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A290090 a(n) is the number of proper divisors of n that are odious (A000069). 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

If n is odd and k >= 1, then a(2^k*n) = (k+1)*n+k if n is in A000069 and (k+1)*n if n is not in A000069. - Robert Israel, Oct 03 2017

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..16385

Index entries for sequences related to binary expansion of n

FORMULA

a(n) = Sum_{d|n, d<n} A010060(d).

a(n) = A227872(n) - A010060(n).

a(n) = A007814(A293231(n)).

A000035(a(n)) = A000035(A292257(n)). [Parity-wise equivalent with A292257.]

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:

map(f, [$1..200]); # Robert Israel, Oct 03 2017

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));

CROSSREFS

Cf. A000005, A000069, A010060, A227872, A292257, A293231.

Sequence in context: A305052 A305079 A319841 * A066075 A072347 A318831

Adjacent sequences:  A290087 A290088 A290089 * A290091 A290092 A290093

KEYWORD

nonn,base

AUTHOR

Antti Karttunen, Oct 03 2017

STATUS

approved

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Last modified January 22 23:00 EST 2019. Contains 319365 sequences. (Running on oeis4.)