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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A335008 Weird numbers (A006037) with more divisors than any smaller weird number.

Original entry on oeis.org

70, 836, 4030, 45356, 1713592, 15126992, 29465852, 1550860550
Offset: 1

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Author

Amiram Eldar, May 20 2020

Keywords

Comments

The corresponding numbers of divisors are 8, 12, 16, 24, 32, 40, 48, 96, ...

Examples

			The first 3 weird numbers, 70, 836 and 4030, have an increasing number of divisors, 8, 12 and 16. The least weird number with more than 16 divisors is the 94th weird number, 45356, which has 24 divisors.
		

Crossrefs

A335030 Numbers m that are not practical and have an abundancy index sigma(m)/m which is larger than that of any smaller number that is not practical.

Original entry on oeis.org

3, 9, 10, 44, 70, 102, 350, 372, 1608, 3492, 6096, 10380, 44040, 100260, 180240, 425160, 1744560, 2425080, 5509980, 10048080, 23614920, 97639920, 396315360, 900229680, 2519017200, 3113704440, 12870562320, 52307529120
Offset: 1

Views

Author

Amiram Eldar, May 20 2020

Keywords

Comments

None of the terms are superabundant (A004394) since all the superabundant numbers are practical numbers (A005153).
The least term m that is k-abundant (having sigma(m)/m > k) for k = 2, 3, ... is A005101(14) = 70, A068403(896) = 44040, A068404(792087) = 3113704440, ...
What is the least 5-abundant number (A215264) that is not practical?

Examples

			The first 5 numbers that are not practical are m = 3, 5, 7, 9, 10. Their abundancy indices sigma(m)/m are 1.333..., 1.2, 1.142..., 1.444..., 1.8. The record values occur at 3, 9 and 10.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^(e + 1) - 1)/(p - 1); pracQ[fct_] := (ind = Position[fct[[;; , 1]]/(1 + FoldList[Times, 1, f @@@ Most@fct]), _?(# > 1 &)]) == {}; seq = {}; rm = 1; Do[fct = FactorInteger[n]; r = Times@@((First/@fct^ (1+Last/@ fct)-1)/(First/@fct-1))/n; If[r > rm && !pracQ[fct], rm = r; AppendTo[seq, n]], {n, 3, 10^5}]; seq
Showing 1-2 of 2 results.