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A324392
a(n) is the number of divisors d of n such that A000120(d) divides n, where A000120(d) gives the binary weight of d.
2
1, 2, 1, 3, 1, 4, 1, 4, 1, 4, 1, 6, 1, 2, 1, 5, 1, 6, 1, 6, 3, 2, 1, 8, 1, 2, 1, 3, 1, 6, 1, 6, 2, 4, 1, 9, 1, 2, 2, 8, 1, 8, 1, 3, 1, 2, 1, 10, 1, 4, 1, 3, 1, 6, 2, 4, 2, 2, 1, 12, 1, 2, 3, 7, 1, 8, 1, 6, 2, 4, 1, 12, 1, 2, 2, 3, 1, 6, 1, 10, 2, 2, 1, 12, 1, 2, 1, 4, 1, 8, 1, 6, 1, 2, 1, 12, 1, 2, 2, 6, 1, 6, 1, 4, 4
OFFSET
1,2
COMMENTS
Number of such positive integers k that both k and A000120(k) [the Hamming weight of k] divide n.
FORMULA
a(n) = Sum_{d|n} [A000120(d) does divide n], where [ ] is the Iverson bracket.
a(n) = A000005(n) - A324393(n).
a(p) = 1 for all odd primes p.
MATHEMATICA
a[n_] := DivisorSum[n, 1 &, Divisible[n, DigitCount[#, 2, 1]] &]; Array[a, 100] (* Amiram Eldar, Dec 04 2020 *)
PROG
(PARI) A324392(n) = sumdiv(n, d, !(n%hammingweight(d)));
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Antti Karttunen, Mar 05 2019
STATUS
approved