%I #14 May 20 2017 21:44:36
%S 1,2,2,4,2,3,2,8,4,3,2,6,2,3,3,16,2,6,2,6,3,3,2,12,4,3,8,6,2,5,2,32,3,
%T 3,3,9,2,3,3,12,2,5,2,6,6,3,2,24,4,6,3,6,2,12,3,12,3,3,2,10,2,3,6,64,
%U 3,5,2,6,3,5,2,18,2,3,6,6,3,5,2,24,16,3,2
%N a(1) = 1; for n > 1, if the first integer with the same prime signature as n is factorized into primorials as Product A002110(i)^e(i), then a(n) = Product prime(i)^e(i).
%C Alternate definition: a(1) = 1; for n > 1, if row n of table A238744 is {k(1), k(2),...,k(A051903(n))}, then a(n) = Product {i = 1 to A051903(n)} prime(k(i)).
%C Since the first integer of each prime signature (A025487) is always a product of primorials (A002110), there is always a value for a(n). Every positive integer appears in the sequence.
%C a(m) = a(n) iff m and n have the same prime signature. If the prime signatures of m and n are conjugate to each other when they are viewed as partitions, then a(n) = A181819(m) and a(m) = A181819(n).
%H Antti Karttunen, <a href="/A238745/b238745.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A181819(A124859(n)).
%F a(n) = A122111(A181819(n)).
%e The first integer with the same prime signature as 40 is 24 = 2^3*3. Since the factorization of 24 into primorials is 24 = 2^2*6 = A002110(1)^2*A002110(2), a(24) = a(40) = prime(1)^2*prime(2) = 2^2*3 = 12.
%t f[n_] := Block[{k = 1, d, a}, While[n - Times @@ Prime@ Range[k + 1] >= 0, k++]; If[n == Product[Prime@ i, {i, k}], Prime@ k, d = Select[Reverse@ FoldList[#1 #2 &, Prime@ Range@ k], Divisible[n, #] &]; If[AllTrue[#, IntegerQ], Times @@ Map[(FactorInteger[#1][[-1, 1]])^#2 & @@ # &, Reverse@ Tally@ #], False] &@ Rest@ NestWhileList[Function[P, {#1/P, P}]@ SelectFirst[d, Function[k, Divisible[#1, k]]] & @@ # &, {n, 1}, First@ # > 1 &][[All, -1]]]]; Table[f@ Apply[Times, MapIndexed[Prime[First@ #2]^#1 &, Sort[FactorInteger[n][[All, -1]], Greater]]] - Boole[n == 1], {n, 83}] (* _Michael De Vlieger_, May 19 2017, Version 10.2 *)
%Y Cf. A181815, A181819, A238744.
%K nonn
%O 1,2
%A _Matthew Vandermast_, Apr 28 2014
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