Displaying 11-15 of 15 results found.
a(n) is the number of proper divisors of n that are odious ( A000069).
+10
4
0, 1, 1, 2, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 4, 1, 2, 1, 3, 2, 3, 1, 4, 1, 3, 1, 5, 1, 2, 1, 5, 2, 2, 2, 3, 1, 3, 2, 4, 1, 5, 1, 5, 1, 2, 1, 5, 2, 3, 1, 5, 1, 2, 2, 7, 2, 2, 1, 3, 1, 3, 3, 6, 2, 4, 1, 3, 1, 5, 1, 4, 1, 3, 2, 5, 3, 4, 1, 5, 1, 3, 1, 8, 1, 2, 1, 7, 1, 2, 3, 3, 2, 3, 2, 6, 1, 5, 2, 5, 1, 2, 1, 7, 4, 2, 1, 3, 1, 5, 2, 9, 1, 4, 1, 3, 2, 3, 2, 4
EXAMPLE
For n = 55 whose proper divisors are 1, 5 and 11 (in binary "1", "101" and "1011"), only 1 and 11 have an odd number of 1's in their binary representations, thus a(55) = 2.
MAPLE
f:= proc(n) nops(select(t -> convert(convert(t, base, 2), `+`)::odd, numtheory:-divisors(n) minus {n})) end proc:
MATHEMATICA
Table[DivisorSum[n, 1 &, And[OddQ@ DigitCount[#, 2, 1], # < n] &], {n, 120}] (* Michael De Vlieger, Oct 03 2017 *)
PROG
(PARI) A290090(n) = sumdiv(n, d, (d<n)*(hammingweight(d)%2));
0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, -2, 0, 3, 2, 0, 0, 1, 0, 2, 3, 7, 0, -8, 2, 9, 3, 4, 0, 2, 0, 0, 7, 14, 5, -10, 0, 15, 9, -2, 0, 9, 0, 12, 7, 18, 0, -22, 3, 18, 14, 16, 0, 6, 9, 1, 15, 24, 0, -24, 0, 25, 13, 0, 11, 26, 0, 26, 18, 25, 0, -45, 0, 33, 20, 28, 10, 32, 0, -14, 13, 37, 0, -15, 16, 38, 24, 13, 0, -8, 12, 34, 25, 41, 17, -52, 0
PROG
(PARI)
A005187(n) = { my(s=n); while(n>>=1, s+=n); s; };
a(n) is the number of such divisors d of n that A000120(d) does not divide n, where A000120(d) gives the binary weight of d.
+10
3
0, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 2, 3, 0, 1, 0, 1, 0, 1, 2, 1, 0, 2, 2, 3, 3, 1, 2, 1, 0, 2, 0, 3, 0, 1, 2, 2, 0, 1, 0, 1, 3, 5, 2, 1, 0, 2, 2, 3, 3, 1, 2, 2, 4, 2, 2, 1, 0, 1, 2, 3, 0, 3, 0, 1, 0, 2, 4, 1, 0, 1, 2, 4, 3, 3, 2, 1, 0, 3, 2, 1, 0, 3, 2, 3, 4, 1, 4, 3, 0, 3, 2, 3, 0, 1, 4, 4, 3, 1, 2, 1, 4, 4
COMMENTS
Number of such positive integers k that divide n but A000120(k) [the Hamming weight of k] does not divide n.
FORMULA
a(n) = Sum_{d|n} [ A000120(d) does not divide n], where [ ] is the Iverson bracket. 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) A324393(n) = sumdiv(n, d, !!(n%hammingweight(d)));
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.
+10
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
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(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)));
50, 75, 100, 125, 150, 175, 190, 200, 222, 225, 238, 242, 246, 250, 275, 285, 300, 325, 333, 338, 350, 357, 363, 369, 374, 375, 380, 400, 425, 438, 442, 444, 450, 475, 476, 484, 492, 494, 500, 507, 525, 550, 555, 561, 570, 575, 578, 595, 600, 605, 615, 625
PROG
a321516(n) = my(d=divisors(n), i=0); for(k=2, #d-1, if(!ispseudoprime(d[k]), i++)); i
is(n) = a294902(n)!=a321516(n)
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