A036460 Erroneous version of A009287 (the 2 should be omitted).
2, 3, 4, 6, 12, 60, 5040, 293318625600
Offset: 0
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From _Gus Wiseman_, May 13 2018: (Start) Like A001462 the following sequence of multisets whose Heinz numbers belong to this sequence is a run-length describing sequence, as the number of k's in row n + 1 is equal to the k-th term of row n. {2} {1,1} {1,2} {1,1,2} {1,1,2,3} {1,1,1,2,2,3,4} {1,1,1,1,2,2,2,3,3,4,4,5,6,7} {1,1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,4,4,4,4,5,5,5,5,6,6,6,7,7,7,8,8,9,9,10,10,11,12,13,14} (End)
Prepend[Function[m,Times@@Prime/@m]/@NestList[Join@@Table[Table[i,{Reverse[#][[i]]}],{i,Length[#]}]&,{2},8],1] (* Gus Wiseman, May 13 2018 *)
a(3) = 12 and tau(12) = 6 = 3!.
For n=7, a(7) = 293318625600 -> 5040 -> 60 -> 12 -> 6 -> 4 -> 3 -> 2. For n=8, a(8) -> 1111523212800 -> 5040 -> 60 -> 12 -> 6 -> 4 -> 3 -> 2.
a036459(n) = {if (n<=2, return(0)); nb = 1; while ((nd = numdiv(n)) > 2, n = nd; nb++); nb;} a(n) = {k = 1; while (a036459(k) != n, k++); k;} \\ Michel Marcus, Oct 28 2015
lista(nn) = {a=8; print1(a, ", "); for (n=2, nn, k=1; while(numdiv(k) != a, k++); print1(k, ", "); a = k;);} \\ Michel Marcus, Feb 17 2015
Since 4 is the HCN with 3 divisors, we have tau(4) = 3 and therefore a(1)=4; the HCN with 4 divisors is 6, so that tau(6) = 4 and hence a(2)=6; the HCN with 6 divisors is 12 so that tau(12) = 6, implying a(3)=12, ...
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