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a(n) is the smallest number which can be written in n different ways as an ordered product of prime factors.
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%I #80 Nov 23 2022 21:58:53

%S 1,6,12,24,48,30,192,384,768,72,3072,60,12288,24576,144,98304,196608,

%T 393216,786432,120,288,6291456,12582912,210,50331648,100663296,

%U 201326592,576,805306368,180,3221225472,6442450944,12884901888,25769803776,432,1152,206158430208,412316860416,824633720832,1649267441664

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

%C 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.

%C 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).

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

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

%H Amiram Eldar, <a href="/A304938/b304938.txt">Table of n, a(n) for n = 1..136</a>

%H The ACM-ICPC International Collegiate Programming Contest, <a href="https://icpc.baylor.edu/download/worldfinals/problems/icpc2013.pdf">ICPC 2013 problems</a>

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

%F 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

%F a(n) = min { k : A008480(k) = n }. - _Alois P. Heinz_, May 26 2018

%e 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.

%e a(2) = 6 = 3 * 2 = 2 * 3 because none of 3, 4, 5 can be written in two different ways.

%e 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).

%e 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).

%t uv=Table[Length[Permutations[Join@@ConstantArray@@@FactorInteger[n]]],{n,1,1000}];

%t Table[Position[uv,k][[1,1]],{k,Min@@Complement[Range[Max@@uv],uv]-1}] (* _Gus Wiseman_, Nov 22 2022 *)

%o (PARI) a008480(n) = my(f=factor(n)); sum(k=1, #f~, f[k,2])!/prod(k=1, #f~, f[k,2]!);

%o a(n) = {my(k=2); while (a008480(k) !=n, k++); k;} \\ _Michel Marcus_, May 23 2018

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

%Y Subsequence of A025487.

%Y The sorted version is A358526.

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

%K nonn

%O 1,2

%A _Vincent Champain_, May 21 2018