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%I #11 Jul 18 2019 12:11:56
%S 1,2,6,4,30,12,28,8,36,60,330,24,156,56,180,16,476,72,152,120,168,660,
%T 828,48,900,312,216,112,1740,360,10230,32,1980,952,840,144,888,304,
%U 936,240,6396,336,2408,1320,1080,1656,8460,96,784,1800,2856,624,848,432
%N a(n) = unique m such that m/A003963(m) = n, where A003963 is product of prime indices.
%C Every positive integer has a unique factorization into factors q(i) = prime(i)/i, i > 0 given by the rows of A324924. Then a(n) is the number obtained by encoding this factorization as a standard factorization into prime numbers (A112798).
%H Rémy Sigrist, <a href="/A324922/b324922.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = Product_t mg(t) where the product is over all (not necessarily distinct) terminal subtrees of the rooted tree with Matula-Goebel number n, and mg(t) is the Matula-Goebel number of t.
%F Completely multiplicative with a(prime(n)) = prime(n) * a(n). - _Rémy Sigrist_, Jul 18 2019
%t primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
%t difac[n_]:=If[n==1,{},With[{m=Product[Prime[i]/i,{i,primeMS[n]}]},Sort[Join[primeMS[n],difac[n/m]]]]];
%t Table[Times@@Prime/@difac[n],{n,30}]
%o (PARI) a(n) = my (f=factor(n)); prod (i=1, #f~, (f[i,1] * a(primepi(f[i,1])))^f[i,2]) \\ _Rémy Sigrist_, Jul 18 2019
%Y Sorting the sequence gives A324850.
%Y Cf. A000081, A003963, A061775, A109129, A112798, A120383, A196050, A317713.
%Y Cf. A324848, A324849, A324923, A324924, A324925, A324931, A324934.
%K nonn,mult
%O 1,2
%A _Gus Wiseman_, Mar 20 2019
%E Keyword mult added by _Rémy Sigrist_, Jul 18 2019
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