A072081 Numbers divisible by the square of the sum of their digits in base 10.

1, 10, 20, 50, 81, 100, 112, 162, 200, 243, 324, 392, 400, 405, 500, 512, 605, 648, 810, 972, 1000, 1053, 1100, 1120, 1134, 1183, 1215, 1296, 1400, 1620, 1701, 1900, 1944, 2000, 2025, 2106, 2156, 2240, 2268, 2300, 2401, 2430, 2511, 2592, 2704, 2800, 2916

Offset: 1

Author

Labos Elemer, Jun 14 2002

Keywords

• base
• nonn
• easy

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^(42 * k - 40) +1, k >= 1, are divisible by 7^2 = digsum(m)^2. Also, the numbers s = 491 * 10^(42 * k - 8) + 3, k >= 1, are divisible by 17^2 = digsum(s)^2. - Marius A. Burtea, Mar 19 2020

The numbers 2^A095412(n), n >= 5, are terms. - Marius A. Burtea, Apr 02 2020

Examples

			k=9477, sumdigits(9477)=27, q=9477=27*27*13.
		

Links

• Donovan Johnson, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A003132, A005349, A003634.

Programs

Magma

[k:k in [1..3000]| k mod &+Intseq(k)^2 eq 0]; // Marius A. Burtea, Mar 19 2020

Mathematica

sud[x_] := Apply[Plus, IntegerDigits[x]] Do[s=sud[n]^2; If[IntegerQ[n/s], Print[n]], {n, 1, 10000}]
Select[Range[3000],Divisible[#,Total[IntegerDigits[#]]^2]&] (* Harvey P. Dale, May 04 2011 *)

PARI

for(n=1,10^4,s=sumdigits(n);if(!(n%s^2),print1(n,", "))) \\ Derek Orr, Apr 29 2015

Python

def ok(n): return n and n%sum(di for di in map(int, str(n)))**2 == 0
print([k for k in range(3000) if ok(k)]) # Michael S. Branicky, Jan 10 2025