A118970 -id:A118970 - cp's OEIS frontend

A143546 G.f. A(x) satisfies A(x) = 1 + xA(x)^3A(-x)^2.

1, 1, 1, 3, 5, 18, 35, 136, 285, 1155, 2530, 10530, 23751, 100688, 231880, 996336, 2330445, 10116873, 23950355, 104819165, 250543370, 1103722620, 2658968130, 11777187240, 28558343775, 127067830773, 309831575760, 1383914371728, 3390416787880, 15194457001440

Offset: 0

Paul D. Hanna, Aug 23 2008

Number of achiral polyominoes composed of n hexagonal cells of the hyperbolic regular tiling with Schläfli symbol {6,oo}. A stereographic projection of the {6,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 23 2024

Number of achiral noncrossing partitions composed of n blocks of size 5. - Andrew Howroyd, Feb 08 2024

			G.f.: A(x) = 1 + x + x^2 + 3*x^3 + 5*x^4 + 18*x^5 + 35*x^6 + 136*x^7 + ...
A(x) = 1 + x*A(x)^3*A(-x)^2 where
A(x)^3 = 1 + 3x + 6x^2 + 16x^3 + 39x^4 + 114x^5 + 304x^6 + 936x^7 + ...
A(-x)^2 = 1 - 2x + 3x^2 - 8x^3 + 17x^4 - 52x^5 + 125x^6 - 408x^7 + ...
Also, A(x) = G(x^2) + x*G(x^2)^3 where
G(x) = 1 + x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + 23751*x^6 + ...
G(x)^3 = 1 + 3*x + 18*x^2 + 136*x^3 + 1155*x^4 + 10530*x^5 + ...

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Michel Bousquet and Cédric Lamathe, On symmetric structures of order two, Discrete Math. Theor. Comput. Sci. 10 (2008), 153-176. See Table 1. - From _N. J. A. Sloane_, Jul 12 2011
Malin Christensson, Make hyperbolic tilings of images, web page, 2019.

Column k=5 of A369929 and k=6 of A370062.
Cf. A118970.
Polyominoes: A221184(n-1) (oriented), A004127 (unoriented), A369473 (chiral), A002294 (rooted), A047749 {4,oo}, A369472 {5,oo}.

terms = 28;
A[] = 1; Do[A[x] = 1 + x A[x]^3 A[-x]^2 + O[x]^terms // Normal, {terms}];
CoefficientList[A[x], x] (* Jean-François Alcover, Jul 24 2018 *)
p=6; Table[If[EvenQ[n],Binomial[(p-1)n/2,n/2]/((p-2)n/2+1),If[OddQ[p],(p-1)Binomial[(p-1)n/2-1,(n-1)/2]/((p-2)n+1),p Binomial[(p-1)n/2-1/2,(n-1)/2]/((p-2)n+2)]],{n,0,35}] (* Robert A. Russell, Jan 23 2024 *)

{a(n)=my(A=1+O(x^(n+1)));for(i=0,n,A=1+x*A^3*subst(A^2,x,-x));polcoef(A,n)}

{a(n)=my(m=n\2,p=2*(n%2)+1);binomial(5*m+p-1,m)*p/(4*m+p)}

G.f.: A(x) = G(x^2) + x*G(x^2)^3 where G(x) = 1 + x*G(x)^5 is the g.f. of A002294.
a(2n) = binomial(5*n,n)/(4*n+1); a(2n+1) = binomial(5*n+2,n)*3/(4*n+3).
From Robert A. Russell, Jan 23 2024: (Start)
a(n+2)/a(n) ~ 3125/256. a(2m+1)/a(2m) ~ 75/16; a(2m)/a(2m-1) ~ 125/48.
a(n) = 2*A004127(n) - A221184(n-1) = A221184(n-1) - 2*A369473(n) = A004127(n) - A369473(n). (End)
a(2m) = A002294(m) ~ (5^5/4^4)^m*sqrt(5/(2*Pi*(4*m)^3)). - Robert A. Russell, Jul 15 2024
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)^5 is the g.f. of A002294.
a(0) = 1; a(n) = Sum_{i, j, k>=0 and i+2*j+2*k=n-1} a(i) * a(2*j) * a(2*k). (End)
a(0) = 1; a(n) = Sum_{i, j, k, l, m>=0 and i+j+k+l+m=n-1} (-1)^(i+j) * a(i) * a(j) * a(k) * a(l) * a(m). - Seiichi Manyama, Jul 08 2025

A234465 a(n) = 3binomial(8n+6,n)/(4*n+3).

Original entry on oeis.org

1, 6, 63, 812, 11655, 178794, 2869685, 47593176, 809172936, 14028048650, 247039158366, 4406956913268, 79470057050020, 1446283758823470, 26529603944225670, 489989612605050800, 9104498753815680600, 170073237411754811568, 3192081704235788729043

Offset: 0

Tim Fulford, Dec 26 2013

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p = 8, r = 6.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
J-C. Aval, Multivariate Fuss-Catalan Numbers, arXiv:0711.0906v1, Discrete Math., 308 (2008), 4660-4669.
Thomas A. Dowling, Catalan Numbers Chapter 7
Wojciech Mlotkowski, Fuss-Catalan Numbers in Noncommutative Probability, Docum. Mathm. 15: 939-955.
Wikipedia, Fuss-Catalan number

Cf. A000108, A007556, A234461, A234462, A234463, A234464, A234466, A234467, A230390, A007556, A069271, A118970, A212073, A233834, A234510, A234571, A235339.

[3*Binomial(8*n+6, n)/(4*n+3): n in [0..30]]; // Vincenzo Librandi, Dec 26 2013

Table[3 Binomial[8 n + 6, n]/(4 n + 3), {n, 0, 40}] (* Vincenzo Librandi, Dec 26 2013 *)

PARI
```
a(n) = 3*binomial(8*n+6,n)/(4*n+3);
```

{a(n)=local(B=1); for(i=0, n, B=(1+x*B^(4/3))^6+x*O(x^n)); polcoeff(B, n)}

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p = 8, r = 6.
O.g.f. A(x) = 1/x * series reversion (x/C(x)^6), where C(x) is the o.g.f. for the Catalan numbers A000108. A(x)^(1/6) is the o.g.f. for A007556. - Peter Bala, Oct 14 2015
D-finite with recurrence: 7*n*(7*n+4)*(7*n+1)*(7*n+5)*(7*n+2)*(7*n+6)*(7*n+3)*a(n) -128*(8*n+3)*(4*n-1)*(8*n+1)*(2*n+1)*(8*n-1)*(4*n+1)*(8*n+5)*a(n-1)=0. - R. J. Mathar, Feb 21 2020

A212073 G.f. satisfies: A(x) = (1 + x*A(x)^(3/2))^4.

Original entry on oeis.org

1, 4, 30, 280, 2925, 32736, 383838, 4654320, 57887550, 734405100, 9467075926, 123648163392, 1632743088275, 21761329287600, 292362576381900, 3955219615609056, 53834425161872586, 736687428853685400, 10129401435828605700, 139876690363085200200

Offset: 0

Paul D. Hanna, Apr 29 2012

Fuss-Catalan sequence is a(n,p,r) = r*binomial(p*n + r, n)/(p*n + r); this is the case p = 6, r = 4. The o.g.f. B(x) of the Fuss_catalan sequence a(n,p,r) satisfies B(x) = {1 + x*B(x)^(p/r)}^r. - Peter Bala, Oct 14 2015

			G.f.: A(x) = 1 + 4*x + 30*x^2 + 280*x^3 + 2925*x^4 + 32736*x^5 +...
Related expansions:
A(x)^(3/2) = 1 + 6*x + 51*x^2 + 506*x^3 + 5481*x^4 +...+ A002295(n+1)*x^n +...
A(x)^(1/4) = 1 + x + 6*x^2 + 51*x^3 + 506*x^4 +...+ A002295(n)*x^n +...

Wikipedia, Fuss-Catalan number

Cf. A002295, A212071, A212072, A130564, A069271, A118970, A233834, A234465, A234510, A234571, A235339.

m = 20; A[_] = 0;
Do[A[x_] = (1 + x*A[x]^(3/2))^4 + O[x]^m, {m}];
CoefficientList[A[x], x] (* Jean-François Alcover, Oct 20 2019 *)

{a(n)=binomial(6*n+4,n) * 4/(6*n+4)}
for(n=0, 40, print1(a(n), ", "))

{a(n)=local(A=1+4*x); for(i=1, n, A=(1+x*A^(3/2))^4+x*O(x^n)); polcoeff(A, n)}

a(n) = 4*binomial(6*n+4,n)/(6*n+4).
G.f. A(x) = G(x)^4 where G(x) = 1 + x*G(x)^6 is the g.f. of A002295.
O.g.f. A(x) = 1/x * series reversion (x/C(x)^4), where C(x) is the o.g.f. for the Catalan numbers A000108. - Peter Bala, Oct 14 2015
D-finite with recurrence 5*n*(5*n+1)*(5*n+2)*(5*n+3)*(5*n+4)*a(n) -72*(6*n-1)*(3*n-1)*(2*n+1)*(3*n+1)*(6*n+1)*a(n-1)=0. - R. J. Mathar, Nov 22 2024

A234571 a(n) = 4binomial(10n+8,n)/(5*n+4).

Original entry on oeis.org

1, 8, 108, 1776, 32430, 632016, 12876864, 270964320, 5843355957, 128462407840, 2868356980060, 64869895026144, 1482877843096650, 34207542810153216, 795318309360948240, 18617396126132233920, 438423206616057162258, 10379232525028947311160, 246878659984195222962220

Offset: 0

Tim Fulford, Dec 28 2013

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), where p = 10, r = 8.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
J-C. Aval, Multivariate Fuss-Catalan Numbers, arXiv:0711.0906v1, Discrete Math., 308 (2008), 4660-4669.
Thomas A. Dowling, Catalan Numbers Chapter 7
Wojciech Mlotkowski, Fuss-Catalan Numbers in Noncommutative Probability, Docum. Mathm. 15: 939-955.
Wikipedia, Fuss-Catalan number

Cf. A059968, A234525, A234526, A234527, A234528, A234529, A234570, A234573, A059968, A069271, A118970, A212073, A233834, A234465, A234510, A235339.

[4*Binomial(10*n+8, n)/(5*n+4): n in [0..30]];

Table[4 Binomial[10 n + 8, n]/(5 n + 4), {n, 0, 30}]

PARI
```
a(n) = 4*binomial(10*n+8,n)/(5*n+4);
```

{a(n)=local(B=1); for(i=0, n, B=(1+x*B^(5/4))^8+x*O(x^n)); polcoeff(B, n)}

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p = 10, r = 8.

O.g.f. A(x) = 1/x * series reversion (x/C(x)^8), where C(x) is the o.g.f. for the Catalan numbers A000108. A(x)^(1/8) is the o.g.f. for A059968. - Peter Bala, Oct 14 2015

A234510 a(n) = 7binomial(9n+7,n)/(9*n+7).

Original entry on oeis.org

1, 7, 84, 1232, 20090, 349860, 6371764, 119877472, 2311664355, 45448324110, 907580289616, 18358110017520, 375353605696524, 7744997102466932, 161070300819384000, 3372697621463787456, 71046594621639707245, 1504569659175026591805

Offset: 0

Tim Fulford, Dec 27 2013

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), where p = 9, r = 7.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
J-C. Aval, Multivariate Fuss-Catalan Numbers, arXiv:0711.0906v1, Discrete Math., 308 (2008), 4660-4669.
Thomas A. Dowling, Catalan Numbers Chapter 7
Wojciech Mlotkowski, Fuss-Catalan Numbers in Noncommutative Probability, Docum. Mathm. 15: 939-955.
Wikipedia, Fuss-Catalan number

Cf. A000108, A143554, A234505, A234506, A234507, A234508, A234509, A234513, A232265, A062994, A069271, A118970, A212073, A233834, A234465, A234571, A235339.

[7*Binomial(9*n+7, n)/(9*n+7): n in [0..30]]; // Vincenzo Librandi, Dec 27 2013

Table[7 Binomial[9 n + 7, n]/(9 n + 7), {n, 0, 40}] (* Vincenzo Librandi, Dec 27 2013 *)

PARI
```
a(n) = 7*binomial(9*n+7,n)/(9*n+7);
```

{a(n)=local(B=1); for(i=0, n, B=(1+x*B^(9/7))^7+x*O(x^n)); polcoeff(B, n)}

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p = 9, r = 7.

O.g.f. A(x) = 1/x * series reversion (x/C(x)^7), where C(x) is the o.g.f. for the Catalan numbers A000108. A(x)^(1/7) is the o.g.f. for A062994. - Peter Bala, Oct 14 2015

A233834 a(n) = 5binomial(7n+5,n)/(7*n+5).

Original entry on oeis.org

1, 5, 45, 500, 6200, 82251, 1142295, 16398200, 241379325, 3623534200, 55262073757, 853814730600, 13335836817420, 210225027967325, 3340362288091500, 53443628421286320, 860246972339613855, 13921016318025200505, 226352372251889455000, 3696160728052814340000

Offset: 0

Tim Fulford, Dec 16 2013

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p = 7, r = 5.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
J-C. Aval, Multivariate Fuss-Catalan Numbers, arXiv:0711.0906v1, Discrete Math., 308 (2008), 4660-4669.
Thomas A. Dowling, Catalan Numbers Chapter 7
Wojciech Mlotkowski, Fuss-Catalan Numbers in Noncommutative Probability, Docum. Mathm. 15: 939-955.
Wikipedia, Fuss-Catalan number

Cf. A000108, A002296, A233832, A233833, A143547, A130565, A233835, A233907, A233908, A002296, A069271, A118970, A212073, A234465, A234510, A234571, A235339.

[5*Binomial(7*n+5, n)/(7*n+5): n in [0..30]];

Table[5 Binomial[7 n + 5, n]/(7 n + 5), {n, 0, 30}]

PARI
```
a(n) = 5*binomial(7*n+5,n)/(7*n+5);
```

{a(n)=local(B=1); for(i=0, n, B=(1+x*B^(7/5))^5+x*O(x^n)); polcoeff(B, n)}

G.f. satisfies: A(x) = {1 + x*A(x)^(p/r)}^r, where p = 7, r = 5.

O.g.f. A(x) = 1/x * series reversion (x/C(x)^5), where C(x) is the o.g.f. for the Catalan numbers A000108. A(x)^(1/5) is the o.g.f. for A002296. - Peter Bala, Oct 14 2015

A118968 a(4n+k) = (k+1)*binomial(5n+k,n)/(4n+k+1), k=0..3.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 4, 5, 11, 18, 26, 35, 80, 136, 204, 285, 665, 1155, 1771, 2530, 5980, 10530, 16380, 23751, 56637, 100688, 158224, 231880, 556512, 996336, 1577532, 2330445, 5620485, 10116873, 16112057, 23950355, 57985070, 104819165, 167710664, 250543370, 608462470

Offset: 0

Paul Barry, May 07 2006

Row sums of Riordan array (1,x(1-x^4))^(-1).

F. Hering et al., The enumeration of stack polytopes and simplicial clusters, Discrete Math., 40 (1982), 203-217.

Cf. A124753, A002294, A118969, A118970, A118971.

Table[k=Mod[n,4];(k+1)Binomial[(5n-k)/4,(n-k)/4]/(n+1),{n,0,40}] (* Robert A. Russell, Mar 14 2024 *)

{a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=1+x*A^2*subst(A,x,-x)*subst(A,x,I*x)*subst(A,x,-I*x));polcoeff(A,n)} \\ Paul D. Hanna, Jun 04 2012

{a(n)=local(A=1+x);for(i=1,n,A=1+x*A*exp(sum(m=1,n\4,4*polcoeff(log(A+x*O(x^n)),4*m)*x^(4*m))+x*O(x^n)));polcoeff(A,n)} \\ Paul D. Hanna, Jun 04 2012

apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
a(n) = apr(n\4, 5, n%4+1); \\ Seiichi Manyama, Jul 20 2025

a(4n) = A002294(n), a(4n+1) = A118969(n), a(4n+2) = A118970(n), a(4n+3) = A118971(n).
G.f. satisfies: A(x) = 1 + x*A(x)^2*A(-x)*A(I*x)*A(-I*x). - Paul D. Hanna, Jun 04 2012
G.f. satisfies: A(x) = 1 + x*A(x)*G(x^4) where G(x) = 1 + x*G(x)^5 is the g.f. of A002294. - Paul D. Hanna, Jun 04 2012
From Robert A. Russell, Mar 14 2024: (Start)
G.f.: G(z^4) + z*G(z^4)^2 + z^2*G(z^4)^3 + z^3*G(z^4)^4, where G(z) = 1 + z*G(z)^5 is the g.f. for A002294.
G.f.: E(1)(t*E(5)(t^4)) (fifth entry in Table 3), where E(d)(t) is defined in formula 3 of Hering link. (End)
a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/4)} a(4*k) * a(n-1-4*k). - Seiichi Manyama, Jul 07 2025

A235339 a(n) = 9binomial(11n+9,n)/(11*n+9).

Original entry on oeis.org

1, 9, 135, 2460, 49725, 1072197, 24163146, 562311720, 13409091540, 325949656825, 8046743477058, 201198155083200, 5084704634041305, 129673310477725350, 3332952595603387800, 86250038091202771344, 2245329811618166111985

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 = 9.

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
Wojciech Mlotkowski, Fuss-Catalan Numbers in Noncommutative Probability, Docum. Mathm. 15: 939-955.
Wikipedia, Fuss-Catalan number

Cf. A230388, A234868, A234869, A234870, A234871, A234872, A234873, A235338, A235340, A069271, A118970, A212073, A230388, A233834, A234465, A234510, A234571.

[9*Binomial(11*n+9, n)/(11*n+9): n in [0..30]];

Table[9 Binomial[11 n + 9, n]/(11 n + 9), {n, 0, 30}]

PARI
```
a(n) = 9*binomial(11*n+9,n)/(11*n+9);
```

{a(n)=local(B=1); for(i=0, n, B=(1+x*B^(11/9))^9+x*O(x^n)); polcoeff(B, n)}

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, here p = 11, r = 9.

O.g.f. A(x) = 1/x * series reversion (x/C(x)^9), where C(x) is the o.g.f. for the Catalan numbers A000108. A(x)^(1/9) is the o.g.f. for A230388. - Peter Bala, Oct 14 2015

A163456 a(n) = binomial(5*n,n)/5.

Original entry on oeis.org

1, 9, 91, 969, 10626, 118755, 1344904, 15380937, 177232627, 2054455634, 23930713170, 279871768995, 3284214703056, 38650751381832, 456002537343216, 5391644226101705, 63871405575418665, 757929628541719755

Offset: 1

Zak Seidov, Jul 28 2009

For prime p, a(p) == 1 (mod p). - Gary Detlefs, Aug 03 2013
In fact, a(p) == 1 (mod p^3) for prime p >= 5. See Mestrovic, Section 3. - Peter Bala, Oct 09 2015
From Robert Israel, Jul 12 2016: (Start)
a(p+1) == 5 (mod p) for primes p >= 5.
a(p^(k+1)) == a(p^k) mod p^(3(k+1)) for primes p >= 5. (End)

Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, 2nd ed. 1994.

Robert Israel, Table of n, a(n) for n = 1..920
Romeo Meštrović, Lucas’ theorem: its generalizations, extensions and applications (1878-2014), arXiv:1409.3820v1 [math.NT], 2014.
N. J. Wildberger and Dean Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly (2025). See section 12.

Cf. A000108, A000245, A001764, A002293, A025174, A118970, A163455, A224274, A227726.

seq(binomial(5*n,n)/5, n=1..20); # Robert Israel, Jul 12 2016

Array[Binomial[5 #, #]/5 &, {18}] (* Michael De Vlieger, Oct 09 2015 *)

a(n) = binomial(5*n,n)/5 \\ Altug Alkan, Oct 09 2015

a(n) = (5*n-1)!/(4*n!*(4*n-1)!) = A001449(n)/5 = A163455(n)/4.
a(n) = binomial(5*n,n)/5. - Gary Detlefs, Aug 03 2013
From Peter Bala, Oct 08 2015: (Start)
a(n) = (1/3)*[x^n] (C(x)^3)^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A224274.
exp( 3*Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 18*x^2 + 136*x^3 + ... is the o.g.f. for A118970. (End)
From Peter Bala,Jul 12 2016: (Start)
a(n) = 1/6*[x^n] (1 + x)/(1 - x)^(4*n + 1).
a(n) = 1/6*[x^n] ( 1/C(-x)^6 )^n. Cf. A227726. (End)
a(n) ~ 2^(-8*n-3/2)*5^(5*n-1/2)*n^(-1/2)/sqrt(Pi). - Ilya Gutkovskiy, Jul 12 2016
From Robert Israel, Jul 12 2016: (Start)
G.f.: x*hypergeom([1, 6/5, 7/5, 8/5, 9/5], [5/4, 3/2, 7/4, 2], (3125/256)*x).
a(n) = 5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)/(8*n*(4*n-3)*(2*n-1)*(4*n-1)). (End)
O.g.f.: f(x)/(1 - 4*f(x)), where f(x) = series reversion (x/(1 + x)^5) = x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + ... is the o.g.f. of A002294 with the initial term omitted. Cf. A025174. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/4)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+4)*n+k-1,k) = C(5*n,n)/5 and (1/5)*Sum_{k = 0..n} (-1)^k*C(x*n,n-k)*C((x-5)*n+k-1,k) = C(5*n,n)/5, both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
Right-hand side of the identity (1/4)*Sum_{k = 0..2*n} (-1)^k*binomial(6*n-k-1,2*n-k)*binomial(4*n+k-1,k) = binomial(5*n,n)/5, for n >= 1. - Peter Bala, Mar 09 2022
a(n) = (1/2)* [x*n] F(x)^(2*n) = [x^n] G(x)^n for n >= 1, where F(x) = Sum_{k >= 0} 1/(2*k + 1)*binomial(3*k,k)*x^k is the o.g.f. of A001764 and G(x) = Sum_{k >= 0} 1/(3*k + 1)*binomial(4*k,k)*x^k is the o.g.f. of A002293 (apply Concrete Mathematics, equation 5.60, p. 201). - Peter Bala, Apr 26 2023

Renamed by Peter Bala, Oct 08 2015

A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 4, 34, 337, 3554, 38754, 431521, 4874377, 55639010, 640177033, 7412165034, 86256322816, 1007980394849, 11820510331777, 139032549536551, 1639506780365337, 19376785465043938, 229458302589724067, 2721958273545613513, 32339465512495259708, 384758834631081248554

Offset: 0

Peter Bala, Mar 15 2020

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(3*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^3 = 1 + 3*x + O(x^2)
  n = 2: c(x)^6 = 1 + 6*x + 27*x^2 + O(x^3)
  n = 3: c(x)^9 = 1 + 9*x + 54*x^2 + 273*x^3 + O(x^4)
  n = 4: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + 2907*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 27 = 34, a(3) = 1 + 9 + 54 + 273 = 337 and a(4) = 1 + 12 + 90 + 544 + 2907 = 3554.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
                                              row sums
  n = 0 |    1                                    1
  n = 1 |    3       1                            4
  n = 2 |   27       6    1                      34
  n = 3 |  273      54    9   1                 337
  n = 4 | 2907     544   90  12   1            3554
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 11820510331777 - 4 = 3*11*(13^3)*(43^2)*88177 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 4583419703934987639046 - 337 = (3^2)*(7^4)*2441* 86893477573061 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 93266278848727959965820004 - 38754 = 2*(5^7)*19* 31416009717466260199 == 0 ( mod 5^6 ).

Cf. A000108, A118970, A333090 through A333097.

seq(add(3*n/(3*n+k)*binomial(3*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(3*n), x, 101):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[3*Binomial[5*n-1, n] * HypergeometricPFQ[{1, -4*n, -n}, {1/2 - 5*n/2, 1 - 5*n/2}, 1/4]/4, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 3*n/(3*n+k)*binomial(3*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^3(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 3*x + 18*x^2 + 136*x^3 + 1155*x^4 + ... = (1/x)*Revert( x/c^3(x) ) is the o.g.f. of A118970.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 5^(5*n + 3/2) / (7 * 2^(8*n + 3/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 3*n/(3*n+2*k)*binomial(3*n+2*k, k) for n >= 1. - Peter Bala, May 03 2024

cp's OEIS Frontend

A143546 G.f. A(x) satisfies A(x) = 1 + x*A(x)^3*A(-x)^2.

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A234465 a(n) = 3*binomial(8*n+6,n)/(4*n+3).

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A212073 G.f. satisfies: A(x) = (1 + x*A(x)^(3/2))^4.

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A234571 a(n) = 4*binomial(10*n+8,n)/(5*n+4).

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A234510 a(n) = 7*binomial(9*n+7,n)/(9*n+7).

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A233834 a(n) = 5*binomial(7*n+5,n)/(7*n+5).

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A118968 a(4n+k) = (k+1)*binomial(5n+k,n)/(4n+k+1), k=0..3.

A143546 G.f. A(x) satisfies A(x) = 1 + xA(x)^3A(-x)^2.

A234465 a(n) = 3binomial(8n+6,n)/(4*n+3).

A234571 a(n) = 4binomial(10n+8,n)/(5*n+4).

A234510 a(n) = 7binomial(9n+7,n)/(9*n+7).

A233834 a(n) = 5binomial(7n+5,n)/(7*n+5).

A235339 a(n) = 9binomial(11n+9,n)/(11*n+9).

A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.