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 1-3 of 3 results.

A345288 Decimal expansion of Product_{p primes} sqrt(1 + 1/(4*p*(p-1))).

Original entry on oeis.org

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

Views

Author

Vaclav Kotesovec, Jun 13 2021

Keywords

Examples

			1.09698311911435715092514804937509666493621167605436728776543452986946366666...

Links

  • Steven R. Finch, Mathematical Constants II, Encyclopedia of Mathematics and Its Applications, Cambridge University Press, Cambridge, 2018, p. 50.

Crossrefs

Cf. A034444, A083281, A345231.

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; Clear[f]; f[p_] := Sqrt[1 + 1/(4*p*(p-1))]; Do[cc = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, m}], x, m + 1]]; Print[f[2] * Exp[N[Sum[Indexed[cc, n] * (PrimeZetaP[n] - 1/2^n), {n, 2, m}], 110]]], {m, 100, 500, 100}]
  • PARI
    sqrt(prodeulerrat(1 + 1/(4*p*(p-1)))) \\ Amiram Eldar, Jun 13 2021

Formula

Equals sqrt(Pi) * lim_{n->infinity} sqrt(log(n))/n * Sum_{k=1..n} 1/A034444(k).

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.

Links

Crossrefs

Cf. A000005, A001248, A051903, A118914, A308043, A345231, A361062, A383156, A383158 (denominators).

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... .

A083281 Decimal expansion of h = Product_{p prime}(sqrt(p(p-1))*log(1/(1-1/p))).

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Jun 02 2003

Keywords

Comments

Arises in formulas like: Sum_{k<=x} 1/tau(kd) = hx/sqrt(Pi*log(x))*{ g(d)+O((3/4)^omega(d)/log(x)) } where g satisfies Sum_{d<=x} g(d))=x/h/sqrt(Pi*log(x))*{ 1+O(1/log(x)) }.
The logarithm of the value has an expansion -P(2)/24 -P(3)/24 -109*P(4)/2880 -49*P(5)/1440-... in terms of the prime zeta functions P(.). - R. J. Mathar, Jan 31 2009
The average order of 1/tau(k) (where tau(k) is the number of divisors of k, A000005), Sum_{k<=x} 1/tau(k) ~ h*x/sqrt(Pi*log(x)), was found by Ramanujan in 1916 and was proven by Wilson in 1923. - Amiram Eldar, Jun 19 2019

Examples

			0.96927694382749163071695371472090732266213688638491621816178588751950570028...

References

  • G. Tenenbaum, Introduction à la théorie analytique et probabiliste des nombres, collection SMF no. 1, 1995, p. 210.

Links

Crossrefs

Cf. A000005, A345231, A345288.

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; m = 1000; f[p_] := Sqrt[p*(p - 1)]*Log[p/(p - 1)]; c = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, m}], x]]; RealDigits[Exp[NSum[Indexed[c, k]*PrimeZetaP[k], {k, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 100][[1]] (* Amiram Eldar, Jun 19 2019 *)
  • PARI
    prod(k=1,40000,sqrt(prime(k)*(prime(k)-1))*log(1/(1-1/prime(k))))

Formula

Equals A345231 * sqrt(Pi). - Vaclav Kotesovec, Jun 13 2021

Extensions

10 more digits from R. J. Mathar, Jan 31 2009
More terms from Amiram Eldar, Jun 19 2019
More digits from Vaclav Kotesovec, Jun 13 2021
Showing 1-3 of 3 results.

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