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

1,2

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

Links

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

Example

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

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 *)

Prog

(PARI) for(n=1, 10^4, s=sumdigits(n); if(!(n%s^2), print1(n, ", "))) \\ Derek Orr, Apr 29 2015
(Magma) [k:k in [1..3000]| k mod &+Intseq(k)^2 eq 0]; // Marius A. Burtea, Mar 19 2020
(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

Crossrefs

Cf. A003132, A005349, A003634.

Sequence in context: A327692 A269234 A160517 * A034087 A117562 A169663

Adjacent sequences: A072078 A072079 A072080 * A072082 A072083 A072084

Keyword

base,nonn,easy,changed

Author

Labos Elemer, Jun 14 2002

Status

approved