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%I #31 Sep 11 2018 21:11:52
%S 1,1,1,2,1,2,1,2,2,2,1,2,2,2,1,3,2,2,2,1,3,2,2,2,2,1,2,3,2,2,2,2,2,1,
%T 2,3,3,2,2,2,2,2,1,2,3,3,2,2,2,2,2,2,1,2,3,3,2,2,2,2,2,2,1,3,2,3,3,2,
%U 2,2,2,2,2,1,3,2,3,3,2,2,3,2,2,2,2,1,3
%N Number of leaves in the free pure symmetric multifunction (with empty expressions allowed) with e-number n.
%C If n = 1 let e(n) be the leaf symbol "o". Given a positive integer n > 1 we construct a unique free pure symmetric multifunction e(n) with one atom by expressing n as a power of a number that is not a perfect power to a product of prime numbers: n = rad(x)^(prime(y_1) * ... * prime(y_k)) where rad = A007916. Then e(n) = e(x)[e(y_1), ..., e(y_k)]. For example, e(21025) = o[o[o]][o] because 21025 = rad(rad(1)^prime(rad(1)^prime(1)))^prime(1).
%F a(rad(x)^(prime(y_1) * ... * prime(y_k)) = a(x) + a(y_1) + ... + a(y_k) where rad = A007916.
%e e(21025) = o[o[o]][o] has 4 leaves so a(21025) = 4.
%t nn=1000;
%t radQ[n_]:=If[n==1,False,GCD@@FactorInteger[n][[All,2]]==1];
%t rad[n_]:=rad[n]=If[n==0,1,NestWhile[#+1&,rad[n-1]+1,Not[radQ[#]]&]];
%t Clear[radPi];Set@@@Array[radPi[rad[#]]==#&,nn];
%t a[n_]:=If[n==1,1,With[{g=GCD@@FactorInteger[n][[All,2]]},a[radPi[Power[n,1/g]]]+Sum[a[PrimePi[pr[[1]]]]*pr[[2]],{pr,If[g==1,{},FactorInteger[g]]}]]];
%t Table[a[n],{n,100}]
%Y Cf. A007916, A052409, A052410, A109129, A277576, A277996, A300626, A316112, A317056, A317658, A317765, A317994.
%K nonn
%O 1,4
%A _Gus Wiseman_, Aug 18 2018
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