A304938 a(n) is the smallest number which can be written in n different ways as an ordered product of prime factors.

1, 6, 12, 24, 48, 30, 192, 384, 768, 72, 3072, 60, 12288, 24576, 144, 98304, 196608, 393216, 786432, 120, 288, 6291456, 12582912, 210, 50331648, 100663296, 201326592, 576, 805306368, 180, 3221225472, 6442450944, 12884901888, 25769803776, 432, 1152, 206158430208, 412316860416, 824633720832, 1649267441664

Offset: 1

Author

Vincent Champain, May 21 2018

Keywords

• nonn

Comments

It can be easily demonstrated that a(n) exists for all n and is less than or equal to 2^(n-1)*3 since 2^(n-1)*3 can be written in n different ways.

If n is a prime number then a(n) = 2^(n-1)*3, but there are also nonprime numbers n with this property (e.g., 9 and 14).

If n=k! then a(n) is the product of the first k prime numbers.

Finding the terms up to 2^64 was the focus of the 4th question of the ICPC coding challenge in 2013.

Examples

			a(1) = 1 because only a prime power or the empty product (which equals 1) can be written in just one way, and no prime power is smaller than 1.
a(2) = 6 = 3 * 2 = 2 * 3 because none of 3, 4, 5 can be written in two different ways.
a(3) = 12 = 3 * 2 * 2 = 2 * 3 * 2 = 2 * 2 * 3 (each of 7, 8, 9, 10, 11 can be written in at most 2 ways).
a(4) = 24 = 2 * 2 * 2 * 3 (each of 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23 can be written in at most 3 ways).
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..136
• The ACM-ICPC International Collegiate Programming Contest, ICPC 2013 problems

Crossrefs

Cf. A000961, A007283 (2^n*3), A008480 (number of ordered prime factorizations of n).

Subsequence of A025487.

The sorted version is A358526.

Cf. A001222, A027746, A206487, A344606, A358508.

Programs

Mathematica

uv=Table[Length[Permutations[Join@@ConstantArray@@@FactorInteger[n]]],{n,1,1000}];
Table[Position[uv,k][[1,1]],{k,Min@@Complement[Range[Max@@uv],uv]-1}] (* Gus Wiseman, Nov 22 2022 *)

PARI

a008480(n) = my(f=factor(n)); sum(k=1, #f~, f[k,2])!/prod(k=1, #f~, f[k,2]!);
a(n) = {my(k=2); while (a008480(k) !=n, k++); k;} \\ Michel Marcus, May 23 2018

Formula

a(p) = 2^(p-1)*3 if p is a prime.

a(k!) = prime(k)# is the k-th primorial number. So for no m < k!, prime(k) | a(m). - David A. Corneth, May 24 2018

a(n) = min { k : A008480(k) = n }. - Alois P. Heinz, May 26 2018