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.

A072082 Numbers divisible by the cube of the sum of their digits in base 10.

Original entry on oeis.org

1, 10, 100, 200, 500, 512, 1000, 2000, 2401, 4913, 5000, 5103, 5120, 5832, 10000, 10206, 11000, 11200, 11664, 13122, 14000, 17576, 19000, 19683, 20000, 20412, 21141, 23000, 23328, 24010, 28000, 29160, 32000, 37000, 39366, 40000, 40824
Offset: 1

Views

Author

Labos Elemer, Jun 14 2002

Keywords

Comments

If k is a term, then 10 * k is a term. There are an infinite number of terms that are not divisible by 10. The numbers m = 24 * 10^(294 * k - 292) +1 are divisible by 7^3 = digsum(m)^3. Also, the numbers s = 491 * 10^(4624 * k - 4623) + 3, k >= 1, are divisible by 17^3 = digsum(s)^3. - Marius A. Burtea, Mar 18 2020

Examples

			k=98415: sumdigits(98415)=27, q=98415=5*27*27*27.
		

Crossrefs

Programs

  • Magma
    [k:k in [1..41000]| k mod &+Intseq(k)^3 eq 0]; // Marius A. Burtea, Mar 18 2020
    
  • Mathematica
    sud[x_] := Apply[Plus, IntegerDigits[x]] Do[s=sud[n]^3; If[IntegerQ[n/s], Print[n]], {n, 1, 10000}]
    Select[Range[50000],Divisible[#,Total[IntegerDigits[#]]^3]&] (* Harvey P. Dale, Mar 22 2016 *)
  • PARI
    is(n)=n%sumdigits(n)^3==0 \\ Charles R Greathouse IV, Mar 19 2020