A320632 Numbers k such that there exists a pair of factorizations of k into factors > 1 where no factor of one divides any factor of the other.

36, 60, 72, 84, 90, 100, 108, 120, 126, 132, 140, 144, 150, 156, 168, 180, 196, 198, 200, 204, 210, 216, 220, 225, 228, 234, 240, 252, 260, 264, 270, 276, 280, 288, 294, 300, 306, 308, 312, 315, 324, 330, 336, 340, 342, 348, 350, 360, 364, 372, 378, 380, 390

Offset: 1

Author

Gus Wiseman, Dec 09 2018

Keywords

• nonn

Comments

Positions of nonzero terms in A322437 or A322438.

Mats Granvik has conjectured that these are all the positive integers k such that sigma_0(k) - 2 > (bigomega(k) - 1) * omega(k), where sigma_0 = A000005, omega = A001221, and bigomega = A001222. - Gus Wiseman, Nov 12 2019

Numbers with more semiprime divisors than prime divisors. - Wesley Ivan Hurt, Jun 10 2021

Examples

			An example of such a pair for 36 is (4*9)|(6*6).
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..23437
• Christophe Cordero, Factorizations of Cyclic Groups and Bayonet Codes, arXiv:2301.13566 [math.CO], 2023, p. 20.

Crossrefs

Cf. A001055, A050336, A285572, A303362, A305149, A305193, A317144, A322435, A322437, A322439, A322440, A322441, A322442.

The following are additional cross-references relating to Granvik's conjecture.

bigomega(n) * omega(n) is A113901(n).

(bigomega(n) - 1) * omega(n) is A307409(n).

sigma_0(n) - bigomega(n) * omega(n) is A328958(n).

sigma_0(n) - 2 - (omega(n) - 1) * nu(n) is A328959(n).

Cf. A060687, A070175, A124010, A323023, A328956, A328957, A328960, A328961, A328963.

Programs

Mathematica

facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
Select[Range[100],Select[Subsets[facs[#],{2}],And[!Or@@Divisible@@@Tuples[#],!Or@@Divisible@@@Reverse/@Tuples[#]]&]!={}&]

PARI

factorizations(n, m=n, f=List([]), z=List([])) = if(1==n, listput(z,Vec(f)); z, my(newf); fordiv(n, d, if((d>1)&&(d<=m), newf = List(f); listput(newf,d); z = factorizations(n/d, d, newf, z))); (z));
is_ndf_pair(fac1,fac2) = { for(i=1,#fac1,for(j=1,#fac2,if(!(fac1[i]%fac2[j])||!(fac2[j]%fac1[i]),return(0)))); (1); };
has_at_least_one_ndfpair(z) = { for(i=1,#z,for(j=i+1,#z,if(is_ndf_pair(z[i],z[j]),return(1)))); (0); };
isA320632(n) = has_at_least_one_ndfpair(Vec(factorizations(n))); \\ Antti Karttunen, Dec 10 2020