A355732 Least k such that there are exactly n ways to choose a sequence of divisors, one of each element of the multiset of prime indices of k (with multiplicity).
1, 3, 7, 9, 53, 21, 311, 27, 49, 159, 8161, 63, 38873, 933, 371, 81, 147, 477, 2177, 24483, 189, 2809, 343, 2799, 1113, 243, 57127, 16483, 441, 1431, 6531, 73449, 2597, 567, 96721, 8427, 1029, 8397, 3339, 15239, 729, 49449, 1323, 19663, 4293, 2401, 19593, 7791
Offset: 1
Keywords
Examples
The terms together with their prime indices begin: 1: {} 3: {2} 7: {4} 9: {2,2} 53: {16} 21: {2,4} 311: {64} 27: {2,2,2} 49: {4,4} 159: {2,16} 8161: {1024} 63: {2,2,4} For example, the choices for a(12) = 63 are: (1,1,1) (1,2,2) (2,1,4) (1,1,2) (1,2,4) (2,2,1) (1,1,4) (2,1,1) (2,2,2) (1,2,1) (2,1,2) (2,2,4)
Crossrefs
Programs
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Mathematica
primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]; mnrm[s_]:=If[Min@@s==1,mnrm[DeleteCases[s-1,0]]+1,0]; az=Table[Times@@Length/@Divisors/@primeMS[n],{n,1000}]; Table[Position[az,k][[1,1]],{k,mnrm[az]}]
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