login
A195199
Smallest multiple of n with more than twice as many divisors as n.
4
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
1,1
FORMULA
a(n) = Min_{A000005(k*n) > 2*A000005(n)} k*n.
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
Cf. A000005.
Sequence in context: A005886 A096442 A211437 * A294628 A323188 A284126
KEYWORD
nonn
AUTHOR
J. Lowell, Oct 12 2011
STATUS
approved