A294080 -id:A294080 - cp's OEIS frontend

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

Gus Wiseman, Feb 07 2018

nonn

By convention a(1) = 0.

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

			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)

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

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

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

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

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

A301367 Regular triangle where T(n,k) is the number of orderless same-trees of weight n with k leaves.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 1, 2, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 3, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 3, 4, 4, 3, 5, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 1, 0, 0, 1, 2, 1, 1, 1, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 4, 5, 10, 11, 14, 12, 14, 7, 13, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1

Offset: 1

Gus Wiseman, Mar 19 2018

An orderless same-tree of weight n > 0 is either a single node of weight n, or a finite multiset of two or more orderless same-trees whose weights are all the same and sum to n.

			Triangle begins:
1
1   1
1   0   1
1   1   1   2
1   0   0   0   1
1   1   1   2   1   3
1   0   0   0   0   0   1
1   1   1   3   4   4   3   5
1   0   1   0   1   0   1   0   2
1   1   0   0   1   2   1   1   1   3
1   0   0   0   0   0   0   0   0   0   1
1   1   2   4   5  10  11  14  12  14   7  13
1   0   0   0   0   0   0   0   0   0   0   0   1
1   1   0   0   0   0   1   2   1   1   1   1   1   3
The T(8,5) = 4 orderless same-trees: (4((11)(11))), (4(1111)), ((22)(2(11))), (222(11)).

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Last entries of each row give A289079. Row sums are A289078.

Cf. A003238, A006241, A008284, A055277, A063834, A273873, A281145, A289501, A294080, A298422, A298426, A299201, A299203, A301343, A301364-A301368.

olstrees[n_]:=Prepend[Join@@Table[Select[Tuples[olstrees/@ptn],OrderedQ],{ptn,Select[IntegerPartitions[n],Length[#]>1&&SameQ@@#&]}],n];
Table[Length[Select[olstrees[n],Count[#,_Integer,{-1}]===k&]],{n,14},{k,n}]

S(g, k)={polcoef(exp(sum(i=1, k, x^i*subst(g, y, y^i)/i) + O(x*x^k)), k)}
A(n)={my(v=vector(n)); for(n=1, n, v[n] = y + sumdiv(n, d, S(v[n/d], d))); apply(p -> Vecrev(p/y), v)}
{ my(v=A(16)); for(n=1, #v, print(v[n])) } \\ Andrew Howroyd, Aug 20 2018

A301366 Regular triangle where T(n,k) is the number of same-trees of weight n with k leaves.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 2, 2, 1, 0, 0, 0, 1, 1, 1, 1, 5, 3, 3, 1, 0, 0, 0, 0, 0, 1, 1, 1, 2, 6, 12, 14, 12, 6, 1, 0, 1, 0, 3, 0, 3, 0, 2, 1, 1, 0, 0, 1, 7, 10, 10, 5, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 3, 7, 21, 41, 58, 100, 100, 94, 48, 20

Offset: 1

Gus Wiseman, Mar 19 2018

A same-tree of weight n > 0 is either a single node of weight n, or a finite sequence of two or more same-trees whose weights are all the same and sum to n.

			Triangle begins:
1
1   1
1   0   1
1   1   2   2
1   0   0   0   1
1   1   1   5   3   3
1   0   0   0   0   0   1
1   1   2   6  12  14  12   6
1   0   1   0   3   0   3   0   2
1   1   0   0   1   7  10  10   5   3
1   0   0   0   0   0   0   0   0   0   1
1   1   3   7  21  41  58 100 100  94  48  20
The T(8,4) = 6 same-trees: (4(2(11))), (4((11)2)), ((22)(22)), ((2(11))4), (((11)2)4), (2222).

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Last entries of each row give A006241. Row sums are A281145.

Cf. A003238, A008284, A055277, A063834, A273873, A289501, A294080, A298422, A298426, A299201, A299203, A300442, A300443, A301343, A301364-A301368.

sametrees[n_]:=Prepend[Join@@Table[Tuples[sametrees/@ptn],{ptn,Select[IntegerPartitions[n],Length[#]>1&&SameQ@@#&]}],n];
Table[Length[Select[sametrees[n],Count[#,_Integer,{-1}]===k&]],{n,12},{k,n}]

A(n)={my(v=vector(n)); for(n=1, n, v[n] = x + sumdiv(n, d, v[n/d]^d)); apply(p -> Vecrev(p/x), v)}
{my(v=A(16)); for(n=1, #v, print(v[n]))} \\ Andrew Howroyd, Aug 20 2018

cp's OEIS Frontend

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

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A301367 Regular triangle where T(n,k) is the number of orderless same-trees of weight n with k leaves.

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Author

Keywords

Comments

Examples

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Programs

Mathematica

PARI

A301366 Regular triangle where T(n,k) is the number of same-trees of weight n with k leaves.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI