2)) = 0.76780883634140395932... and the asymptotic standard deviation is 0.29730736888962774256... .

Examples

			0.85620050793747714939728108959516040498849031584132...
		

Crossrefs

Similar constants: A307869, A308042, A308043, A358659, A361059, A361060, A361061, A361062, A366586 (mean of the inverse ratio).

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; m = 1000; c = LinearRecurrence[{0, 1/3}, {0, -(2/3)}, m]; RealDigits[Exp[NSum[Indexed[c, n] * PrimeZetaP[n]/n, {n, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 105][[1]]
  • PARI
    prodeulerrat(1 - 1/(3*p^2))

Formula

Equals lim_{m->oo} (1/m) * Sum_{k=1..m} A034444(k)/A073184(k).
Equals Product_{p prime} (1 - 1/(3*p^2)).
In general, the asymptotic mean of the ratio between the number of k-free divisors and the number of (k-1)-free divisors, for k >= 3, is Product_{p prime} (1 - 1/(k*p^2)).

A380601 Decimal expansion of the asymptotic mean of the ratio A322483(k)/A000005(k).

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Jan 27 2025

Keywords

Examples

			0.85980677933034363311244767594941832466515809551385...
		

Crossrefs

Cf. A000005, A322483, A380602 (mean of the inverse ratio).
Similar constants: A308043, A361060, A361062.

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; m = 1000; f[x_] := (2*x + 3*(x-1)*Log[1 - x] + (x-1)*Log[1+x])/(4*x); c = Rest[CoefficientList[Series[Log[f[x]], {x, 0, m}], x]]; RealDigits[Exp[NSum[Indexed[c, k]*PrimeZetaP[k], {k, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 120][[1]]
  • PARI
    default(realprecision, 120); default(parisize, 30000000);
    my(m = 1024, x = 'x + O('x^m), v); v = Vec((2*x + 3*(x-1)*log(1-x) + (x-1)*log(1+x))/(4*x)); prodeulerrat(sum(i=1, #v, v[i]/p^(i-1)))

Formula

Equals lim_{m->oo} (1/m) * Sum_{k=1..m} A322483(k)/A000005(k).
Equals Product_{p prime} (1/2 - ((p-1)/4) * (3*log(1-1/p) + log(1+1/p))).
Showing 1-6 of 6 results.