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A294019 Number of same-trees whose leaves are the parts of the integer partition with Heinz number n.

%I A294019 #14 Sep 24 2018 08:54:07
%S A294019 0,1,1,1,1,0,1,1,1,0,1,2,1,0,0,2,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,1,0,0,
%T A294019 0,3,1,0,0,2,1,0,1,0,0,0,1,3,1,0,0,0,1,0,0,0,0,0,1,0,1,0,2,3,0,0,1,0,
%U A294019 0,0,1,0,1,0,0,0,0,0,1,0,2,0,1,4,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,8
%N A294019 Number of same-trees whose leaves are the parts of the integer partition with Heinz number n.
%C A294019 By convention a(1) = 0.
%C A294019 The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).
%H A294019 Antti Karttunen, <a href="/A294019/b294019.txt">Table of n, a(n) for n = 1..65537</a>
%F A294019 A281145(n) = Sum_{i=1..A000041(n)} a(A215366(n,i)).
%F A294019 a(p^n) = A006241(n) for any prime p and exponent n >= 1. - _Antti Karttunen_, Sep 22 2018
%e A294019 The a(108) = 8 same-trees: ((22)(2(11))), ((22)((11)2)), ((2(11))(22)), (((11)2)(22)), (222(11)), (22(11)2), (2(11)22), ((11)222).
%e A294019 From _Antti Karttunen_, Sep 22 2018: (Start)
%e A294019 For 12 = prime(1)^2 * prime(2)^1, we have the following two cases: 2(11) and (11)2, thus a(12) = 2.
%e A294019 For 36 = prime(1)^2 * prime(2)^2, we have the following cases: (11)22, 2(11)2, 22(11), thus a(36) = 3.
%e A294019 For 144  = prime(1)^4 * prime(2)^2, we have the following 14 cases: (1111)(22), (22)(1111); ((11)(11))(22), (22)((11)(11)); (11)(11)22, (11)2(11)2, (11)22(11), 2(11)2(11), 2(11)(11)2, 22(11)(11); ((11)2)(11(2)), ((11)2)(2(11)), (2(11))((11)2), (2(11))(2(11)), thus a(144) = 14.
%e A294019 For n = 8775 = 3^3 * 5^2 * 13^1 = prime(2)^3 * prime(3)^2 * prime(6)^1, we have the following six cases: (222)(33)6, (222)6(33), (33)(222)6, (33)6(222), 6(222)(33), 6(33)(222), thus a(8775) = 6.
%e A294019 (End)
%t A294019 nn=120;
%t A294019 ptns=Table[If[n===1,{},Join@@Cases[FactorInteger[n]//Reverse,{p_,k_}:>Table[PrimePi[p],{k}]]],{n,nn}];
%t A294019 tris=Join@@Map[Tuples[IntegerPartitions/@#]&,ptns];
%t A294019 qci[y_]:=qci[y]=If[Length[y]===1,1,Sum[Times@@qci/@t,{t,Select[tris,And[Length[#]>1,Sort[Join@@#,Greater]===y,SameQ@@Total/@#]&]}]];
%t A294019 qci/@ptns
%o A294019 (PARI)
%o A294019 A056239(n) = { my(f); if(1==n, 0, f=factor(n); sum(i=1, #f~, f[i,2] * primepi(f[i,1]))); }
%o A294019 productifbalancedfactorization(v) = if(!#v, 1, my(pw=A056239(v[1]), m=1); for(i=1,#v,if(A056239(v[i])!=pw,return(0), m *= A294019(v[i]))); (m));
%o A294019 A294019aux(n, m, facs) = if(1==n, productifbalancedfactorization(Vec(facs)), my(s=0, newfacs); fordiv(n, d, if((d>1)&&(d<=m), newfacs = List(facs); listput(newfacs,d); s += A294019aux(n/d, m, newfacs))); (s));
%o A294019 A294019(n) = if(1==n,0,if(isprime(n),1,A294019aux(n, n-1, List([]))));
%o A294019 \\ A memoized implementation:
%o A294019 map294019 = Map();
%o A294019 A294019(n) = if(1==n,0,if(isprime(n),1,if(mapisdefined(map294019,n), mapget(map294019,n), my(v=A294019aux(n, n-1, List([]))); mapput(map294019,n,v); (v)))); \\ _Antti Karttunen_, Sep 22 2018
%Y A294019 Cf. A000005, A000041, A000720, A001222, A006241, A056239, A063834, A196545, A215366, A273873, A281145, A289501, A296150, A299200, A299201, A299202, A299203, A294018, A294080.
%K A294019 nonn
%O A294019 1,12
%A A294019 _Gus Wiseman_, Feb 07 2018