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.

A344733 Numbers k such that k divides A327573(k).

Original entry on oeis.org

1, 25, 387, 6063, 1416379, 1416403, 1416411, 331362359, 5068450527
Offset: 1

Views

Author

Amiram Eldar, May 27 2021

Keywords

Comments

The corresponding quotients A327573(k)/k are 1, 3, 5, 7, 11, 11, 11, 15, 17, ...
a(10) > 7.5*10^10, if it exists.

Examples

			a(1) = 1 since A327573(1) = 1 is divisible by 1.
a(2) = 25 since A327573(25) = 75 = 3 * 25 is divisible by 25.

Crossrefs

Cf. A037445, A327573.
The infinitary version of A050226.
Similar sequences: A064610, A344731, A344732.

Programs

  • Mathematica
    f[p_, e_] := 2^DigitCount[e, 2, 1]; s[1] = 1; s[n_] := s[n] = s[n - 1] + Times @@ f @@@ FactorInteger[n]; Select[Range[1.5*10^6], Divisible[s[#], #] &]

A327576 Decimal expansion of the constant that appears in the asymptotic formula for average order of the number of infinitary divisors function (A037445).

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Sep 17 2019

Keywords

Examples

			0.366625276945381909556532720687001563033612155971009...

References

  • Steven R. Finch, Mathematical Constants II, Cambridge University Press, 2018, section 1.7.5, pp. 53-54.

Links

Crossrefs

Cf. A037445, A050376, A327573.
Cf. A059956 (corresponding constant for unitary divisors), A306071 (bi-unitary).

Programs

  • Mathematica
    m = 1000; em = 10; f[x_] := Sum[Log[1 - 1/(1 + 1/x^(2^e))^2], {e, 0, em}]; c = Rest[CoefficientList[Series[f[x], {x, 0, m}], x]*Range[0, m]]; $MaxExtraPrecision = 1500; RealDigits[(1/2)*Exp[f[1/2] + f[1/3]]* Exp[NSum[Indexed[c, k]*(PrimeZetaP[k] - (1/2)^k - (1/3)^k)/k, {k, 2, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 100][[1]]

Formula

Equals Limit_{n->oo} A327573(n)/(2 * n * log(n)). [Corrected by Amiram Eldar, May 07 2021]
Equals (1/2) * Product_{P} (1 - 1/(P+1)^2), where P are numbers of the form p^(2^k) where p is prime and k >= 0 (A050376).

A344274 a(n) is the least k such that the average number of infinitary divisors of {1..k} is >= n.

Original entry on oeis.org

1, 6, 24, 105, 385, 1554, 6063, 23688, 92610, 362112, 1416360, 5539296, 21663378, 84725487, 331362185, 1295952084, 5068450464, 19822658688
Offset: 1

Views

Author

Amiram Eldar, May 13 2021

Keywords

Examples

			a(2) = 6 since the average number of infinitary divisors of {1..6} is A327573(6)/6 = 13/6 > 2.

Crossrefs

Cf. A037445, A327573, A327576.
The infinitary version of A085829.
Similar sequences: A328331, A336304, A338891, A338943, A344272, A344273.

Programs

  • Mathematica
    f[p_, e_] := 2^DigitCount[e, 2, 1]; idivnum[1] = 1; idivnum[n_] := Times @@ (f @@@ FactorInteger[n]); seq={}; s = 0; k = 1; Do[While[s = s + idivnum[k]; s < k*n, k++]; AppendTo[seq, k]; k++, {n, 1, 10}]; seq

Formula

Lim_{n->oo} a(n+1)/a(n) = exp(1/(2*c)) = 3.9109891037..., where c is A327576.
Showing 1-3 of 3 results.

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