A367099 Least positive integer such that the number of divisors having two distinct prime factors is n.

1, 6, 12, 24, 36, 60, 72, 120, 144, 216, 288, 360, 432, 960, 720, 864, 1296, 1440, 1728, 2160, 2592, 3456, 7560, 4320, 5184, 7776, 10800, 8640, 10368, 12960, 15552, 17280, 20736, 40320, 25920, 31104, 41472, 60480, 64800, 51840, 62208, 77760, 93312

Offset: 0

Author

Gus Wiseman, Nov 09 2023

Keywords

• nonn

Comments

Does this contain every power of six, namely 1, 6, 36, 216, 1296, 7776, ...?

Yes, every power of six is a term, since 6^k = 2^k * 3^k is the least positive integer having n = tau(6^k) - (2k+1) divisors with two distinct prime factors. - Ivan N. Ianakiev, Nov 11 2023

Examples

			The divisors of 60 having two distinct prime factors are: 6, 10, 12, 15, 20. Since 60 is the first number having five such divisors, we have a(5) = 60.
The terms together with their prime indices begin:
     1: {}
     6: {1,2}
    12: {1,1,2}
    24: {1,1,1,2}
    36: {1,1,2,2}
    60: {1,1,2,3}
    72: {1,1,1,2,2}
   120: {1,1,1,2,3}
   144: {1,1,1,1,2,2}
   216: {1,1,1,2,2,2}
   288: {1,1,1,1,1,2,2}
   360: {1,1,1,2,2,3}
   432: {1,1,1,1,2,2,2}
   960: {1,1,1,1,1,1,2,3}
   720: {1,1,1,1,2,2,3}
   864: {1,1,1,1,1,2,2,2}
		

Links

• Amiram Eldar, Table of n, a(n) for n = 0..4469

Crossrefs

The version for all divisors is A005179 (firsts of A000005).

For all prime factors (A001222) we have A220264, firsts of A086971.

Positions of first appearances in A367098 (counts divisors in A007774).

A000961 lists prime powers, complement A024619.

A001221 counts distinct prime factors.

A001358 lists semiprimes, squarefree A006881, complement A100959.

A367096 lists semiprime divisors, sum A076290.

Cf. A001248, A054753, A056170, A079275, A146289, A366740, A367093.

Programs

Mathematica

nn=1000;
w=Table[Length[Select[Divisors[n],PrimeNu[#]==2&]],{n,nn}];
spnm[y_]:=Max@@NestWhile[Most,y,Union[#]!=Range[0,Max@@#]&];
Table[Position[w,k][[1,1]],{k,0,spnm[w]}]

PARI

a(n) = my(k=1); while (sumdiv(k, d, omega(d)==2) != n, k++); k; \\ Michel Marcus, Nov 11 2023