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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A048698 Nonprime numbers k such that sum of aliquot divisors of k is a cube.

Original entry on oeis.org

1, 10, 49, 56, 69, 76, 122, 133, 568, 578, 1001, 1018, 1243, 1324, 1431, 1611, 1685, 1819, 1994, 2296, 2323, 3344, 3403, 3627, 3641, 3763, 3981, 4336, 5482, 8186, 9077, 9641, 10113, 10688, 13471, 14188, 14509, 14727, 15940, 16697, 17141, 17619, 19241, 19637
Offset: 1

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Author

Keywords

Comments

The sum of the aliquot divisors of a prime is exactly 1. - Martin Renner, Oct 13 2011

Examples

			a(4) = 56: the aliquot divisors 1,2,4,7,8,14,28 sum to 64, a cube.
		

Crossrefs

Programs

  • Maple
    a := []; for n from 1 to 1000 do if sigma(n) <> n+1 and type( simplify((sigma(n)-n)^(1/3)), `integer`) then a := [op(a), n]; fi; od: a;
  • Mathematica
    Select[Range[20000], !PrimeQ[#] && IntegerQ @ Surd[DivisorSigma[1, #] - #, 3] &] (* Amiram Eldar, Feb 23 2020 *)
  • PARI
    c=0; for(n=1, 13127239, if(isprime(n)==0, if(ispower(sigma(n)-n, 3), c++; write("b048698.txt", c " " n)))) /* Donovan Johnson, Mar 10 2013 */

A244429 Nonprimes n such that the sum of proper divisors of n and the product of proper divisors of n are both perfect cubes.

Original entry on oeis.org

1, 56, 568, 1001, 1431, 3344, 10688, 17619, 24099, 37432, 40797, 46096, 50571, 52687, 55581, 62375, 75221, 88863, 97273, 98752, 111224, 134672, 235495, 247033, 251403, 266176, 269072, 271099, 275077, 299576, 320489, 333888, 364067, 372331, 407319, 534413, 561008, 614465, 646691
Offset: 1

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Author

Derek Orr, Jun 27 2014

Keywords

Comments

Primes trivially satisfy this property and are therefore not included in this sequence.
These numbers are in the intersection of A194948 and A229972.

Examples

			The proper divisors of 56 are {1, 2, 4, 7, 8, 14, 28}. 1*2*4*7*8*14*28 = 175616 = 56^3. 1+2+4+7+8+14+28 = 64 = 4^3. Since both the product of proper divisors and the sum of proper divisors are cubes, 56 is a member of this sequence.
		

Crossrefs

Programs

  • PARI
    for(n=1,10^7,d=divisors(n);s=sum(i=1,#d-1,d[i]);p=prod(j=1,#d-1,d[j]);if((p!=1||s!=1),if(ispower(s,3)&&ispower(p,3),print1(n,", "))))
Showing 1-2 of 2 results.