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 1-8 of 8 results.

A054357 Number of unlabeled 2-ary cacti having n polygons. Also number of bicolored plane trees with n edges.

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 28, 63, 190, 546, 1708, 5346, 17428, 57148, 191280, 646363, 2210670, 7626166, 26538292, 93013854, 328215300, 1165060668, 4158330416, 14915635378, 53746119972, 194477856100, 706437056648, 2575316704200, 9419571138368
Offset: 0

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Keywords

Comments

a(n) = the number of inequivalent non-crossing partitions of n points (equally spaced) on a circle, under rotations of the circle. This may be considered the number of non-crossing partitions of n unlabeled points on a circle, so this sequence has the same relation to the Catalan numbers (A000108) as the number of partitions of an integer (A000041) has to the Bell numbers (A000110). - Len Smiley, Sep 06 2005

Crossrefs

Column k=2 of A303912.
Row sums of A209805.

Programs

  • Mathematica
    a[n_] := If[n == 0, 1, (Binomial[2*n, n]/(n + 1) + DivisorSum[n, Binomial[2*#, #]*EulerPhi[n/#]*Boole[# < n] & ])/n]; Table[a[n], {n, 0, 28}] (* Jean-François Alcover, Jul 17 2017 *)
  • PARI
    a(n)=if(n==0, 1, (binomial(2*n, n)/(n + 1) + sumdiv(n, d, binomial(2*d, d)*eulerphi(n/d)*(dIndranil Ghosh, Jul 17 2017
    
  • PARI
    a(n) = if(n==0, 1, sumdiv(n, d, eulerphi(n/d)*binomial(2*d, d))/n - binomial(2*n, n)/(n+1)) \\ Andrew Howroyd, May 02 2018
    
  • Python
    from sympy import binomial, divisors, totient
    def a(n): return 1 if n==0 else (binomial(2*n, n)//(n + 1) + sum(binomial(2*d, d)*totient(n//d)*(dIndranil Ghosh, Jul 17 2017

Formula

a(n) = (1/n)*(Sum_{d|n} phi(n/d)*binomial(2*d, d)) - binomial(2*n, n)/(n+1) for n > 0. - Andrew Howroyd, May 02 2018
a(n) ~ 2^(2*n) / (sqrt(Pi) * n^(5/2)). - Vaclav Kotesovec, Jul 17 2017

Extensions

More terms from Len Smiley, Sep 06 2005
More terms from Vladeta Jovovic, Oct 04 2007

A303913 Array read by antidiagonals: T(n,k) is the number of (planar) unlabeled asymmetric k-ary cacti having n polygons.

Original entry on oeis.org

1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 3, 2, 0, 1, 1, 0, 6, 10, 8, 0, 1, 1, 0, 10, 28, 54, 18, 0, 1, 1, 0, 15, 60, 193, 222, 61, 0, 1, 1, 0, 21, 110, 505, 1140, 1107, 170, 0, 1, 1, 0, 28, 182, 1095, 3876, 7688, 5346, 538, 0, 1, 1, 0, 36, 280, 2093, 10326, 33125, 52364, 27399, 1654, 0
Offset: 0

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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. - Andrew Howroyd, Feb 18 2020

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  | 0   0     0      0       0        0        0         0 ...
3  | 0   1     3      6      10       15       21        28 ...
4  | 0   2    10     28      60      110      182       280 ...
5  | 0   8    54    193     505     1095     2093      3654 ...
6  | 0  18   222   1140    3876    10326    23394     47208 ...
7  | 0  61  1107   7688   33125   107056   285383    662620 ...
8  | 0 170  5346  52364  290700  1149126  3621150   9702008 ...
9  | 0 538 27399 373560 2661100 12845166 47813367 147765409 ...
...
		

Crossrefs

Columns k=2..7 are A054358, A054422, A052395, A054364, A054367, A054370.

Programs

  • Mathematica
    T[0, _] = 1;
    T[n_, k_] := DivisorSum[n, MoebiusMu[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, moebius(n/d)*binomial(k*d, d))/n - (k-1)*binomial(k*n, n)/((k-1)*n+1))}

Formula

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

A361236 Array read by antidiagonals: T(n,k) is the number of noncrossing k-gonal cacti with n polygons up to rotation.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 5, 11, 1, 1, 1, 1, 8, 33, 49, 1, 1, 1, 1, 9, 63, 230, 204, 1, 1, 1, 1, 12, 105, 664, 1827, 984, 1, 1, 1, 1, 13, 159, 1419, 7462, 15466, 4807, 1, 1, 1, 1, 16, 221, 2637, 21085, 90896, 137085, 24739, 1
Offset: 0

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Author

Andrew Howroyd, Mar 05 2023

Keywords

Comments

The number of noncrossing k-gonal cacti is given by column 2*(k-1) of A070914. This sequence enumerates these cacti with rotations being considered equivalent.
Equivalently, T(n,k) is the number of connected acyclic k-uniform noncrossing antichains with n blocks covering (k-1)*n+1 nodes where the intersection of two blocks is at most 1 node modulo cyclic rotation of the nodes.
Noncrossing trees correspond to the case of k = 2.

Examples

			=====================================================
n\k | 1     2       3        4        5         6 ...
----+------------------------------------------------
  0 | 1     1       1        1        1         1 ...
  1 | 1     1       1        1        1         1 ...
  2 | 1     1       1        1        1         1 ...
  3 | 1     4       5        8        9        12 ...
  4 | 1    11      33       63      105       159 ...
  5 | 1    49     230      664     1419      2637 ...
  6 | 1   204    1827     7462    21085     48048 ...
  7 | 1   984   15466    90896   334707    941100 ...
  8 | 1  4807  137085  1159587  5579961  19354687 ...
  9 | 1 24739 1260545 15369761 96589350 413533260 ...
  ...
		

Crossrefs

Columns k=1..4 are A000012, A296532, A361237, A361238.
Row n=3 is A042948.

Programs

  • PARI
    \\ here u is Fuss-Catalan sequence with p = 2*k-1.
    u(n,k,r) = {r*binomial(n*(2*k-1) + r, n)/(n*(2*k-1) + r)}
    T(n,k) = if(n==0, 1, u(n, k, 1)/((k-1)*n+1) + sumdiv(gcd(k,n-1), d, if(d>1, eulerphi(d)*u((n-1)/d, k, 2*k/d)/k)))

Formula

T(0,k) = T(1,k) = T(2,k) = 1.

A052393 Number of unlabeled ternary cacti having n triangles.

Original entry on oeis.org

1, 1, 3, 6, 19, 57, 258, 1110, 5475, 27429, 143379, 764970, 4173906, 23140816, 130205922, 741650802, 4270593219, 24825326199, 145535320383, 859627488966, 5112006997539, 30586307211945, 184023410798910, 1112800162657902, 6760426635625170
Offset: 0

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Crossrefs

Column k=3 of A303912.

Programs

  • Mathematica
    a[n_] := If[n == 0, 1, (Binomial[3*n, n]/(2*n+1) + DivisorSum[n, Binomial[ 3*#, #]*EulerPhi[n/#]*Boole[# < n]&])/n]; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Jul 17 2017 *)
  • PARI
    a(n) = if(n==0, 1, sumdiv(n, d, eulerphi(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} phi(n/d)*binomial(3*d, d)) - 2*binomial(3*n, n)/(2*n+1) for n > 0. - Andrew Howroyd, May 02 2018
a(n) ~ 3^(3*n + 1/2) / (sqrt(Pi) * 2^(2*n + 2) * n^(5/2)). - Vaclav Kotesovec, Jul 17 2017

Extensions

More terms from Jean-François Alcover, Jul 17 2017

A052394 Number of unlabeled 4-ary cacti having n polygons.

Original entry on oeis.org

1, 1, 4, 10, 44, 197, 1228, 7692, 52828, 373636, 2735952, 20506258, 156922676, 1221179926, 9642496488, 77092885016, 623121750844, 5085013101164, 41850600967984, 347060754685888, 2897800158952304, 24344668688424333, 205667187527660076, 1746375819789491996, 14898241072028602276
Offset: 0

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Crossrefs

Column k=4 of A303912.

Programs

  • Mathematica
    a[n_] := If[n == 0, 1, (Binomial[4*n, n]/(3*n + 1) + DivisorSum[n, Binomial[4*#, #]*EulerPhi[n/#]*Boole[# < n] & ])/n]; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Jul 17 2017 *)
  • PARI
    a(n) = if(n==0, 1, sumdiv(n, d, eulerphi(n/d)*binomial(4*d, d))/n - 3*binomial(4*n, n)/(3*n+1)) \\ Andrew Howroyd, May 02 2018

Formula

a(n) = (1/n)*(Sum_{d|n} phi(n/d)*binomial(4*d, d)) - 3*binomial(4*n, n)/(3*n+1) for n > 0. - Andrew Howroyd, May 02 2018
a(n) ~ 2^(8*n + 1/2) / (sqrt(Pi) * n^(5/2) * 3^(3*n + 3/2)). - Vaclav Kotesovec, Jul 17 2017

Extensions

More terms from Jean-François Alcover, Jul 17 2017

A054363 Number of unlabeled 5-ary cacti having n polygons.

Original entry on oeis.org

1, 1, 5, 15, 85, 510, 4051, 33130, 291925, 2661255, 25059670, 241724380, 2379912355, 23833198140, 242173108050, 2491817151160, 25921371278805, 272256630756265, 2884054952424115, 30784716141936525, 330853932861650870, 3577823885433087690, 38907658120970944700
Offset: 0

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Crossrefs

Column k=5 of A303912.

Programs

  • Mathematica
    a[n_] := If[n == 0, 1, (Binomial[5*n, n]/(4*n + 1) + DivisorSum[n, Binomial[5*#, #]*EulerPhi[n/#]*Boole[# < n] & ])/n]; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Jul 17 2017 *)
  • PARI
    a(n) = if(n==0, 1, sumdiv(n, d, eulerphi(n/d)*binomial(5*d, d))/n - 4*binomial(5*n, n)/(4*n+1)) \\ Andrew Howroyd, May 02 2018

Formula

a(n) = (1/n)*(Sum_{d|n} phi(n/d)*binomial(5*d, d)) - 4*binomial(5*n, n)/(4*n+1) for n > 0. - Andrew Howroyd, May 02 2018
a(n) ~ 5^(5*n + 1/2) / (sqrt(Pi) * n^(5/2) * 2^(8*n + 7/2)). - Vaclav Kotesovec, Jul 17 2017

Extensions

More terms from Jean-François Alcover, Jul 17 2017

A054366 Number of unlabeled 6-ary cacti having n polygons.

Original entry on oeis.org

1, 1, 6, 21, 146, 1101, 10632, 107062, 1151802, 12845442, 147845706, 1743640908, 20988257544, 256987965379, 3192893716320, 40171643847696, 510997002280522, 6563060603543658, 85017387536789916, 1109744672540225367, 14585261039005676046
Offset: 0

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Crossrefs

Column k=6 of A303912.

Programs

  • Mathematica
    a[n_] := If[n == 0, 1, (Binomial[6*n, n]/(5 n + 1) + DivisorSum[n, Binomial[6*#, #]*EulerPhi[n/#]*Boole[# < n] & ])/n]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jul 17 2017 *)
  • PARI
    a(n) = if(n==0, 1, sumdiv(n, d, eulerphi(n/d)*binomial(6*d, d))/n - 5*binomial(6*n, n)/(5*n+1)) \\ Andrew Howroyd, May 02 2018

Formula

a(n) = (1/n)*(Sum_{d|n} phi(n/d)*binomial(6*d, d)) - 5*binomial(6*n, n)/(5*n+1) for n > 0. - Andrew Howroyd, May 02 2018
a(n) ~ sqrt(3) * 6^(6*n) / (sqrt(Pi) * n^(5/2) * 5^(5*n + 3/2)). - Vaclav Kotesovec, Jul 17 2017

Extensions

More terms from Jean-François Alcover, Jul 17 2017

A054369 Number of unlabeled 7-ary cacti having n polygons.

Original entry on oeis.org

1, 1, 7, 28, 231, 2100, 23884, 285390, 3626295, 47813815, 650367788, 9066061200, 128987761308, 1866877313448, 27417589615234, 407771633434368, 6131640607962135, 93096368350684727, 1425633586192690945, 21998953427963954554, 341803227016091180620
Offset: 0

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Crossrefs

Column k=7 of A303912.

Programs

  • Mathematica
    a[n_] := If[n == 0, 1, (Binomial[7*n, n]/(6 n + 1) + DivisorSum[n, Binomial[7*#, #]*EulerPhi[n/#]*Boole[# < n] & ])/n]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jul 17 2017 *)
  • PARI
    a(n) = if(n==0, 1, sumdiv(n, d, eulerphi(n/d)*binomial(7*d, d))/n - 6*binomial(7*n, n)/(6*n+1)) \\ Andrew Howroyd, May 02 2018

Formula

a(n) = (1/n)*(Sum_{d|n} phi(n/d)*binomial(7*d, d)) - 6*binomial(7*n, n)/(6*n+1) for n > 0. - Andrew Howroyd, May 02 2018
a(n) ~ 7^(7*n + 1/2) / (2 * sqrt(3*Pi) * n^(5/2) * 6^(6*n + 1)). - Vaclav Kotesovec, Jul 17 2017

Extensions

More terms from Jean-François Alcover, Jul 17 2017
Showing 1-8 of 8 results.