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 1-10 of 14 results. Next

A007556 Number of 8-ary trees with n vertices.

Original entry on oeis.org

1, 1, 8, 92, 1240, 18278, 285384, 4638348, 77652024, 1329890705, 23190029720, 410333440536, 7349042994488, 132969010888280, 2426870706415800, 44627576949364104, 826044435409399800, 15378186970730687400, 287756293703544823872, 5409093674555090316300
Offset: 0

Views

Author

Keywords

Comments

Shifts left when convolved three times.
From Wolfdieter Lang, Sep 14 2007: (Start)
a(n), n >= 1, enumerates octic (8-ary) trees (rooted, ordered, incomplete) with n vertices (including the root).
Pfaff-Fuss-Catalan sequence C^{m}_n for m = 8. See the Graham et al. reference, p. 347. eq. 7.66. See also the Pólya-Szegő reference.
Also 8-Raney sequence. See the Graham et al. reference, p. 346-7.
(End)
This is instance k = 8 of the generalized Catalan family {C(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment of A130564. - _Wolfdieter Lang, Feb 05 2024

Examples

			There are a(2) = 8 octic trees (vertex degree less than or equal to 8 and 8 possible branchings) with 2 vertices (one of them the root). Adding one more branch (one more vertex) to these 8 trees yields 8*8 + binomial(8, 2) = 92 = a(3) such trees.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 200, 347.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, Heidelberg, New York, 2 vols., 1972, Vol. 1, problem 211, p. 146 with solution on p. 348.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Seventh column of triangle A062993.
Cf. A130564.

Programs

  • Haskell
    a007556 0 = 1
    a007556 n = a007318' (8 * n) (n - 1) `div` n
    -- Reinhard Zumkeller, Jul 30 2013
    
  • Magma
    [Binomial(8*n, n)/(7*n+1): n in [0..20]]; // Vincenzo Librandi, Apr 02 2015
    
  • Maple
    seq(binomial(8*n+1,n)/(8*n+1),n=0..30); # Robert FERREOL, Apr 01 2015
    n:=30: G:=series(RootOf(g = 1+x*g^8, g),x=0,n+1): seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 01 2015
  • Mathematica
    Table[Binomial[8n, n]/(7n + 1), {n, 0, 20}] (* Harvey P. Dale, Dec 24 2012 *)
  • PARI
    vector(100, n, n--; binomial(8*n, n)/(7*n+1)) \\ Altug Alkan, Oct 14 2015

Formula

a(n) = binomial(8*n, n)/(7*n+1) = binomial(8*n+1, n)/(8*n+1) = A062993(n+6,6).
O.g.f.: A(x) = 1 + x*A(x)^8 = 1/(1-x*A(x)^7).
a(0) = 1; a(n) = Sum_{i1 + i2 + .. i8 = n - 1} a(i1)*a(i2)*...*a(i8) for n >= 1. - Robert FERREOL, Apr 01 2015
a(n) = binomial(8*n, n - 1)/n for n >= 1, a(0) = 1 (from the Lagrange series of the o.g.f. A(x) with its above given implicit equation).
From Karol A. Penson, Mar 26 2015: (Start)
In Maple notation,
e.g.f.: hypergeom([1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8], [2/7, 3/7, 4/7, 5/7, 6/7, 1, 8/7],(2^24/7^7)*z);
o.g.f.: hypergeom([1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8], [2/7, 3/7, 4/7, 5/7, 6/7, 8/7],(2^24/7^7)*z);
a(n) are special values of Jacobi polynomials, in Maple notation:
a(n) = JacobiP(n - 1, 7*n + 1, -n, 1)/n, n = 1, 2, ...
(End)
From Peter Bala, Oct 14 2015: (Start)
A(x)^2 is o.g.f. for A234461; A(x)^3 is o.g.f. for A234462;
A(x)^4 is o.g.f. for A234463; A(x)^5 is o.g.f. for A234464;
A(x)^6 is o.g.f. for A234465; A(x)^7 is o.g.f. for A234466;
A(x)^9 is o.g.f. for A234467. (End)
a(n) ~ 2^(24*n + 1)/(sqrt(Pi)*7^(7*n + 3/2)*n^(3/2)). - Ilya Gutkovskiy, Feb 07 2017
D-finite with recurrence: 7*n*(7*n-3)*(7*n+1)*(7*n-2)*(7*n-5)*(7*n-1)*(7*n-4)*a(n) -128*(8*n-5)*(4*n-1)*(8*n-7)*(2*n-1)*(8*n-1)*(4*n-3)*(8*n-3)*a(n-1)=0. - R. J. Mathar, Feb 20 2020
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^15). - Seiichi Manyama, Jun 16 2025

A118971 a(n) = binomial(5*n+3,n)/(n+1).

Original entry on oeis.org

1, 4, 26, 204, 1771, 16380, 158224, 1577532, 16112057, 167710664, 1772645420, 18974357220, 205263418941, 2240623268512, 24648785802336, 272994644359580, 3041495503591365, 34064252968167180, 383302465665133014
Offset: 0

Views

Author

Paul Barry, May 07 2006

Keywords

Comments

A quadrisection of A118968.
For n >= 1, a(n-1) is the number of lattice paths from (0,0) to (4n,n) using only the steps (1,0) and (0,1) and which stay strictly below the line y = x/4 except at the path's endpoints. - Lucas A. Brown, Aug 21 2020
This is instance k = 4 of the family {c(k, n+1)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment in A130564. - _Wolfdieter Lang, Feb 04 2024

Crossrefs

Cf. A000108, A006013, A006632, A130564, A130565, A234466, A234513, A234573, A235340 (members of the same family).

Programs

  • Mathematica
    Table[4*Binomial[5n+3,n]/(4n+4),{n,0,30}] (* Harvey P. Dale, Apr 09 2012 *)

Formula

G.f.: If the inverse series of y*(1-y)^4 is G(x) then A(x)=G(x)/x.
D-finite with recurrence 8*(4*n+1)*(2*n+1)*(4*n+3)*(n+1)*a(n) -5*(5*n+1)*(5*n+2)*(5*n+3)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Nov 26 2012
a(n) = (4/5)*binomial(5*(n+1),n+1)/(5*(n+1)-1). - Bruno Berselli, Jan 17 2014
E.g.f.: 4F4(4/5,6/5,7/5,8/5; 5/4,3/2,7/4,2; 3125*x/256). - Ilya Gutkovskiy, Jan 23 2018
G.f.: 5F4([4,5,6,7,8]/5, [5,6,7,8]/4; (5^5/4^4)*x) = (4/(5*x))*(1 - 4F3([-1,1,2,3]/5, [1,2,3]/4; (5^5/4^4)*x)). - Wolfdieter Lang, Feb 15 2024

A130564 Member k=5 of a family of generalized Catalan numbers.

Original entry on oeis.org

1, 5, 40, 385, 4095, 46376, 548340, 6690585, 83615350, 1064887395, 13770292256, 180320238280, 2386316821325, 31864803599700, 428798445360120, 5809228810425801, 79168272296871450, 1084567603590147950
Offset: 1

Views

Author

Wolfdieter Lang, Jul 13 2007

Keywords

Comments

The generalized Catalan numbers C(k,n):= binomial(k*n+1,n)/(k*n+1) become for negative k=-|k|, with |k|>=2, ((-1)^(n-1))*binomial((|k|+1)*n-2,n)/(|k|*n-1), n>=0.
The family c(k,n):=binomial((k+1)*n-2,n)/(k*n-1), n>=1, has the members A000108, A006013, A006632, A118971 for k=1,2,3,4, respectively (but the offset there is 0).
The members of the C(k,n) family for positive k are: A000012 (powers of 1), A000108, A001764, A002293, A002294, A002295, A002296, A007556, A062994, for k=1..9.

References

  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1994, pp. 200, 363.

Crossrefs

Programs

  • Mathematica
    Rest@ CoefficientList[InverseSeries[Series[y (1 - y)^5, {y, 0, 18}], x], x] (* Michael De Vlieger, Oct 13 2019 *)

Formula

a(n) = binomial((k+1)*n-2,n)/(k*n-1), with k=5.
G.f.: inverse series of y*(1-y)^5.
a(n) = (5/6)*binomial(6*n,n)/(6*n-1). [Bruno Berselli, Jan 17 2014]
From Wolfdieter Lang, Feb 06 2020: (Start)
G.f.: (5/6)*(1 - hypergeom([-1, 1, 2, 3, 4]/6, [1, 2, 3, 4]/5,(6^6/5^5)*x)).
E.g.f.: (5/6)*(1 - hypergeom([-1, 1, 2, 3, 4]/6, [1, 2, 3, 4, 5]/5,(6^6/5^5)*x)). (End)
D-finite with recurrence 5*n*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n) -72*(6*n-7)*(3*n-1)*(2*n-1)*(3*n-2)*(6*n-5)*a(n-1)=0. - R. J. Mathar, May 07 2021

A234465 a(n) = 3*binomial(8*n+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

Views

Author

Tim Fulford, Dec 26 2013

Keywords

Comments

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

Crossrefs

Programs

  • Magma
    [3*Binomial(8*n+6, n)/(4*n+3): n in [0..30]]; // Vincenzo Librandi, Dec 26 2013
  • Mathematica
    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);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(4/3))^6+x*O(x^n)); polcoeff(B, n)}
    

Formula

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

A234467 a(n) = 9*binomial(8*n + 9,n)/(8*n + 9).

Original entry on oeis.org

1, 9, 108, 1488, 22230, 350244, 5729724, 96395616, 1657248417, 28987537150, 514215324216, 9229030737264, 167283594343320, 3057857090083908, 56305821384711720, 1043424549990820800, 19445145508444588200, 364191559218548917713, 6851518654436447733980
Offset: 0

Views

Author

Tim Fulford, Dec 26 2013

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(n*p + r,n)/(n*p + r); this is the case p = 8, r = 9.

Crossrefs

Cf. A000108, A000245 (k = 3), A006629 (k = 4), A196678 (k = 5), A233668 (k = 6), A233743 (k = 7), A233835 (k = 8), A232265 (k = 10), A229963 (k = 11).

Programs

  • Magma
    [9*Binomial(8*n+9, n)/(8*n+9): n in [0..30]]; // Vincenzo Librandi, Dec 26 2013
  • Mathematica
    Table[9 Binomial[8 n + 9, n]/(8 n + 9), {n, 0, 40}] (* Vincenzo Librandi, Dec 26 2013 *)
  • PARI
    a(n) = 9*binomial(8*n+9,n)/(8*n+9);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(8/9))^9+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: A(x) = {1 + x*A(x)^(p/r)}^r, where p = 8, r = 9.
From Peter Bala, Oct 16 2015: (Start)
O.g.f.: (1/x) * series reversion (x*C(-x)^9), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. See cross-references for other Fuss-Catalan sequences with o.g.f. 1/x * series reversion (x*C(-x)^k), k = 3 through 11.
A(x)^(1/9) is the o.g.f. for A007556. (End)
D-finite with recurrence +7*n*(7*n+3)*(7*n+4)*(7*n+5)*(7*n+6)*(7*n+8)*(7*n+9)*a(n)-128*(2*n+1)*(4*n+1)*(4*n+3)*(8*n+1)*(8*n+3)*(8*n+5)*(8*n+7)*a(n-1) = 0. - R. J. Mathar, Feb 09 2020
E.g.f.: F([9/8, 5/4, 11/8, 3/2, 13/8, 7/4, 15/8], [1, 10/7, 11/7, 12/7, 13/7, 15/7, 16/7], 16777216*x/823543), where F is the generalized hypergeometric function. - Stefano Spezia, Feb 09 2020

A234461 a(n) = binomial(8*n+2,n)/(4*n+1).

Original entry on oeis.org

1, 2, 17, 200, 2728, 40508, 635628, 10368072, 174047640, 2987139122, 52177566870, 924548764752, 16578073731752, 300252605231600, 5484727796499708, 100933398334075824, 1869468985400220600, 34823332479175275600, 651947852922093741585
Offset: 0

Views

Author

Tim Fulford, Dec 26 2013

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(n*p + r,n)/(n*p + r), this is the case p = 8, r = 2.

Crossrefs

Programs

  • Magma
    [Binomial(8*n+2, n)/(4*n+1): n in [0..30]];
  • Mathematica
    Table[Binomial[8 n + 2, n]/(4 n + 1), {n, 0, 30}]
  • PARI
    a(n) = binomial(8*n+2,n)/(4*n+1);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^4)^2+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: A(x) = {1 + x*A(x)^(p/r)}^r, where p = 8, r = 2.
a(n) = 2*binomial(8n+1,n-1)/n for n>0, a(0)=1. [Bruno Berselli, Jan 19 2014]
A(x^3) = 1/x * series reversion (x/C(x^3)^2), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. A(x)^(1/2) is the o.g.f. for A007556. - Peter Bala, Oct 14 2015

A234513 8*binomial(9*n+8,n)/(9*n+8).

Original entry on oeis.org

1, 8, 100, 1496, 24682, 433160, 7932196, 149846840, 2898753715, 57135036024, 1143315429776, 23166186450680, 474347963242860, 9799792252101016, 204022381037886400, 4276098770070159096, 90151561242584838605, 1910564646571462338800
Offset: 0

Views

Author

Tim Fulford, Dec 27 2013

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), where p=9, r=8.

Crossrefs

Programs

  • Magma
    [8*Binomial(9*n+8, n)/(9*n+8): n in [0..30]]; // Vincenzo Librandi, Dec 28 2013
  • Mathematica
    Table[8 Binomial[9 n + 8, n]/(9 n + 8), {n, 0, 30}] (* Vincenzo Librandi, Dec 28 2013 *)
  • PARI
    a(n) = 8*binomial(9*n+8,n)/(9*n+8);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(9/8))^8+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p=9, r=8.
From Wolfdieter Lang, Feb 06 2020: (Start)
G.f.: hypergeom([8, 9, ..., 16]/9, [9, 10, ..., 16]/8, (9^9/8^8)*x).
E,g,f.: hypergeom([8, 10, 11, ..., 16]/9, [9, 10,..., 16]/8, (9^9/8^8)*x). Cf. Ilya Gutkovsky in A118971. (End)
D-finite with recurrence 128*(8*n+3)*(4*n+3)*(8*n+1)*(2*n+1)*(8*n+7)*(4*n+1)*(8*n+5)*(n+1)*a(n) -81*(9*n+2)*(9*n+4)*(3*n+2)*(9*n-1)*(9*n+1)*(3*n+1)*(9*n+5)*(9*n+7)*a(n-1)=0. - R. J. Mathar, Aug 01 2022
From Wolfdieter Lang, Feb 15 2024: (Start)
a(n) = binomial(9*n+7, n+1)/(8*n+7), which is instance k = 8 of c(k, n+1) given in A130564.
The g.f. given above, and called B in the first line above, satisfies B(x)*(1 - x*B(x))^8 = 1. For the analog proof of the equivalence see A234466. x*B(x) is the compositional inverse of y*(1 - y)^8.
Another formula for the g.f. is B(x) = (8/(9*x))*(1 - 8F7([-1,1,2,3,4,5,6.7]/9, [1,2,3,4,5,6.7]/8; (9^9/8^8)*x)). (End)

A234463 Binomial(8*n+4,n)/(2*n+1).

Original entry on oeis.org

1, 4, 38, 468, 6545, 98728, 1566040, 25747128, 434824104, 7498246100, 131477423220, 2337053822012, 42016842044268, 762702138530080, 13959382918289880, 257323577200329904, 4773171937236245400, 89028543731246186400, 1668706597425638149302
Offset: 0

Views

Author

Tim Fulford, Dec 26 2013

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p=8, r=4.

Crossrefs

Programs

  • Magma
    [Binomial(8*n+4, n)/(2*n+1): n in [0..30]]; // Vincenzo Librandi, Dec 26 2013
  • Mathematica
    Table[Binomial[8 n + 4, n]/(2 n + 1), {n, 0, 40}] (* Vincenzo Librandi, Dec 26 2013 *)
  • PARI
    a(n) = binomial(8*n+4,n)/(2*n+1);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^2)^4+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p=8, r=4.

A251578 E.g.f.: exp(8*x*G(x)^7) / G(x)^7 where G(x) = 1 + x*G(x)^8 is the g.f. of A007556.

Original entry on oeis.org

1, 1, 8, 176, 6896, 397888, 30584128, 2948178304, 342418882688, 46582810477568, 7268517454045184, 1279982790328858624, 251155319283837571072, 54344039464582833577984, 12855960226911391575670784, 3301167001281829056285458432, 914476489427649778704952819712
Offset: 0

Views

Author

Paul D. Hanna, Dec 06 2014

Keywords

Examples

			E.g.f.: A(x) = 1 + x + 8*x^2/2! + 176*x^3/3! + 6896*x^4/4! + 397888*x^5/5! +...
such that A(x) = exp(8*x*G(x)^7) / G(x)^7
where G(x) = 1 + x*G(x)^8 is the g.f. of A007556:
G(x) = 1 + x + 8*x^2 + 92*x^3 + 1240*x^4 + 18278*x^5 + 285384*x^6 +...
Note that
A'(x) = exp(8*x*G(x)^7) = 1 + 8*x + 176*x^2/2! + 6896*x^3/3! +...
LOGARITHMIC DERIVATIVE.
The logarithm of the e.g.f. begins:
log(A(x)) = x + 7*x^2/2 + 77*x^3/3 + 1015*x^4/4 + 14763*x^5/5 +...
and so A'(x)/A(x) = G(x)^7.
TABLE OF POWERS OF E.G.F.
Form a table of coefficients of x^k/k! in A(x)^n as follows.
n=1: [1, 1,   8,  176,   6896,  397888,  30584128,  2948178304, ...];
n=2: [1, 2,  18,  400,  15584,  892896,  68217472,  6543183488, ...];
n=3: [1, 3,  30,  678,  26352, 1501344, 114073632, 10890011520, ...];
n=4: [1, 4,  44, 1016,  39512, 2241472, 169479808, 16107837568, ...];
n=5: [1, 5,  60, 1420,  55400, 3133560, 235931200, 22331561600, ...];
n=6: [1, 6,  78, 1896,  74376, 4200048, 315106128, 29713474944, ...];
n=7: [1, 7,  98, 2450,  96824, 5465656, 408881872, 38425052848, ...];
n=8: [1, 8, 120, 3088, 123152, 6957504, 519351232, 48658878080, ...]; ...
in which the main diagonal begins (see A251587):
[1, 2, 30, 1016, 55400, 4200048, 408881872, 48658878080, ...]
and is given by the formula:
[x^n/n!] A(x)^(n+1) = 8^(n-6) * (n+1)^(n-7) * (16807*n^6 + 143031*n^5 + 525875*n^4 + 1074745*n^3 + 1294846*n^2 + 876856*n + 262144) for n>=0.
		

Crossrefs

Programs

  • Mathematica
    Flatten[{1,1,Table[Sum[8^k * n!/k! * Binomial[8*n-k-8, n-k] * (k-1)/(n-1),{k,0,n}],{n,2,20}]}] (* Vaclav Kotesovec, Dec 07 2014 *)
  • PARI
    {a(n) = local(G=1);for(i=1,n,G=1+x*G^8 +x*O(x^n)); n!*polcoeff(exp(8*x*G^7)/G^7, n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    {a(n) = if(n==0|n==1, 1, sum(k=0, n, 8^k * n!/k! * binomial(8*n-k-8,n-k) * (k-1)/(n-1) ))}
    for(n=0, 20, print1(a(n), ", "))

Formula

Let G(x) = 1 + x*G(x)^8 be the g.f. of A007556, then the e.g.f. A(x) of this sequence satisfies:
(1) A'(x)/A(x) = G(x)^7.
(2) A'(x) = exp(8*x*G(x)^7).
(3) A(x) = exp( Integral G(x)^7 dx ).
(4) A(x) = exp( Sum_{n>=1} A234466(n-1)*x^n/n ), where A234466(n-1)(n) = binomial(8*n-2,n)/(7*n-1).
(5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251588.
(6) A(x) = Sum_{n>=0} A251588(n)*(x/A(x))^n/n! and
(7) [x^n/n!] A(x)^(n+1) = (n+1)*A251588(n),
where A251588(n) = 8^(n-6) * (n+1)^(n-8) * (16807*n^6 + 143031*n^5 + 525875*n^4 + 1074745*n^3 + 1294846*n^2 + 876856*n + 262144).
a(n) = Sum_{k=0..n} 8^k * n!/k! * binomial(8*n-k-8, n-k) * (k-1)/(n-1) for n>1.
Recurrence: 7*(7*n-13)*(7*n-12)*(7*n-11)*(7*n-10)*(7*n-9)*(7*n-8)*(4096*n^6 - 66048*n^5 + 446400*n^4 - 1620808*n^3 + 3339890*n^2 - 3711613*n + 1743218)*a(n) = 128*(536870912*n^13 - 14831058944*n^12 + 188986949632*n^11 - 1471608258560*n^10 + 7817645654016*n^9 - 29941451735040*n^8 + 85134250240000*n^7 - 182149348773632*n^6 + 293626158621632*n^5 - 352753169299376*n^4 + 307548490429492*n^3 - 184675145918224*n^2 + 68635535585133*n - 11961900200250)*a(n-1) + 16777216*(4096*n^6 - 41472*n^5 + 177600*n^4 - 413768*n^3 + 556826*n^2 - 414321*n + 135135)*a(n-2). - Vaclav Kotesovec, Dec 07 2014
a(n) ~ 8^(8*(n-1)-1/2) / 7^(7*(n-1)-1/2) * n^(n-2) / exp(n-1). - Vaclav Kotesovec, Dec 07 2014

A234462 a(n) = 3*binomial(8*n+3,n)/(8*n+3).

Original entry on oeis.org

1, 3, 27, 325, 4488, 67158, 1059380, 17346582, 292046040, 5023824887, 87915626370, 1560176040519, 28011228029512, 507874087572600, 9286024289123268, 171026036066072924, 3169969149156895800, 59085490354010508600, 1106795192170066119435
Offset: 0

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Author

Tim Fulford, Dec 26 2013

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(n*p + r, n)/(n*p + r), this is the case p = 8, r = 3.

Crossrefs

Programs

  • Magma
    [3*Binomial(8*n+3, n)/(8*n+3): n in [0..30]]; // Vincenzo Librandi, Dec 26 2013
  • Mathematica
    Table[3 Binomial[8 n + 3, n]/(8 n + 3), {n, 0, 40}] (* Vincenzo Librandi, Dec 26 2013 *)
  • PARI
    a(n) = 3/(8*n+3)*binomial(8*n+3,n);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(8/3))^3+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: A(x) = {1 + x*A(x)^(p/r)}^r, where p = 8, r = 3.
A(x^2) = 1/x * series reversion (x/C(x^2)^3), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. A(x)^(1/3) is the o.g.f. for A007556. - Peter Bala, Oct 14 2015
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