A356040 a(n) is the smallest integer that has exactly n odious divisors (A227872) and n evil divisors (A356018).

3, 6, 12, 24, 48, 96, 192, 210, 252, 528, 3072, 420, 12288, 2112, 1008, 840, 196608, 2016, 786432, 1680, 4032, 33792, 12582912, 3360, 30000, 135168, 16128, 6720, 805306368, 19152, 3221225472, 13440, 64512, 2162688, 120000, 26880, 206158430208, 8650752, 258048, 31920

Offset: 1

Author

Bernard Schott, Jul 24 2022

Keywords

• nonn
• base

Comments

As the number of divisors is even, there is no square in the sequence.

Examples

			48 has ten divisors, five of which are odious {1, 2, 4, 8, 16} as they have an odd number of 1's in their binary expansion: 1, 10, 100, 1000 and 10000; the five other divisors are evil {3, 6, 12, 24, 48} as they have an even number of 1's in their binary expansion: 11, 110, 1100, 11000 and 110000; also, no positive integer smaller than 48 has five divisors that are evil and five divisors that are odious, hence a(5) = 48.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..3322

Crossrefs

Cf. A000069, A001969, A227872, A356018.

Cf. A230851, A005179.

Programs

Mathematica

f[n_] := DivisorSum[n, {1, (-1)^DigitCount[#, 2][[1]]} &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i, d}, While[c < len && n < nmax, i = f[n]; If[i[[2]] == 0, d = i[[1]]/2; If[d <= len && s[[d]] == 0, c++; s[[d]] = n]]; n++]; s]; seq[16, 10^6] (* Amiram Eldar, Jul 24 2022 *)

PARI

a(n) = if(isprime(n), return(2^(n-1)*3)); forfactored(i=1, 2^(n-1)*3, if(numdiv(i[2]) == 2*n, d=divisors(i[2]); if(sum(j=1, #d, isevil(d[j])) == n, return(i[1]))))
isevil(n) = bitand(hammingweight(n), 1) == 0 \\ David A. Corneth, Jul 24 2022

Python

from sympy import divisors, isprime
from itertools import count, islice
def f(n):
    counts = [0, 0]
    for d in divisors(n, generator=True):
        counts[bin(d).count("1")&1] += 1
    return counts[0] if counts[0] == counts[1] else -1
def a(n):
    if isprime(n): return 2**(n-1) * 3
    return next(k for k in count(1) if f(k) == n)
print([a(n) for n in range(1, 34)]) # Michael S. Branicky, Jul 24 2022

Formula

a(n) <= 2^(n-1) * 3.

a(n) = 2^(n-1) * 3 if n is prime. - David A. Corneth, Jul 24 2022

a(n) >= A005179(2*n). - Michael S. Branicky, Jul 24 2022

Extensions

a(9)-a(37) from Amiram Eldar, Jul 24 2022

a(38)-a(40) from David A. Corneth, Jul 24 2022