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A294019
Number of same-trees whose leaves are the parts of the integer partition with Heinz number n.
3
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
1,12
COMMENTS
By convention a(1) = 0.
The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).
LINKS
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
EXAMPLE
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)
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
PROG
(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
KEYWORD
nonn
AUTHOR
Gus Wiseman, Feb 07 2018
STATUS
approved