A230388 -id:A230388 - cp's OEIS frontend

A235340 a(n) = 10binomial(11n+10,n)/(11*n+10).

1, 10, 155, 2870, 58565, 1270752, 28765650, 671650110, 16057800980, 391139588190, 9672348219898, 242182964452000, 6127720969229265, 156431295179478200, 4024231652469275640, 104218796026870015374, 2714941275486017847825

Offset: 0

Tim Fulford, Jan 06 2014

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p=11, r=10.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
J-C. Aval, Multivariate Fuss-Catalan Numbers, arXiv:0711.0906 [math.CO], 2007; Discrete Math., 308 (2008), 4660-4669.
Thomas A. Dowling, Catalan Numbers Chapter 7
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
Wojciech Mlotkowski, Fuss-Catalan Numbers in Noncommutative Probability, Docum. Mathm. 15: 939-955.

Cf. A130564, A230388, A234868, A234869, A234870, A234871, A234872, A234873, A235338, A235339.

[10*Binomial(11*n+10, n)/(11*n+10): n in [0..30]];

Table[10 Binomial[11 n + 10, n]/(11 n + 10), {n, 0, 30}]

a(n) = 10*binomial(11*n+10,n)/(11*n+10);

{a(n)=local(B=1); for(i=0, n, B=(1+x*B^(11/10))^10+x*O(x^n)); polcoeff(B, n)}

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, here p=11, r=10.
From Wolfdieter Lang, Feb 15 2024: (Start)
a(n) = binomial(11*n + 9, n + 1)/(10*n + 9) which is instance k = 10 of c(k, n+1) given in a comment in A130564.
x*B(x), with the g.f. above named B(x), is the compositional inverse of y*(1 - y)^10, hence  B(x)*(1 - x*B(x))^10 = 1.
G.f.: 11F10([10..20]/11, [11..20]/10; (11^11/10^10)*x) =  (10/(11*x))*(1 - 10F9([-1,1,2,3,4,5,6,7,8,9]/11, [1,2,3,4,5,6,7,8,9]/10; (11^11/10^10)*x)).
 (End)

A192893 Number of symmetric 11-ary factorizations of the n-cycle (1,2...n).

Original entry on oeis.org

1, 1, 1, 6, 11, 81, 176, 1406, 3311, 27636, 68211, 585162, 1489488, 13019909, 33870540, 300138696, 793542167, 7105216833, 19022318084, 171717015470, 464333035881, 4219267597578, 11502251937176, 105085831400550, 288417894029200, 2647012241261856, 7306488667126803

Offset: 0

N. J. A. Sloane, Jul 12 2011

nonn

The six sequences displayed in Table 1 of the Bousquet-Lamathe reference are A047749, A143546, A143547, A143554, this sequence, and A192894. From this one should be able to guess a g.f.

Number of achiral noncrossing partitions composed of n blocks of size 11. - Andrew Howroyd, Feb 08 2024

Andrew Howroyd, Table of n, a(n) for n = 0..500
Michel Bousquet and Cédric Lamathe, On symmetric structures of order two, Discrete Math. Theor. Comput. Sci. 10 (2008), 153-176. See Table 1.

Column k=11 of A369929 and k=12 of A370062.

Cf. A143048.

a(n)={my(m=n\2, p=5*(n%2)+1); binomial(11*m+p-1, m)*p/(10*m+p)} \\ Andrew Howroyd, Feb 08 2024

From Andrew Howroyd, Feb 08 2024: (Start)
a(2n) = binomial(11*n,n)/(10*n+1); a(2n+1) = binomial(11*n+5,n)*6/(10*n+6).
G.f. A(x) satisfies A(x) = 1 + x*A(x)^6*A(-x)^5. (End)
From Seiichi Manyama, Jul 07 2025: (Start)
G.f. A(x) satisfies A(x)*A(-x) = (A(x) + A(-x))/2 = G(x^2), where G(x) = 1 + x*G(x)^11 is the g.f. of A230388.
a(0) = 1; a(n) = Sum_{x_1, x_2, ..., x_6>=0 and x_1+2*(x_2+x_3+...+x_6)=n-1} a(x_1) * Product_{k=2..6} a(2*x_k). (End)
a(0) = 1; a(n) = Sum_{x_1, x_2, ..., x_11>=0 and x_1+x_2+...+x_11=n-1} (-1)^(x_1+x_2+x_3+x_4+x_5) * Product_{k=1..11} a(x_k). - Seiichi Manyama, Jul 09 2025

a(11) onwards from Andrew Howroyd, Jan 26 2024

a(0)=1 prepended by Andrew Howroyd, Feb 08 2024

A235534 a(n) = binomial(6n, 2n) / (4*n + 1).

Original entry on oeis.org

1, 3, 55, 1428, 43263, 1430715, 50067108, 1822766520, 68328754959, 2619631042665, 102240109897695, 4048514844039120, 162250238001816900, 6568517413771094628, 268225186597703313816, 11034966795189838872624, 456949965738717944767791

Offset: 0

Bruno Berselli, Jan 12 2014

This is the case l=4, k=2 of binomial((l+k)*n,k*n)/((l*n+1)/gcd(k,l*n+1)), see Theorem 1.1 in Zhi-Wei Sun's paper.

First bisection of A001764.

Zhi-Wei Sun, On Divisibility Of Binomial Coefficients, Journal of the Australian Mathematical Society 93 (2012), p. 189-201.

Cf. similar sequences generated by binomial((l+k)*n,k*n)/(l*n+1), where l is divisible by all the factors of k: A000108 (l=1, k=1), A001764 (l=2, k=1), A002293 (l=3, k=1), A002294 (l=4, k=1), A002295 (l=5, k=1), A002296 (l=6, k=1), A007556 (l=7, k=1), A062994 (l=8, k=1), A059968 (l=9, k=1), A230388 (l=10, k=1), A048990 (l=2, k=2), this sequence (l=4, k=2), A235536 (l=6, k=2), A187357 (l=3, k=3), A235535 (l=6, k=3).

l:=4; k:=2; [Binomial((l+k)*n,k*n)/(l*n+1): n in [0..20]]; /* where l is divisible by all the prime factors of k */

Table[Binomial[6 n, 2 n]/(4 n + 1), {n, 0, 20}]

a(n) = A047749(4*n-2) for n>0.
From Ilya Gutkovskiy, Jun 21 2018: (Start)
G.f.: 4F3(1/6,1/3,2/3,5/6; 1/2,3/4,5/4; 729*x/16).
a(n) ~ 3^(6*n+1/2)/(sqrt(Pi)*2^(4*n+7/2)*n^(3/2)). (End)

A235535 a(n) = binomial(9n, 3n) / (6*n + 1).

Original entry on oeis.org

1, 12, 1428, 246675, 50067108, 11124755664, 2619631042665, 642312451217745, 162250238001816900, 41932353590942745504, 11034966795189838872624, 2946924270225408943665279, 796607831560617902288322405, 217550867863011281855594752680

Offset: 0

Bruno Berselli, Jan 12 2014

This is the case l=6, k=3 of binomial((l+k)*n,k*n)/((l*n+1)/gcd(k,l*n+1)), see Theorem 1.1 in Zhi-Wei Sun's paper.

Also, the sequence follows A002296 and A235536, namely binomial(7*n,n)/(6*n+1) and binomial(8*n,2*n)/(6*n+1); naturally, even binomial(10*n,4*n)/(6*n+1) is always integer.

Robert Israel, Table of n, a(n) for n = 0..403
Zhi-Wei Sun, On Divisibility Of Binomial Coefficients, Journal of the Australian Mathematical Society 93 (2012), p. 189-201.

Cf. similar sequences generated by binomial((l+k)*n,k*n)/(l*n+1), where l is divisible by all the factors of k: A000108 (l=1, k=1), A001764 (l=2, k=1), A002293 (l=3, k=1), A002294 (l=4, k=1), A002295 (l=5, k=1), A002296 (l=6, k=1), A007556 (l=7, k=1), A062994 (l=8, k=1), A059968 (l=9, k=1), A230388 (l=10, k=1), A048990 (l=2, k=2), A235534 (l=4, k=2), A235536 (l=6, k=2), A187357 (l=3, k=3), this sequence (l=6, k=3).

l:=6; k:=3; [Binomial((l+k)*n,k*n)/(l*n+1): n in [0..20]]; /* here l is divisible by all the prime factors of k */

seq(binomial(9*n,3*n)/(6*n+1), n=0..30); # Robert Israel, Feb 15 2021

Table[Binomial[9 n, 3 n]/(6 n + 1), {n, 0, 20}]

a(n) = A001764(3*n) = A047749(6*n).
From Ilya Gutkovskiy, Jun 21 2018: (Start)
G.f.: 6F5(1/9,2/9,4/9,5/9,7/9,8/9; 1/3,1/2,2/3,5/6,7/6; 19683*x/64).
a(n) ~ 3^(9*n-1)/(sqrt(Pi)*4^(3*n+1)*n^(3/2)). (End)
D-finite with recurrence 8*(6*n + 5)*(2*n + 1)*(n + 1)*(3*n + 2)*(3*n + 1)*(6*n + 7)*a(n + 1) = 3*(9*n + 8)*(9*n + 7)*(9*n + 5)*(9*n + 4)*(9*n + 2)*(9*n + 1)*a(n). - Robert Israel, Feb 15 2021

A235338 a(n) = 8binomial(11n+8,n)/(11*n+8).

Original entry on oeis.org

1, 8, 116, 2080, 41650, 892552, 20027112, 464550336, 11050084695, 268070745800, 6607118937848, 164979021222400, 4164615224071926, 106105019316578800, 2724883054841727200, 70462458864489354624, 1833143662625459289495

Offset: 0

Tim Fulford, Jan 06 2014

nonn

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p=11, r=8.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
J-C. Aval, Multivariate Fuss-Catalan Numbers, arXiv:0711.0906 [math.CO], 2007; Discrete Math., 308 (2008), 4660-4669.
Thomas A. Dowling, Catalan Numbers Chapter 7 [broken link]
Wojciech Mlotkowski, Fuss-Catalan Numbers in Noncommutative Probability, Docum. Mathm. 15: 939-955.

Cf. A230388, A234868, A234869, A234870, A234871, A234872, A234873, A235339, A235340.

[8*Binomial(11*n+8, n)/(11*n+8): n in [0..30]];

Table[8 Binomial[11 n + 8, n]/(11 n + 8), {n, 0, 30}]

PARI
```
a(n) = 8*binomial(11*n+8, n)/(11*n+8);
```

{a(n)=local(B=1); for(i=0, n, B=(1+x*B^(11/8))^8+x*O(x^n)); polcoeff(B, n)}

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, here p=11, r=8.

A235536 a(n) = binomial(8n, 2n) / (6*n + 1).

Original entry on oeis.org

1, 4, 140, 7084, 420732, 27343888, 1882933364, 134993766600, 9969937491420, 753310723010608, 57956002331347120, 4524678117939182220, 357557785658996609700, 28545588568201512137904, 2298872717007844035521848, 186533392975795702301759056

Offset: 0

Bruno Berselli, Jan 12 2014

This is the case l=6, k=2 of binomial((l+k)*n,k*n)/((l*n+1)/gcd(k,l*n+1)), see Theorem 1.1 in Zhi-Wei Sun's paper.
First bisection of A002293.
Also, the sequence is between A002296 and A235535.

Zhi-Wei Sun, On Divisibility Of Binomial Coefficients, Journal of the Australian Mathematical Society 93 (2012), p. 189-201.

Cf. similar sequences generated by binomial((l+k)*n,k*n)/(l*n+1), where l is divisible by all the factors of k: A000108 (l=1, k=1), A001764 (l=2, k=1), A002293 (l=3, k=1), A002294 (l=4, k=1), A002295 (l=5, k=1), A002296 (l=6, k=1), A007556 (l=7, k=1), A062994 (l=8, k=1), A059968 (l=9, k=1), A230388 (l=10, k=1), A048990 (l=2, k=2), A235534 (l=4, k=2), this sequence (l=6, k=2), A187357 (l=3, k=3), A235535 (l=6, k=3).

l:=6; k:=2; [Binomial((l+k)*n,k*n)/(l*n+1): n in [0..20]]; /* where l is divisible by all the prime factors of k */

Table[Binomial[8 n, 2 n]/(6 n + 1), {n, 0, 20}]

a(n) = A124753(6*n).
From Ilya Gutkovskiy, Jun 21 2018: (Start)
G.f.: 6F5(1/8,1/4,3/8,5/8,3/4,7/8; 1/3,1/2,2/3,5/6,7/6; 65536*x/729).
a(n) ~ 2^(16*n-1)/(sqrt(Pi)*3^(6*n+3/2)*n^(3/2)). (End)

cp's OEIS Frontend

A235340 a(n) = 10*binomial(11*n+10,n)/(11*n+10).

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A192893 Number of symmetric 11-ary factorizations of the n-cycle (1,2...n).

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A235534 a(n) = binomial(6*n, 2*n) / (4*n + 1).

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A235535 a(n) = binomial(9*n, 3*n) / (6*n + 1).

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A235338 a(n) = 8*binomial(11*n+8,n)/(11*n+8).

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A235536 a(n) = binomial(8*n, 2*n) / (6*n + 1).

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A235340 a(n) = 10binomial(11n+10,n)/(11*n+10).

A235534 a(n) = binomial(6n, 2n) / (4*n + 1).

A235535 a(n) = binomial(9n, 3n) / (6*n + 1).

A235338 a(n) = 8binomial(11n+8,n)/(11*n+8).

A235536 a(n) = binomial(8n, 2n) / (6*n + 1).