OFFSET
1,2
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)
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.
EXAMPLE
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.
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Rémy Sigrist, Apr 27 2017
STATUS
approved