A273873 -id:A273873 - cp's OEIS frontend

A289078 Number of orderless same-trees of weight n.

1, 2, 2, 5, 2, 9, 2, 22, 6, 11, 2, 94, 2, 13, 12, 334, 2, 205, 2, 210, 14, 17, 2, 7218, 8, 19, 68, 443, 2, 1687, 2, 69109, 18, 23, 16, 167873, 2, 25, 20, 89969, 2, 7041, 2, 1548, 644, 29, 2, 36094795, 10, 3078, 24, 2604, 2, 1484102, 20, 1287306, 26, 35, 2

Offset: 1

Gus Wiseman, Jun 23 2017

nonn

An orderless same-tree t is either: (case 1) a positive integer, or (case 2) a finite multiset of two or more orderless same-trees, all having the same weight. The weight of t in case 1 is the number itself, and in case 2 it is the sum of weights of the branches. For example {{{3,{1,1,1}},{2,{1,1},{1,1}}},{{{1,1,1},{1,1,1}},{{1,1},{1,1},{1,1}}}} is an orderless same-tree of weight 24 with 2 branches.

			The a(6)=9 orderless same-trees are: 6, (33), (3(111)), (222), (22(11)), (2(11)(11)), ((11)(11)(11)), ((111)(111)), (111111).

Alois P. Heinz, Table of n, a(n) for n = 1..2500

Cf. A196545, A273873, A275870, A281145, A281146, A289079.

with(numtheory):
a:= proc(n) option remember; 1 + add(
      binomial(a(n/d)+d-1, d), d=divisors(n) minus {1})
    end:
seq(a(n), n=1..60);  # Alois P. Heinz, Jul 05 2017

a[n_]:=If[n===1,1,1+Sum[Binomial[a[n/d]+d-1,d],{d,Rest[Divisors[n]]}]];
Array[a,100]

seq(n)={my(v=vector(n)); for(n=1, n, v[n] = 1 + sumdiv(n, d, binomial(v[n/d]+d-1, d))); v} \\ Andrew Howroyd, Aug 20 2018

a(n) = 1 + Sum_{d|n, d>1} binomial(a(n/d)+d-1, d).

A289079 Number of orderless same-trees of weight n with all leaves equal to 1.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 5, 2, 3, 1, 13, 1, 3, 3, 22, 1, 16, 1, 15, 3, 3, 1, 151, 2, 3, 6, 17, 1, 41, 1, 334, 3, 3, 3, 637, 1, 3, 3, 275, 1, 56, 1, 21, 19, 3, 1, 15591, 2, 27, 3, 23, 1, 902, 3, 516, 3, 3, 1, 7858, 1, 3, 21, 69109, 3, 98, 1, 27, 3, 67, 1, 811756, 1

Offset: 1

Gus Wiseman, Jun 23 2017

nonn

a(n) is also the number of orderless same-trees of weight n with all leaves greater than 1.

			The a(12)=13 orderless same-trees with all leaves greater than 1 are: ((33)(33)), ((33)(222)), ((33)6), ((222)(222)), ((222)6), (66), ((22)(22)(22)), ((22)(22)4), ((22)44), (444), (3333), (222222), 12.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A196545, A273873, A275870, A281145, A281146, A289078.

with(numtheory):
a:= proc(n) option remember; `if`(n=1, 1, add(
      binomial(a(n/d)+d-1, d), d=divisors(n) minus {1}))
    end:
seq(a(n), n=1..80);  # Alois P. Heinz, Jul 05 2017

a[n_]:=If[n===1,1,Sum[Binomial[a[n/d]+d-1,d],{d,Rest[Divisors[n]]}]];
Array[a,100]

seq(n)={my(v=vector(n)); v[1]=1; for(n=2, n, v[n] = sumdiv(n, d, binomial(v[n/d]+d-1, d))); v} \\ Andrew Howroyd, Aug 20 2018

from sympy import divisors, binomial
l=[0, 1]
for n in range(2, 101): l+=[sum([binomial(l[n//d] + d - 1, d) for d in divisors(n)[1:]]), ]
l[1:] # Indranil Ghosh, Jul 06 2017

a(1) = 1, a(n>1) = Sum_{d|n, d>1} binomial(a(n/d)+d-1, d).

A299203 Number of enriched p-trees whose multiset of leaves is the integer partition with Heinz number n.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 5, 1, 3, 1, 3, 1, 1, 1, 11, 1, 1, 2, 3, 1, 5, 1, 12, 1, 1, 1, 15, 1, 1, 1, 11, 1, 4, 1, 3, 3, 1, 1, 38, 1, 3, 1, 3, 1, 9, 1, 9, 1, 1, 1, 21, 1, 1, 4, 34, 1, 4, 1, 3, 1, 5, 1, 54, 1, 1, 3, 3, 1, 4, 1, 33, 5, 1, 1, 23, 1, 1, 1, 9, 1, 20, 1, 3, 1, 1, 1, 117, 1, 3, 3, 12, 1, 4, 1, 9, 4, 1, 1, 57, 1, 4, 1, 34

Offset: 1

Gus Wiseman, Feb 05 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).

			a(54) = 9: (((22)2)1), ((222)1), (((22)1)2), (((21)2)2), ((221)2), ((22)(21)), ((22)21), ((21)22), (2221).
a(40) = 11: ((31)(11)), (((31)1)1), ((3(11))1), ((311)1), (3((11)1)), (3(111)), (((11)1)3), ((111)3), ((31)11), (3(11)1), (3111).
a(36) = 15: ((22)(11)), ((2(11))2), (((11)2)2), (((21)1)2), ((211)2), (((22)1)1), (((21)2)1), ((221)1), ((21)(21)), (22(11)), (2(11)2), ((11)22), ((22)11), ((21)21), (2211).

Cf. A000041, A063834, A112798, A196545, A273873, A281145, A289501, A290261, A296150, A299200, A299201, A299202.

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]&]}]];
qci/@ptns

A061260 G.f.: Product_{k>=1} (1-y*x^k)^(-numbpart(k)), where numbpart(k) = number of partitions of k, cf. A000041.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 5, 6, 2, 1, 7, 11, 6, 2, 1, 11, 23, 15, 6, 2, 1, 15, 40, 32, 15, 6, 2, 1, 22, 73, 67, 37, 15, 6, 2, 1, 30, 120, 134, 79, 37, 15, 6, 2, 1, 42, 202, 255, 172, 85, 37, 15, 6, 2, 1, 56, 320, 470, 348, 187, 85, 37, 15, 6, 2, 1, 77, 511, 848, 697, 397, 194, 85, 37, 15, 6, 2, 1

Offset: 1

Vladeta Jovovic, Apr 23 2001

Multiset transformation of A000041. - R. J. Mathar, Apr 30 2017

Number of orderless twice-partitions of n of length k. A twice-partition of n is a choice of a partition of each part in a partition of n. The T(5,3) = 6 orderless twice-partitions: (3)(1)(1), (21)(1)(1), (111)(1)(1), (2)(2)(1), (2)(11)(1), (11)(11)(1). - Gus Wiseman, Mar 23 2018

			:  1;
:  2,   1;
:  3,   2,   1;
:  5,   6,   2,   1;
:  7,  11,   6,   2,  1;
: 11,  23,  15,   6,  2,  1;
: 15,  40,  32,  15,  6,  2,  1;
: 22,  73,  67,  37, 15,  6,  2, 1;
: 30, 120, 134,  79, 37, 15,  6, 2, 1;
: 42, 202, 255, 172, 85, 37, 15, 6, 2, 1;

Alois P. Heinz, Rows n = 1..141, flattened
Index entries for triangles generated by the Multiset Transformation

Row sums: A001970, first column: A000041.
T(2,n) gives A061261,
Cf. A063834, A119442, A273873, A285229, A289078, A289501, A299200, A299201.

b:= proc(n, i, p) option remember; `if`(p>n, 0, `if`(n=0, 1,
      `if`(min(i, p)<1, 0, add(b(n-i*j, i-1, p-j)*binomial(
       combinat[numbpart](i)+j-1, j), j=0..min(n/i, p)))))
    end:
T:= (n, k)-> b(n$2, k):
seq(seq(T(n, k), k=1..n), n=1..14);  # Alois P. Heinz, Apr 13 2017

b[n_, i_, p_] := b[n, i, p] = If[p > n, 0, If[n == 0, 1, If[Min[i, p] < 1, 0, Sum[b[n - i*j, i - 1, p - j]*Binomial[PartitionsP[i] + j - 1, j], {j, 0, Min[n/i, p]}]]]];
T[n_, k_] := b[n, n, k];
Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 17 2018, after Alois P. Heinz *)

A358830 Number of twice-partitions of n into partitions with all different lengths.

Original entry on oeis.org

1, 1, 2, 4, 9, 15, 31, 53, 105, 178, 330, 555, 1024, 1693, 2991, 5014, 8651, 14242, 24477, 39864, 67078, 109499, 181311, 292764, 483775, 774414, 1260016, 2016427, 3254327, 5162407, 8285796, 13074804, 20812682, 32733603, 51717463, 80904644, 127305773, 198134675, 309677802

Offset: 0

Gus Wiseman, Dec 03 2022

nonn

A twice-partition of n is a sequence of integer partitions, one of each part of an integer partition of n.

			The a(1) = 1 through a(5) = 15 twice-partitions:
  (1)  (2)   (3)      (4)       (5)
       (11)  (21)     (22)      (32)
             (111)    (31)      (41)
             (11)(1)  (211)     (221)
                      (1111)    (311)
                      (11)(2)   (2111)
                      (2)(11)   (11111)
                      (21)(1)   (21)(2)
                      (111)(1)  (22)(1)
                                (3)(11)
                                (31)(1)
                                (111)(2)
                                (211)(1)
                                (111)(11)
                                (1111)(1)

The version for set partitions is A007837.
For sums instead of lengths we have A271619.
For constant instead of distinct lengths we have A306319.
The case of distinct sums also is A358832.
The version for multiset partitions of integer partitions is A358836.
A063834 counts twice-partitions, strict A296122, row-sums of A321449.
A273873 counts strict trees.
Cf. A000009, A000219, A001970, A141199, A279375, A279785, A279790, A336342, A358334, A358831.

twiptn[n_]:=Join@@Table[Tuples[IntegerPartitions/@ptn],{ptn,IntegerPartitions[n]}];
Table[Length[Select[twiptn[n],UnsameQ@@Length/@#&]],{n,0,10}]

seq(n)={ local(Cache=Map());
  my(g=Vec(-1+1/prod(k=1, n, 1 - y*x^k + O(x*x^n))));
  my(F(m,r,b) = my(key=[m,r,b], z); if(!mapisdefined(Cache,key,&z),
  z = if(r<=0||m==0, r==0, self()(m-1, r, b) + sum(k=1, m, my(c=polcoef(g[m],k)); if(!bittest(b,k)&&c, c*self()(min(m,r-m), r-m, bitor(b, 1<Andrew Howroyd, Dec 31 2022

Terms a(26) and beyond from Andrew Howroyd, Dec 31 2022

A273866 Coefficients a(k,m) of polynomials a{k}(h) appearing in the product Product_{k >= 1} (1 - a{k}(h)x^k) = 1 - hx/(1-x).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 3, 5, 5, 3, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 9, 13, 13, 9, 4, 1, 1, 4, 10, 17, 20, 17, 10, 4, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 5, 16, 36, 57, 66, 57, 36, 16, 5, 1

Offset: 1

Gevorg Hmayakyan, Jun 01 2016

The a(k,m) form a table where each row has k-1 elements starting from 2 and the a(1,1) = 1.

			a{1}(h) = h,
a{2}(h) = h,
a{3}(h) = h^2 + h,
a{4}(h) = h^3 + h^2 + h,
a{5}(h) = h^4 + 2*h^3 + 2*h^2 + h,
a{6}(h) = h^5 + 2*h^4 + 2*h^3 + 2*h^2 + h,
a{7}(h) = h^6 + 3*h^5 + 5*h^4 + 5*h^3 + 3*h^2 + h,
a{8}(h) = h^7 + 3*h^6 + 6*h^5 + 7*h^4 + 6*h^3 + 3*h^2 + h,
a{9}(h) = h^8 + 4*h^7 + 9*h^6 + 13*h^5 + 13*h^4 + 9*h^3 + 4*h^2 + h
...
and the corresponding a(k,m) table is:
  1,
  1,
  1,  1,
  1,  1,  1,
  1,  2,  2,  1,
  1,  2,  2,  2,  1,
  1,  3,  5,  5,  3,  1,
  1,  3,  6,  7,  6,  3,  1,
  1,  4,  9, 13, 13,  9,  4,  1,
  ...
a(7,3) = 5 because there are six strict trees contributing positive one {{5,1},1}, {{4,2},1}, {{4,1},2}, {{3,2},2}, {4,{2,1}}, {{3,1},3} and there is one strict tree contributing negative one {4,2,1}. - _Gus Wiseman_, Nov 14 2016

Giedrius Alkauskas, One curious proof of Fermat's little theorem, arXiv:0801.0805 [math.NT], 2008.
Giedrius Alkauskas, A curious proof of Fermat's little theorem, Amer. Math. Monthly 116(4) (2009), 362-364.
H. Gingold, H. W. Gould, and Michael E. Mays, Power Product Expansions, Utilitas Mathematica 34 (1988), 143-161.
H. Gingold and A. Knopfmacher, Analytic properties of power product expansions, Canad. J. Math. 47 (1995), 1219-1239.
W. Lang, Recurrences for the general problem.

Cf. A196545, A220418, A220420, A273866, A273873, A279785, A290261, A290262, A290971, A295632.

with(ListTools), with(numtheory), with(combinat);
L := product(1-a[k]*x^k, k = 1 .. 600);
S := Flatten([seq(-h, i = 1 .. 100)]);
Sabs := Flatten([seq(i, i = 1 .. 100)]);
seq(assign(a[i] = solve(coeff(L, x^i) = `if`(is(i in Sabs), S[Search(i, Sabs)], 0), a[i])), i = 1 .. 20);
map(coeffs, [seq(simplify(a[i]), i = 1 .. 20)]);

strictrees[n_Integer?Positive]:=Prepend[Join@@Function[ptn,Tuples[strictrees/@ptn]]/@Select[IntegerPartitions[n],And[Length[#]>1,UnsameQ@@#]&],n];
Table[Sum[(-1)^(Count[tree,,{0,Infinity}]-1),{tree,Select[strictrees[n],Length[Flatten[{#}]]===m&]}],{n,1,9},{m,1,n-1/.(0->1)}] (* _Gus Wiseman, Nov 14 2016 *)
(* second program *)
A[m_, n_] :=
  A[m, n] =
   Which[m == 1, -h, m > n >= 1, 0, True,
    A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1]];
a[n_] := Expand[-A[n, n]];
a /@ Range[1, 25] (* Petros Hadjicostas, Oct 04 2019, courtesy of Jean-François Alcover *)

a(k,m) = a(k, k-m).
For prime p: Sum_{m = 1..p-1} a(p, m) = (2^p - 2)/p.
a{k}(h) satisfies Sum_{d|k} (1/d)*(a{k/d}(h))^d = ((h+1)^k - 1)/k. [Corrected by Petros Hadjicostas, Oct 04 2019]
For prime p: a{p}(h) = ((h+1)^p - h^p - 1)/p.
See A273873 for the definition of strict tree. Then a(n,m) = Sum_t (-1)^{v(t)-1} where the sum is over all strict trees of weight n with m leaves, and v(t) is the number of nodes in t (including the leaves, which are positive integers). See example 2 and the first Mathematica program. - Gus Wiseman, Nov 14 2016

A281119 Number of complete tree-factorizations of n >= 2.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 5, 1, 3, 1, 3, 1, 1, 1, 9, 1, 1, 2, 3, 1, 4, 1, 12, 1, 1, 1, 12, 1, 1, 1, 9, 1, 4, 1, 3, 3, 1, 1, 29, 1, 3, 1, 3, 1, 9, 1, 9, 1, 1, 1, 17, 1, 1, 3, 34, 1, 4, 1, 3, 1, 4, 1, 44, 1, 1, 3, 3, 1, 4, 1, 29, 5, 1, 1

Offset: 2

Gus Wiseman, Jan 15 2017

nonn

A tree-factorization of n>=2 is either (case 1) the number n or (case 2) a sequence of two or more tree-factorizations, one of each part of a weakly increasing factorization of n into factors greater than 1. A complete (or total) tree-factorization is a tree-factorization whose leaves are all prime numbers.

a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018

			The a(36)=12 complete tree-factorizations of 36 are:
(2(2(33))), (2(3(23))), (2(233)),   (3(2(23))),
(3(3(22))), (3(223)),   ((22)(33)), ((23)(23)),
(22(33)),   (23(23)),   (33(22)),   (2233).

Michael De Vlieger, Table of n, a(n) for n = 2..10000
A. Knopfmacher, M. Mays, Ordered and Unordered Factorizations of Integers, The Mathematica Journal 10(1), 2006.

Cf. A000311, A001055, A063834, A196545, A262673, A273873, A281113, A281118.

postfacs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[postfacs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
treefacs[n_]:=If[n<=1,{{}},Prepend[Join@@Function[q,Tuples[treefacs/@q]]/@DeleteCases[postfacs[n],{n}],n]];
Table[Length[Select[treefacs[n],FreeQ[#,_Integer?(!PrimeQ[#]&)]&]],{n,2,83}]

seq(n)={my(v=vector(n), w=vector(n)); v[1]=1; for(k=2, n, w[k]=v[k]+isprime(k); forstep(j=n\k*k, k, -k, my(i=j, e=0); while(i%k==0, i/=k; e++; v[j]+=w[k]^e*v[i]))); w[2..n]} \\ Andrew Howroyd, Nov 18 2018

a(p^n) = A196545(n) for prime p. - Andrew Howroyd, Nov 18 2018

A301342 Regular triangle where T(n,k) is the number of rooted identity trees with n nodes and k leaves.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 2, 0, 0, 0, 1, 4, 1, 0, 0, 0, 1, 6, 5, 0, 0, 0, 0, 1, 9, 13, 2, 0, 0, 0, 0, 1, 12, 28, 11, 0, 0, 0, 0, 0, 1, 16, 53, 40, 3, 0, 0, 0, 0, 0, 1, 20, 91, 109, 26, 0, 0, 0, 0, 0, 0, 1, 25, 146, 254, 116, 6, 0, 0, 0, 0, 0, 0, 1, 30, 223, 524, 387, 61, 0, 0, 0, 0, 0, 0, 0, 1, 36

Offset: 1

Gus Wiseman, Mar 19 2018

			Triangle begins:
1
1   0
1   0   0
1   1   0   0
1   2   0   0   0
1   4   1   0   0   0
1   6   5   0   0   0   0
1   9  13   2   0   0   0   0
1  12  28  11   0   0   0   0   0
1  16  53  40   3   0   0   0   0   0
1  20  91 109  26   0   0   0   0   0   0
1  25 146 254 116   6   0   0   0   0   0   0
1  30 223 524 387  61   0   0   0   0   0   0   0
The T(6,2) = 4 rooted identity trees: (((o(o)))), ((o((o)))), (o(((o)))), ((o)((o))).

A version with the zeroes removed is A055327.

Cf. A000081, A001190, A003238, A004111, A032305, A055277, A273873, A276625, A277098, A290689, A298118, A298422, A298426, A301343, A301344, A301345.

irut[n_]:=irut[n]=If[n===1,{{}},Join@@Function[c,Select[Union[Sort/@Tuples[irut/@c]],UnsameQ@@#&]]/@IntegerPartitions[n-1]];
Table[Length[Select[irut[n],Count[#,{},{-2}]===k&]],{n,8},{k,n}]

A295279 Number of strict tree-factorizations of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 4, 1, 2, 2, 3, 1, 4, 1, 4, 2, 2, 1, 10, 1, 2, 2, 4, 1, 8, 1, 6, 2, 2, 2, 12, 1, 2, 2, 10, 1, 8, 1, 4, 4, 2, 1, 26, 1, 4, 2, 4, 1, 10, 2, 10, 2, 2, 1, 28, 1, 2, 4, 12, 2, 8, 1, 4, 2, 8, 1, 44, 1, 2, 4, 4, 2, 8, 1, 26, 3, 2, 1

Offset: 1

Gus Wiseman, Nov 19 2017

nonn

A strict tree-factorization of n is either (case 1) the number n itself or (case 2) a set of two or more strict tree-factorizations, one of each factor in a factorization of n into distinct factors greater than one.

a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018

			The a(30) = 8 strict tree-factorizations are: 30, (2*3*5), (2*15), (2*(3*5)), (3*10), (3*(2*5)), (5*6), (5*(2*3)).
The a(36) = 12 strict tree-factorizations are: 36, (2*3*6), (2*3*(2*3)), (2*18), (2*(2*9)), (2*(3*6)), (2*(3*(2*3))), (3*12), (3*(2*6)), (3*(2*(2*3))), (3*(3*4)), (4*9).

Andrew Howroyd, Table of n, a(n) for n = 1..10000

Cf. A005804, A045778, A273873, A281113 A281118, A292504, A294786, A295281.

sfs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[sfs[n/d],Min@@#>d&]],{d,Rest[Divisors[n]]}]];
sft[n_]:=1+Total[Function[fac,Times@@sft/@fac]/@Select[sfs[n],Length[#]>1&]];
Array[sft,100]

seq(n)={my(v=vector(n), w=vector(n)); w[1]=v[1]=1; for(k=2, n, w[k]=v[k]+1; forstep(j=n\k*k, k, -k, v[j]+=w[k]*v[j/k])); w} \\ Andrew Howroyd, Nov 18 2018

a(product of n distinct primes) = A005804(n).
a(prime^n) = A273873(n).
Dirichlet g.f.: (Zeta(s) + Product_{n >= 2}(1 + a(n)/n^s))/2.

A298204 Number of unlabeled rooted trees with n nodes in which all outdegrees are either 0, 1, or 3.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 9, 16, 29, 55, 104, 200, 389, 763, 1507, 3002, 6010, 12102, 24484, 49751, 101475, 207723, 426542, 878451, 1813945, 3754918, 7790326, 16196629, 33739335, 70410401, 147187513, 308171861, 646188276, 1356847388, 2852809425, 6005542176

Offset: 1

Gus Wiseman, Jan 14 2018

nonn

			The a(7) = 9 trees: ((((((o)))))), ((((ooo)))), (((oo(o)))), ((oo((o)))), ((o(o)(o))), (oo(((o)))), (oo(ooo)), (o(o)((o))), ((o)(o)(o)).

Alois P. Heinz, Table of n, a(n) for n = 1..1000
Gus Wiseman, Illustration of the first 8 terms.

Cf. A000081, A000598, A001190, A001399, A004111, A014591, A032305, A273873, A292050, A295461, A298118, A298205.

b:= proc(n, i, v) option remember; `if`(n=0,
      `if`(v=0, 1, 0), `if`(i<1 or v<1 or n `if`(n<2, n, add(b(n-1$2, j), j=[1, 3])):
seq(a(n), n=1..40);  # Alois P. Heinz, Jan 30 2018

multing[n_,k_]:=Binomial[n+k-1,k];
a[n_]:=a[n]=If[n===1,1,Sum[Product[multing[a[x],Count[ptn,x]],{x,Union[ptn]}],{ptn,Select[IntegerPartitions[n-1],MemberQ[{1,3},Length[#]]&]}]];
Table[a[n],{n,40}]
(* Second program: *)
b[n_, i_, v_] := b[n, i, v] = If[n == 0,
     If[v == 0, 1, 0], If[i < 1 || v < 1 || n < v, 0,
     If[n == v, 1, Sum[Binomial[a[i] + j - 1, j]*
     b[n - i*j, i - 1, v - j], {j, 0, Min[n/i, v]}]]]];
a[n_] := If[n < 2, n, Sum[b[n - 1, n - 1, j], {j, {1, 3}}]];
Array[a, 40] (* Jean-François Alcover, May 10 2021, after Alois P. Heinz *)

cp's OEIS Frontend

A289078 Number of orderless same-trees of weight n.

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A289079 Number of orderless same-trees of weight n with all leaves equal to 1.

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A299203 Number of enriched p-trees whose multiset of leaves is the integer partition with Heinz number n.

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A061260 G.f.: Product_{k>=1} (1-y*x^k)^(-numbpart(k)), where numbpart(k) = number of partitions of k, cf. A000041.

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A358830 Number of twice-partitions of n into partitions with all different lengths.

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A273866 Coefficients a(k,m) of polynomials a{k}(h) appearing in the product Product_{k >= 1} (1 - a{k}(h)*x^k) = 1 - h*x/(1-x).

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A281119 Number of complete tree-factorizations of n >= 2.

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A273866 Coefficients a(k,m) of polynomials a{k}(h) appearing in the product Product_{k >= 1} (1 - a{k}(h)x^k) = 1 - hx/(1-x).