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A285815
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Numbers k such that, for any divisor d of k, the digital sum of d divides k.
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2
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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
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OFFSET
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1,2
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COMMENTS
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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).
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)
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LINKS
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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