A195199 Smallest multiple of n with more than twice as many divisors as n.

4, 12, 12, 24, 20, 36, 28, 48, 36, 60, 44, 120, 52, 84, 60, 96, 68, 144, 76, 120, 84, 132, 92, 240, 100, 156, 108, 168, 116, 180, 124, 192, 132, 204, 140, 360, 148, 228, 156, 240, 164, 252, 172, 264, 180, 276, 188, 480, 196, 300, 204, 312, 212, 432, 220, 336

Offset: 1

Author

J. Lowell, Oct 12 2011

Keywords

• nonn

Examples

			a(4) must have more than 6 divisors because 4 has 3 divisors and 3*2=6. Therefore, it cannot be 16 because 16 has only 5 divisors.
		

Crossrefs

Cf. A000005.

Programs

Maple

A195199 := proc(n)
        for k from 2 do
                if numtheory[tau](k*n) > 2*numtheory[tau](n) then
                        return k*n ;
                end if;
        end do:
end proc: # R. J. Mathar, Oct 21 2011

Mathematica

Table[d = DivisorSigma[0, n]; m = 1; While[DivisorSigma[0, m*n] <= 2*d, m++]; m*n, {n, 100}] (* T. D. Noe, Oct 21 2011 *)

PARI

a(n) = my(m=n, d=numdiv(n)); while(numdiv(m)<=2*d, m+=n); m; \\ Michel Marcus, Jan 08 2022

Python

from sympy import divisor_count
def a(n):
    dtarget, m = 2*divisor_count(n), 2*n
    while divisor_count(m) <= dtarget: m += n
    return m
print([a(n) for n in range(1, 57)]) # Michael S. Branicky, Jan 08 2022

Python

from math import prod
from itertools import count
from collections import Counter
from sympy import factorint
def A195199(n):
    f = Counter(factorint(n))
    d = prod(e+1 for e in f.values())
    for m in count(2):
        if prod(e+1 for e in (f+Counter(factorint(m))).values()) > 2*d:
            return m*n # Chai Wah Wu, Feb 28 2022

Formula

a(n) = Min_{A000005(k*n) > 2*A000005(n)} k*n.