A182378
G.f. satisfies A(x) = 1 + x*cycle_index(Sym(7), A(x)).
Original entry on oeis.org
1, 1, 1, 2, 4, 9, 20, 48, 115, 285, 716, 1833, 4740, 12410, 32754, 87176, 233547, 629540, 1705809, 4644231, 12697500, 34848694, 95973026, 265142431, 734606478, 2040683413, 5682634446, 15859800889, 44355531103, 124290064228, 348904212741, 981082979409
Offset: 0
Cf.
A000081,
A036717,
A036718,
A036719,
A036720,
A036721,
A036722,
A244372,
A292553,
A292554,
A292555,
A292556.
-
b:= proc(n, i, t, k) option remember; `if`(n=0, 1,
`if`(i<1, 0, add(binomial(b((i-1)$2, k$2)+j-1, j)*
b(n-i*j, i-1, t-j, k), j=0..min(t, n/i))))
end:
a:= n-> `if`(n=0, 1, b(n-1$2, 7$2)):
seq(a(n), n=0..35); # Alois P. Heinz, Sep 20 2017
-
b[n_, i_, t_, k_] := b[n, i, t, k] = If[n == 0, 1, If[i < 1, 0, Sum[ Binomial[ b[i-1, i-1, k, k] + j - 1, j]*b[n-i*j, i-1, t-j, k], {j, 0, Min[t, n/i]}]]];
a[n_] := If[n == 0, 1, b[n-1, n-1, 7, 7]];
Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Jan 15 2018, after Alois P. Heinz *)
A299039
Number of rooted trees with 2n nodes where each node has at most n children.
Original entry on oeis.org
1, 1, 3, 17, 106, 693, 4690, 32754, 234746, 1719325, 12820920, 97039824, 743680508, 5759507657, 45006692668, 354425763797, 2809931206626, 22409524536076, 179655903886571, 1447023307374888, 11703779855021636, 95020085240320710, 774088021528328920
Offset: 0
a(2) = 3:
o o o
| | / \
o o o o
| / \ |
o o o o
|
o
-
b:= proc(n, i, t, k) option remember; `if`(n=0, 1,
`if`(i<1, 0, add(binomial(b((i-1)$2, k$2)+j-1, j)*
b(n-i*j, i-1, t-j, k), j=0..min(t, n/i))))
end:
a:= n-> `if`(n=0, 1, b(2*n-1$2, n$2)):
seq(a(n), n=0..25);
-
b[n_, i_, t_, k_] := b[n, i, t, k] = If[n == 0, 1, If[i < 1, 0, Sum[ Binomial[b[i - 1, i - 1, k, k] + j - 1, j]*b[n - i*j, i - 1, t - j, k], {j, 0, Min[t, n/i]}]]];
a[n_] := If[n == 0, 1, b[2n - 1, 2n - 1, n, n]];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jun 04 2018, from Maple *)
A127119
Triangle read by rows: T(n,k) = number of endofunctions on a set with n elements, where the maximum indegree is k.
Original entry on oeis.org
1, 2, 1, 3, 3, 1, 5, 10, 3, 1, 7, 24, 12, 3, 1, 11, 64, 39, 12, 3, 1, 15, 149, 122, 41, 12, 3, 1, 22, 366, 368, 138, 41, 12, 3, 1, 30, 857, 1092, 439, 140, 41, 12, 3, 1, 42, 2050, 3179, 1395, 455, 140, 41, 12, 3, 1, 56, 4828, 9160, 4326, 1467, 457, 140, 41, 12, 3, 1
Offset: 1
For n = 3, the 7 endofunctions are (1,2,3) -> (1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,2,3), (1,3,2) and (2,3,1). In the first, node 1 has indegree 3, the next 3 have node 1 with indegree 2 and the final 3 are permutations, each node having indegree 1. So row 3 of the triangle is 3,3,1.
The triangle starts:
1
2 1
3 3 1
5 10 3 1
7 24 12 3 1
-
\\ Here R(n,k) gives column k of A299038 as series.
EulerT(v)={Vec(exp(x*Ser(dirmul(v, vector(#v, n, 1/n))))-1, -#v)}
MSetUptoK(g, k)={my(n=serprec(g,x)); polcoef(if(k==0, 1, exp( sum(i=1, k, (y^i + O(y*y^k))*subst(g + O(x*x^(n\i)), x, x^i)/i )))/(1 - y) + O(y*y^k), k, y) + O(x^n)}
CIK(p,n)={sum(d=1, n, eulerphi(d)/d*log(subst(1/(1+O(x*x^(n\d))-p), x, x^d)))}
R(n,k)={my(p=O(x)); for(n=1, n, p=x*MSetUptoK(p, k)); p}
F(n)={my(M=Mat(vector(n, k, EulerT(Vec(CIK(x*MSetUptoK(R(n,k), k-1), n)))~))); M-matconcat([vectorv(#M), M[, 1..n-1]])}
{ my(A=F(12)); for(n=1, #A, print(A[n, 1..n])) } \\ Andrew Howroyd, Feb 21 2020
A352460
Triangle read by rows: T(n,k), 2 <= k < n is the number of n-element k-ary unlabeled rooted trees where a subtree consisting of h + 1 nodes has exactly min{h,k} subtrees.
Original entry on oeis.org
1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 4, 3, 2, 1, 1, 5, 4, 3, 2, 1, 1, 9, 6, 5, 3, 2, 1, 1, 13, 10, 6, 5, 3, 2, 1, 1, 23, 15, 10, 7, 5, 3, 2, 1, 1, 35, 24, 14, 10, 7, 5, 3, 2, 1, 1, 61, 39, 23, 14, 11, 7, 5, 3, 2, 1, 1, 98, 63, 34, 21, 14, 11, 7, 5, 3, 2, 1, 1
Offset: 3
Triangle begins:
1;
1, 1;
2, 1, 1;
2, 2, 1, 1;
4, 3, 2, 1, 1;
5, 4, 3, 2, 1, 1;
9, 6, 5, 3, 2, 1, 1;
13, 10, 6, 5, 3, 2, 1, 1;
23, 15, 10, 7, 5, 3, 2, 1, 1;
35, 24, 14, 10, 7, 5, 3, 2, 1, 1;
61, 39, 23, 14, 11, 7, 5, 3, 2, 1, 1;
98, 63, 34, 21, 14, 11, 7, 5, 3, 2, 1, 1;
In particular, the rooted trees counted in the first three rows of the triangle are shown by using the Hasse diagram as follows:
---------
o o
\ /
o
----------------------
o |
| |
o o | o o o
\ / | \ | /
o | o
------------------------------------------------------
o o o o | o |
\ / | | | | |
o o o o | o o o | o o o o
\ / \ / | \ | / | \ \ / /
o o | o | o
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