cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A303912 Array read by antidiagonals: T(n,k) is the number of (planar) unlabeled k-ary cacti having n polygons.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 4, 6, 6, 1, 1, 1, 5, 10, 19, 10, 1, 1, 1, 6, 15, 44, 57, 28, 1, 1, 1, 7, 21, 85, 197, 258, 63, 1, 1, 1, 8, 28, 146, 510, 1228, 1110, 190, 1, 1, 1, 9, 36, 231, 1101, 4051, 7692, 5475, 546, 1, 1, 1, 10, 45, 344, 2100, 10632, 33130, 52828, 27429, 1708, 1
Offset: 0

Views

Author

Andrew Howroyd, May 02 2018

Keywords

Comments

A k-ary cactus is a planar k-gonal cactus with vertices on each polygon numbered 1..k counterclockwise with shared vertices having the same number. In total there are always exactly k ways to number a given cactus since all polygons are connected. See the reference for a precise definition.

Examples

			Array begins:
===============================================================
n\k| 1   2     3      4       5        6        7         8
---+-----------------------------------------------------------
0  | 1   1     1      1       1        1        1         1 ...
1  | 1   1     1      1       1        1        1         1 ...
2  | 1   2     3      4       5        6        7         8 ...
3  | 1   3     6     10      15       21       28        36 ...
4  | 1   6    19     44      85      146      231       344 ...
5  | 1  10    57    197     510     1101     2100      3662 ...
6  | 1  28   258   1228    4051    10632    23884     47944 ...
7  | 1  63  1110   7692   33130   107062   285390    662628 ...
8  | 1 190  5475  52828  291925  1151802  3626295   9711032 ...
9  | 1 546 27429 373636 2661255 12845442 47813815 147766089 ...
...
		

Crossrefs

Programs

  • Mathematica
    T[0, _] = 1;
    T[n_, k_] := DivisorSum[n, EulerPhi[n/#] Binomial[k #, #]&]/n - (k-1) Binomial[n k, n]/((k-1) n + 1);
    Table[T[n-k, k], {n, 0, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, May 22 2018 *)
  • PARI
    T(n,k)={if(n==0, 1, sumdiv(n, d, eulerphi(n/d)*binomial(k*d, d))/n - (k-1)*binomial(k*n, n)/((k-1)*n+1))}

Formula

T(n,k) = (Sum_{d|n} phi(n/d)*binomial(k*d, d))/n - (k-1)*binomial(k*n, n)/((k-1)*n+1) for n > 0.
T(n,k) ~ A070914(n,k-1)/n for fixed k > 1.

A054423 Number of unlabeled 3-gonal cacti having n triangles.

Original entry on oeis.org

1, 1, 1, 2, 7, 19, 86, 372, 1825, 9143, 47801, 254990, 1391302, 7713642, 43401974, 247216934, 1423531255, 8275108733, 48511773461, 286542497274, 1704002332513, 10195435737315, 61341136938138, 370933387552634, 2253475545208390, 13748639775492766, 84211761819147696
Offset: 0

Views

Author

Simon Plouffe, Mar 15 2000

Keywords

Comments

Also, the number of noncrossing partitions up to rotation composed of n blocks of size 3. - Andrew Howroyd, May 04 2018

Crossrefs

Column k=3 of A303694.

Programs

  • Maple
    with(combinat): with(numtheory): m := 3: for p from 1 to 40 do s1 := 0: s2 := 0:
    for d from 1 to p do if p mod d = 0 then s1 := s1+phi(p/d)*binomial(m*d,d) fi: od:
    for d from 1 to p-1 do if gcd(m, p-1) mod d = 0 then s2 := s2+phi(d)*binomial((p*m)/d, (p-1)/d) fi: od:
    printf(`%d,`, (s1+s2)/(m*p)-binomial(m*p,p)/(p*(m-1)+1)) od: # James Sellers, Mar 17 2000
  • Mathematica
    a[0] = 1;
    a[n_] := (DivisorSum[n, EulerPhi[n/#] Binomial[3 #, #]&] + DivisorSum[GCD[n - 1, 3], EulerPhi[#] Binomial[3n/#, (n-1)/#]&])/(3n) - Binomial[3n, n]/ (2n + 1);
    Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Jul 02 2018, after Andrew Howroyd *)
  • PARI
    a(n) = {if(n==0, 1, (sumdiv(n, d, eulerphi(n/d)*binomial(3*d, d)) + sumdiv(gcd(n-1, 3), d, eulerphi(d)*binomial(3*n/d, (n-1)/d)))/(3*n) - binomial(3*n, n)/(2*n+1))} \\ Andrew Howroyd, May 04 2018

Formula

a(n) = ((Sum_{d|n} phi(n/d)*binomial(3*d, d)) + (Sum_{d|gcd(n-1, 3)} phi(d)*binomial(3*n/d, (n-1)/d)))/(3*n) - binomial(3*n, n)/(2*n+1) for n > 0. - Andrew Howroyd, May 04 2018
a(n) ~ 3^(3*n - 1/2) / (sqrt(Pi) * n^(5/2) * 2^(2*n + 2)). - Vaclav Kotesovec, Jun 01 2022

Extensions

More terms from James Sellers, Mar 17 2000
Terms a(24) and beyond from Andrew Howroyd, May 04 2018

A054422 Number of unlabeled asymmetric ternary cacti having n triangles.

Original entry on oeis.org

1, 1, 0, 3, 10, 54, 222, 1107, 5346, 27399, 142770, 764967, 4170672, 23140813, 130189302, 741650172, 4270501218, 24825326196, 145534796520, 859627488963, 5112003992610, 30586307195304, 184023393204654, 1112800162657899
Offset: 0

Views

Author

Simon Plouffe, Mar 15 2000

Keywords

Crossrefs

Column k=3 of A303913.

Programs

  • Mathematica
    a[0] = 1; a[n_] := (1/n) Sum[MoebiusMu[n/d] Binomial[3d, d], {d, Divisors[n] } ] - 2 Binomial[3n, n]/(2n + 1);
    Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Jul 24 2018, after Andrew Howroyd *)
  • PARI
    a(n) = if(n==0, 1, sumdiv(n, d, moebius(n/d)*binomial(3*d, d))/n - 2*binomial(3*n, n)/(2*n+1)) \\ Andrew Howroyd, May 02 2018

Formula

a(n) = (1/n)*(Sum_{d|n} mu(n/d)*binomial(3*d, d)) - 2*binomial(3*n, n)/(2*n+1) for n > 0. - Andrew Howroyd, May 02 2018

Extensions

More terms from James Sellers, Mar 16 2000
Showing 1-3 of 3 results.