A347046 Greatest divisor of n with exactly half as many prime factors (counting multiplicity) as n, or 1 if there are none.

1, 1, 1, 2, 1, 3, 1, 1, 3, 5, 1, 1, 1, 7, 5, 4, 1, 1, 1, 1, 7, 11, 1, 6, 5, 13, 1, 1, 1, 1, 1, 1, 11, 17, 7, 9, 1, 19, 13, 10, 1, 1, 1, 1, 1, 23, 1, 1, 7, 1, 17, 1, 1, 9, 11, 14, 19, 29, 1, 15, 1, 31, 1, 8, 13, 1, 1, 1, 23, 1, 1, 1, 1, 37, 1, 1, 11, 1, 1, 1, 9

Offset: 1

Author

Gus Wiseman, Aug 16 2021

Keywords

• nonn

Comments

Problem: What are the positions of last appearances > 1?

Examples

			The divisors of 90 with half bigomega are: 6, 9, 10, 15, so a(90) = 15.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

The greatest divisor without the condition is A006530 (smallest: A020639).

Positions of 1's are A026424.

The case of powers of 2 is A072345.

Positions of first appearances are A123667 (sorted: A123666).

Divisors of this type are counted by A345957 (rounded: A096825).

The rounded version is A347044.

The smallest divisor of this is A347045 (rounded: A347043).

A000005 counts divisors.

A001221 counts distinct prime factors.

A001222 counts all prime factors (also called bigomega).

A056239 adds up prime indices, row sums of A112798.

A207375 lists central divisors (min: A033676, max: A033677).

A340387 lists numbers whose sum of prime indices is twice bigomega.

A340609 lists numbers whose maximum prime index divides bigomega.

A340610 lists numbers whose maximum prime index is divisible by bigomega.

A347042 counts divisors d|n such that bigomega(d) divides bigomega(n).

Cf. A000720, A027746, A038548, A060775, A106529, A244991, A335433, A335448.

Programs

Mathematica

Table[If[#=={},1,Max[#]]&@Select[Divisors[n], PrimeOmega[#]==PrimeOmega[n]/2&],{n,100}]
a[n_] := Module[{p = Flatten[Table[#[[1]], {#[[2]]}] & /@ FactorInteger[n]], np}, np = Length[p]; If[OddQ[np], 1, Times @@ p[[np/2+1 ;; np]]]]; Array[a, 100] (* Amiram Eldar, Nov 02 2024 *)

Python

from sympy import divisors, factorint
def a(n):
    npf = len(factorint(n, multiple=True))
    for d in divisors(n)[-1:0:-1]:
        if 2*len(factorint(d, multiple=True)) == npf: return d
    return 1
print([a(n) for n in range(1, 82)]) # Michael S. Branicky, Aug 18 2021

Python

from math import prod
from sympy import factorint
def A347046(n):
    fs = factorint(n,multiple=True)
    q, r = divmod(len(fs),2)
    return 1 if r else prod(fs[q:]) # Chai Wah Wu, Aug 20 2021

Formula

a(n) = Product_{k=A001222(n)/2+1..A001222(n)} A027746(n,k) if A001222(n) is even, and 1 otherwise. - Amiram Eldar, Nov 02 2024