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.

Previous Showing 11-20 of 30 results. Next

A062994 Eighth column of triangle A062993 (without leading zeros). A Pfaff-Fuss or 9-Raney sequence.

Original entry on oeis.org

1, 1, 9, 117, 1785, 29799, 527085, 9706503, 184138713, 3573805950, 70625252863, 1416298046436, 28748759731965, 589546754316126, 12195537924351375, 254184908607118800, 5332692942907262361
Offset: 0

Views

Author

Wolfdieter Lang, Jul 12 2001

Keywords

Comments

See Graham et al., Hilton and Pedersen, Hoggat and Bicknell, Frey and Sellers references given in A062993.
Essentially the same as A059967. a(n), n>=1, enumerates 9-ary trees (rooted, ordered, incomplete) with n vertices (including the root).
These numbers appear in a formula on p. 24 of Gross et al. for b = -2 or 4. For b = -1 or 3, see A002293.- Tom Copeland, Dec 24 2019
This is instance k = 9 of the generalized Catalan family {C(k, n)}_{n>=0} given in a comment of A130564. - Wolfdieter Lang, Feb 05 2024

Examples

			There are a(2)=9 9-ary trees (vertex degree <=9 and 9 possible branchings) with 2 vertices (one of them the root). Adding one more branch (one more vertex) to these 9 trees yields 9*9 + binomial(9,2) = 117 = a(3) such trees.
		

References

  • 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.

Crossrefs

Programs

  • Maple
    seq(binomial(9*k+1,k)/(8*k+1),k=0..30);
    n:=30: G:=series(RootOf(g = 1+x*g^9, g),x=0,n+1): seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 01 2015
  • Mathematica
    Table[Binomial[9n,n]/(8n+1),{n,0,30}] (* Harvey P. Dale, Oct 28 2012 *)
  • PARI
    { for (n=0, 100, write("b062994.txt", n, " ", binomial(9*n, n)/(8*n + 1)) ) } \\ Harry J. Smith, Aug 15 2009

Formula

a(n) = A062993(n+7, 7) = binomial(9*n, n)/(8*n+1).
G.f.: RootOf((_Z^9)*x-_Z+1) (Maple notation, from ECS, see links for A007556).
Recurrence: a(0) = 1; a(n) = Sum_{i1+i2+..+i9=n-1} a(i1)*a(i2)*...*a(i9) for n>=1. - Robert FERREOL, Apr 01 2015
From Ilya Gutkovskiy, Jan 16 2017: (Start)
O.g.f.: 8F7(1/9,2/9,1/3,4/9,5/9,2/3,7/9,8/9; 1/4,3/8,1/2,5/8,3/4,7/8,9/8; 387420489*x/16777216).
E.g.f.: 8F8(1/9,2/9,1/3,4/9,5/9,2/3,7/9,8/9; 1/4,3/8,1/2,5/8,3/4,7/8,1,9/8; 387420489*x/16777216).
a(n) ~ 3^(18*n+1)/(sqrt(Pi)*2^(24*n+5)*n^(3/2)). (End)
D-finite with recurrence: 128*n*(8*n-5)*(4*n-1)*(8*n+1)*(2*n-1)*(8*n-1)*(4*n-3)*(8*n-3)*a(n) -81*(9*n-7)*(9*n-5)*(3*n-1)*(9*n-1)*(9*n-8)*(3*n-2)*(9*n-4)*(9*n-2)*a(n-1)=0. - R. J. Mathar, Feb 20 2020
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^17). - Seiichi Manyama, Jun 16 2025

Extensions

9-ary tree comments and Pólya and G. Szegő reference from Wolfdieter Lang, Sep 14 2007

A230388 a(n) = binomial(11*n+1,n)/(11*n+1).

Original entry on oeis.org

1, 1, 11, 176, 3311, 68211, 1489488, 33870540, 793542167, 19022318084, 464333035881, 11502251937176, 288417894029200, 7306488667126803, 186719056586568660, 4807757550367267056, 124609430032127192295, 3248403420844673986345
Offset: 0

Views

Author

Tim Fulford, Jan 01 2014

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p=11, r=1. Interesting property when r=1, a(n+1,p,1) = a(n,p,p) for n>=0.
This is also instance k = 10 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

Crossrefs

Programs

  • Magma
    [Binomial(11*n+1,n)/(11*n+1): n in [0..30]]; // Vincenzo Librandi, Jan 01 2014
  • Maple
    seq(binomial(11*k+1,k)/(11*k+1),k=0..30); # Robert FERREOL, Apr 01 2015
    n:=30:G:=series(RootOf(g = 1+x*g^11, g),x=0,n+1):seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 01 2015
  • Mathematica
    Table[Binomial[11 n + 1, n]/(11 n + 1), {n, 0, 30}] (* Vincenzo Librandi, Jan 01 2014 *)
  • PARI
    a(n) = binomial(11*n+1,n)/(11*n+1);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(11))+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, with p=11, r=1.
From Robert FERREOL, Apr 01 2015: (Start)
a(n) = binomial(11*n,n)/(10*n+1) = A062993(n+9, 9).
a(0) = 1; a(n) = Sum_{i1+i2+..i11=n-1} a(i1)*a(i2)*...*a(i11) for n>=1.
(End)
O.g.f.: hypergeometric([1,...,10]/11,[2,...,9,11]/10,(11^11/10^10)*x). For the e.g.f. put an extra 1 = 10/10 into the second part. - Wolfdieter Lang, Feb 05 2024
a(n) ~ (11^11/10^10)^n*sqrt(11/(2*Pi*(10*n)^3)). - Robert A. Russell, Jul 15 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^21). - Seiichi Manyama, Jun 16 2025

A234466 a(n) = 7*binomial(8*n+7,n)/(8*n+7).

Original entry on oeis.org

1, 7, 77, 1015, 14763, 228459, 3689595, 61474519, 1048927880, 18236463245, 321899509386, 5753527081211, 103922382296180, 1893943017506925, 34783258504651434, 643111366544129175, 11960812088346090200, 223614812152492437432, 4200107505573406222425
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=7.

Crossrefs

Programs

  • Magma
    [7*Binomial(8*n+7, n)/(8*n+7): n in [0..30]]; // Vincenzo Librandi, Dec 26 2013
  • Mathematica
    Table[7 Binomial[8 n + 7, n]/(8 n + 7), {n, 0, 40}] (* Vincenzo Librandi, Dec 26 2013 *)
  • PARI
    a(n) = 7*binomial(8*n+7,n)/(8*n+7);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(8/7))^7+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p=8, r=7.
E.g.f.: hypergeom([7, 9, 10, 11, 12, 13, 14]/8, [8, 9, 10, 11, 12, 13, 14]/7, (8^8/7^7)*x). Cf.: Ilya Gutkovskiy in A118971. - Wolfdieter Lang, Feb 06 2020
D-finite with recurrence: +7*(7*n+4)*(7*n+1)*(7*n+5)*(7*n+2)*(7*n+6)*(7*n+3)*(n+1)*a(n) -128*(8*n+3)*(4*n+3)*(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
From Wolfdieter Lang, Feb 15 2024: (Start)
a(n) = binomial(8*n + 6, n+1)/(7*n + 6). This is instance k = 7 of c(k, n+1) given in a comment in A130564.
The compositional inverse of y*(1 - y)^7 is x*G(x), where G is the o.g.f.. That is, G(x)*(1 - x*G(x))^7 = 1. This is equivalent to the formula of the first line above with B = G. Take A = B^(1/7) then A*(1 - x*B) = 1 or B*(1 - x*B)^7 = 1.
The o.g.f is G(x) = 8F7([7..14]/8, [8..14]/7; (8^8/7^7)*x) = (7/(8*x))*(1 - 7F6([-1,1,2,3,4,5,6]/8, [1,2,3,4,5,6]/7; (8^8/7^7)*x)). See the e.g.f. above.(End)

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

Views

Author

Paul D. Hanna, Apr 29 2012

Keywords

Comments

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

Examples

			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 +...
		

Crossrefs

Programs

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

Formula

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

A212071 G.f. satisfies: A(x) = (1 + x*A(x)^3)^2.

Original entry on oeis.org

1, 2, 13, 114, 1150, 12586, 145299, 1741844, 21475146, 270570300, 3468352701, 45089941936, 593082894768, 7878407177270, 105542811922950, 1424267372100456, 19343105144742098, 264182048662182420, 3626176386241346070, 49995713597946235350, 692084935397470961346
Offset: 0

Views

Author

Paul D. Hanna, Apr 29 2012

Keywords

Comments

The two parameter Fuss-Catalan sequence is A(n,p,r) := r*binomial(n*p + r, n)/(n*p + r), with o.g.f. G(p,r,x) = G(x) satisfying G(x) = {1 + x*G(x)^(p/r)}^r ; this is the case p = 6, r = 2. - Peter Bala, Oct 14 2015

Examples

			G.f.: A(x) = 1 + 2*x + 13*x^2 + 114*x^3 + 1150*x^4 + 12586*x^5 +...
Related expansions:
A(x)^3 = 1 + 6*x + 51*x^2 + 506*x^3 + 5481*x^4 +...+ A002295(n+1)*x^n +...
A(x)^(1/2) = 1 + x + 6*x^2 + 51*x^3 + 506*x^4 +...+ A002295(n)*x^n +...
		

Crossrefs

Programs

  • Mathematica
    Table[c=6n+2;(2*Binomial[c,n])/c,{n,0,20}] (* Harvey P. Dale, Oct 14 2013 *)
  • PARI
    {a(n)=binomial(6*n+2,n) * 2/(6*n+2)}
    for(n=0, 40, print1(a(n), ", "))
    
  • PARI
    {a(n)=local(A=1+2*x); for(i=1, n, A=(1+x*A^3)^2+x*O(x^n)); polcoeff(A, n)}

Formula

a(n) = 2*binomial(6*n+2,n)/(6*n+2).
G.f.: A(x) = G(x)^2 where G(x) = 1 + x*G(x)^6 is the g.f. of A002295.
a(n) = 2*binomial(6n+1, n-1)/n for n>0, a(0)=1. [Bruno Berselli, Jan 19 2014]
A(x^2) = 1/x * series reversion (x/C(x^2)^2), where C(x) = (1 - sqrt( 1 - 4*x))/(2*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-2)*(5*n-1)*a(n) -72*(6*n-1)*(3*n-1)*(2*n-1)*(3*n-2)*(6*n+1)*a(n-1)=0. - R. J. Mathar, Nov 22 2024

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)

A234573 a(n) = 9*binomial(10*n+9,n)/(10*n+9).

Original entry on oeis.org

1, 9, 126, 2109, 38916, 763686, 15636192, 330237765, 7141879503, 157366449604, 3520256293710, 79735912636302, 1825080422272800, 42148579533938784, 980892581545169496, 22980848343194476245, 541581608172776494554, 12829884648994115426295, 305349921559399354716430
Offset: 0

Views

Author

Tim Fulford, Dec 28 2013

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), where p=10, r=9.

Crossrefs

Programs

  • Magma
    [9*Binomial(10*n+9, n)/(10*n+9): n in [0..30]];
  • Mathematica
    Table[9 Binomial[10 n + 9, n]/(10 n + 9), {n, 0, 30}]
  • PARI
    a(n) = 9*binomial(10*n+9,n)/(10*n+9);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(10/9))^9+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p=10, r=9.
From Wolfdieter Lang, Feb 06 2020: (Start)
G.f.: hypergeom([9, 10, ..., 18]/10, [10, 11, ..., 18]/9, (10^10/9^9)*x).
E.g.f.: hypergeom([9, 11, 12, ..., 18]/10, [10, 11, ..., 18]/9, (10^10/9^9) * x). Cf. Ilya Gutkovsky in A118971. (End)
a(n) = binomial(10*n + 8 , n+1)/(9*n + 8) which is instance k = 9 of c(k, n+1) given in a comment in A130564. x*B(x), with the above given g.f. B(x), is the compositional inverse of y*(1 - y)^9, hence B(x)*(1 - x*B(x))^9 = 1. For another formula for B(x) involving the hypergeometric function 9F8 see the analog formula in A234513. - Wolfdieter Lang, Feb 15 2024

A130565 Member k=6 of a family of generalized Catalan numbers.

Original entry on oeis.org

1, 6, 57, 650, 8184, 109668, 1533939, 22137570, 327203085, 4928006512, 75357373305, 1166880131820, 18259838103852, 288308609783760, 4587430875645660, 73484989079268690, 1184104656043939071
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.
For the members of the family C(k,n), k=2..9, see A130564.
The family c(k,n):=binomial((k+1)*n-2,n)/(k*n-1), n>=1, has the members A006013, A006632, A118971,for k=2,3,4 respectively (but the offset there is 0) and A130564 for k=5.

Crossrefs

Cf. k=5 member A130564. A006013, A006632, A118971,

Programs

  • Mathematica
    Table[Binomial[7n-2,n]/(6n-1),{n,20}] (* Harvey P. Dale, Feb 25 2013 *)

Formula

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

A251576 E.g.f.: exp(6*x*G(x)^5) / G(x)^5 where G(x) = 1 + x*G(x)^6 is the g.f. of A002295.

Original entry on oeis.org

1, 1, 6, 96, 2736, 115056, 6455376, 454666176, 38610711936, 3842344221696, 438721154343936, 56549927146392576, 8123473514799876096, 1287034084022760677376, 222964032114987212998656, 41930788886197036399190016, 8507629742037888427727486976, 1852490637585980898960109142016
Offset: 0

Views

Author

Paul D. Hanna, Dec 06 2014

Keywords

Examples

			E.g.f.: A(x) = 1 + x + 6*x^2/2! + 96*x^3/3! + 2736*x^4/4! + 115056*x^5/5! +...
such that A(x) = exp(6*x*G(x)^5) / G(x)^5
where G(x) = 1 + x*G(x)^6 is the g.f. of A002295:
G(x) = 1 + x + 6*x^2 + 51*x^3 + 506*x^4 + 5481*x^5 + 62832*x^6 +...
Note that
A'(x) = exp(6*x*G(x)^5) = 1 + 6*x + 96*x^2/2! + 2736*x^3/3! +...
LOGARITHMIC DERIVATIVE.
The logarithm of the e.g.f. begins:
log(A(x)) = x + 5*x^2/2 + 40*x^3/3 + 385*x^4/4 + 4095*x^5/5 + 46376*x^6/6 +...
and so A'(x)/A(x) = G(x)^5.
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,   6,   96,  2736,  115056,   6455376,  454666176, ...];
n=2: [1, 2,  14,  228,  6456,  268992,  14968224, 1047087648, ...];
n=3: [1, 3,  24,  402, 11376,  470808,  26011584, 1808151552, ...];
n=4: [1, 4,  36,  624, 17736,  730944,  40143456, 2774490624, ...];
n=5: [1, 5,  50,  900, 25800, 1061400,  58017600, 3989340000, ...];
n=6: [1, 6,  66, 1236, 35856, 1475856,  80395056, 5503484736, ...];
n=7: [1, 7,  84, 1638, 48216, 1989792, 108156384, 7376303088, ...];
n=8: [1, 8, 104, 2112, 63216, 2620608, 142314624, 9676910592, ...]; ...
in which the main diagonal begins (see A251586):
[1, 2, 24, 624, 25800, 1475856, 108156384, 9676910592, ...]
and is given by the formula:
[x^n/n!] A(x)^(n+1) = 6^(n-4) * (n+1)^(n-5) * (125*n^4 + 810*n^3 + 2095*n^2 + 2586*n + 1296) for n>=0.
		

Crossrefs

Programs

  • Mathematica
    Flatten[{1,1,Table[Sum[6^k * n!/k! * Binomial[6*n-k-6, 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^6 +x*O(x^n)); n!*polcoeff(exp(6*x*G^5)/G^5, n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    {a(n) = if(n==0||n==1, 1, sum(k=0, n, 6^k * n!/k! * binomial(6*n-k-6,n-k) * (k-1)/(n-1) ))}
    for(n=0, 20, print1(a(n), ", "))

Formula

Let G(x) = 1 + x*G(x)^6 be the g.f. of A002295, then the e.g.f. A(x) of this sequence satisfies:
(1) A'(x)/A(x) = G(x)^5.
(2) A'(x) = exp(6*x*G(x)^5).
(3) A(x) = exp( Integral G(x)^5 dx ).
(4) A(x) = exp( Sum_{n>=1} A130564(n)*x^n/n ), where A130564(n) = binomial(6*n-2,n)/(5*n-1).
(5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251586.
(6) A(x) = Sum_{n>=0} A251586(n)*(x/A(x))^n/n! and
(7) [x^n/n!] A(x)^(n+1) = (n+1)*A251586(n),
where A251586(n) = 6^(n-4) * (n+1)^(n-6) * (125*n^4 + 810*n^3 + 2095*n^2 + 2586*n + 1296).
a(n) = Sum_{k=0..n} 6^k * n!/k! * binomial(6*n-k-6, n-k) * (k-1)/(n-1) for n>1.
Recurrence: 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(9*n^4 - 99*n^3 + 413*n^2 - 777*n + 559)*a(n) = 72*(5832*n^9 - 113724*n^8 + 986580*n^7 - 5003586*n^6 + 16373448*n^5 - 35916483*n^4 + 52931854*n^3 - 50678109*n^2 + 28701206*n - 7357350)*a(n-1) + 46656*(9*n^4 - 63*n^3 + 170*n^2 - 212*n + 105)*a(n-2). - Vaclav Kotesovec, Dec 07 2014
a(n) ~ 6^(6*(n-1)-1/2) / 5^(5*(n-1)-1/2) * n^(n-2) / exp(n-1). - Vaclav Kotesovec, Dec 07 2014

A233743 a(n) = 7*binomial(6*n + 7, n)/(6*n + 7).

Original entry on oeis.org

1, 7, 63, 644, 7105, 82467, 992446, 12271512, 154962990, 1990038435, 25909892008, 341225775072, 4537563627415, 60842326873230, 821692714673340, 11167153485624304, 152610018401940330, 2095863415900961490, 28910564819681953485, 400379714692751795820
Offset: 0

Views

Author

Tim Fulford, Dec 15 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 = 6, r = 7.

Crossrefs

Cf. A000245 (k = 3), A006629 (k = 4), A196678 (k = 5), A233668 (k = 6), A233835 (k = 8), A234467 (k = 9), A232265 (k = 10), A229963 (k = 11).

Programs

  • Magma
    [7*Binomial(6*n+7, n)/(6*n+7): n in [0..30]]; // Vincenzo Librandi, Dec 16 2013
  • Mathematica
    Table[7 Binomial[6 n + 7, n]/(6 n + 7), {n, 0, 40}] (* Vincenzo Librandi, Dec 16 2013 *)
  • PARI
    a(n) = 7*binomial(6*n+7,n)/(6*n+7);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(6/7))^7+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: A(x) = {1 + x*A(x)^(p/r)}^r, where p = 6, r = 7.
From _Peter Bala, Oct 16 2015: (Start)
O.g.f. A(x) = 1/x * series reversion (x*C(-x)^7), 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)^(1/7) is the o.g.f. for A002295. (End)

Extensions

More terms from Vincenzo Librandi, Dec 16 2013
Previous Showing 11-20 of 30 results. Next