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

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, 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, 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

Offset: 1

Author

Gus Wiseman, Feb 07 2018

Keywords

• nonn

Comments

By convention a(1) = 0.

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

Examples

			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).
From _Antti Karttunen_, Sep 22 2018: (Start)
For 12 = prime(1)^2 * prime(2)^1, we have the following two cases: 2(11) and (11)2, thus a(12) = 2.
For 36 = prime(1)^2 * prime(2)^2, we have the following cases: (11)22, 2(11)2, 22(11), thus a(36) = 3.
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.
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.
(End)
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A000005, A000041, A000720, A001222, A006241, A056239, A063834, A196545, A215366, A273873, A281145, A289501, A296150, A299200, A299201, A299202, A299203, A294018, A294080.

Programs

Mathematica

nn=120;
ptns=Table[If[n===1,{},Join@@Cases[FactorInteger[n]//Reverse,{p_,k_}:>Table[PrimePi[p],{k}]]],{n,nn}];
tris=Join@@Map[Tuples[IntegerPartitions/@#]&,ptns];
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/@#]&]}]];
qci/@ptns

PARI

A056239(n) = { my(f); if(1==n, 0, f=factor(n); sum(i=1, #f~, f[i,2] * primepi(f[i,1]))); }
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));
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));
A294019(n) = if(1==n,0,if(isprime(n),1,A294019aux(n, n-1, List([]))));
\\ A memoized implementation:
map294019 = Map();
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

Formula

A281145(n) = Sum_{i=1..A000041(n)} a(A215366(n,i)).

a(p^n) = A006241(n) for any prime p and exponent n >= 1. - Antti Karttunen, Sep 22 2018