A285815 Numbers k such that, for any divisor d of k, the digital sum of d divides k.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 54, 60, 63, 72, 81, 90, 108, 120, 162, 180, 216, 243, 270, 324, 360, 486, 540, 648, 810, 972, 1080, 1458, 1620, 1944, 2430, 2916, 3240, 4374, 4860, 5832, 7290, 8748, 9720, 13122, 14580, 17496

Offset: 1

Author

Rémy Sigrist, Apr 27 2017

Keywords

• nonn
• base

Comments

All terms are Niven numbers (A005349).

All terms > 1 have a prime divisor < 10.

Is this sequence infinite?

Some families of terms:

- 2*3^k with 0 <= k <= 12,

- 2*3^k*5 with 0 <= k <= 10,

- 2^2*3^k with 0 <= k <= 13,

- 2^2*3^k*5 with 0 <= k <= 22,

- 2^3*3^k with 0 <= k <= 13,

- 2^3*3^k*5 with 0 <= k <= 22,

- 3^k with 0 <= k <= 5.

The first 99 terms are 7-smooth (A002473).

From David A. Corneth, Apr 20 2021: (Start)

Let k be a term. If 11|k then (1+1)=2|k so 22|k. Similarily if 22|k then 44|k. If 44|k then 88|k. If 88|k then 176|k. If 176|k then (1+7+6) = 14|k so lcm(176, 14) = 1232. Repeating this a few times we see k > 10^43.

Can we use this to prove if p|k then p <= 7 where p is a prime and k is a term?

(End)

Examples

			The divisors of 243 are: 1, 3, 9, 27, 81, 243; their digital sums are: 1, 3, 9, 9, 9, 9, all divisors of 243; hence 243 is in the sequence.
14 divides 42, but its digital sum, 5, does not divide 42; hence 42 is not in the sequence.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..99
• David A. Corneth, Conjectured full sequence with 108 terms. All 9973-smooth terms <= 10^30.

Crossrefs

Cf. A002473, A005349.

Programs

PARI

is(n) = fordiv(n, d, if (n % sumdigits(d), return (0))); return (1)

Python

from sympy import divisors
from sympy.ntheory.factor_ import digits
def ok(n):
   return all(n%sum(digits(d)[1:])==0 for d in divisors(n))
print([n for n in range(1, 20001) if ok(n)]) # Indranil Ghosh, Apr 28 2017