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 10 results.

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

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

A234571 a(n) = 4*binomial(10*n+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

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 = 8.

Crossrefs

Programs

  • Magma
    [4*Binomial(10*n+8, n)/(5*n+4): n in [0..30]];
  • Mathematica
    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);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(5/4))^8+x*O(x^n)); polcoeff(B, n)}
    

Formula

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) = 7*binomial(9*n+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

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 = 7.

Crossrefs

Programs

  • Magma
    [7*Binomial(9*n+7, n)/(9*n+7): n in [0..30]]; // Vincenzo Librandi, Dec 27 2013
  • Mathematica
    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);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(9/7))^7+x*O(x^n)); polcoeff(B, n)}
    

Formula

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

A233835 a(n) = 8*binomial(7*n + 8, n)/(7*n + 8).

Original entry on oeis.org

1, 8, 84, 1008, 13090, 179088, 2542512, 37106784, 553270671, 8391423040, 129058047580, 2008018827360, 31550226597162, 499892684834368, 7978140653296800, 128138773298754240, 2069603881026760323, 33593111381834512200, 547698081896206040800, 8965330544164089648000, 147285313888568167177866
Offset: 0

Views

Author

Tim Fulford, Dec 16 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 = 7, r = 8.

Crossrefs

Cf. A000245 (k = 3), A006629 (k = 4), A196678 (k = 5), A233668 (k = 6), A233743 (k = 7), A234467 (k = 9), A232265 (k = 10), A229963 (k = 11).

Programs

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

Formula

G.f. satisfies: A(x) = {1 + x*A(x)^(p/r)}^r, where p = 7, r = 8.
From _Peter Bala, Oct 16 2015: (Start)
O.g.f. A(x) = 1/x * series reversion (x*C(-x)^8), 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)^(1/8) is the o.g.f. for A002296. (End)

A235339 a(n) = 9*binomial(11*n+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

Views

Author

Tim Fulford, Jan 06 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 = 9.

Crossrefs

Programs

  • Magma
    [9*Binomial(11*n+9, n)/(11*n+9): n in [0..30]];
  • Mathematica
    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);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(11/9))^9+x*O(x^n)); polcoeff(B, n)}
    

Formula

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

A333097 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(5*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 6, 76, 1101, 16876, 266881, 4305247, 70414133, 1163355884, 19369868385, 324486751951, 5462851474614, 92346622131103, 1566455916243068, 26649562889363259, 454528917186429226, 7769463895152496364, 133064720735632286722, 2282869928179537263601, 39225214245206751480102
Offset: 0

Views

Author

Peter Bala, Mar 15 2020

Keywords

Comments

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), A333095 (m = 3), A333096 (m = 4).
In general, for m > 0 and c(x)^(m*n) is a(n) ~ m * (m+2)^((m+2)*n + 3/2) / (((m+1)*(m+2)+1) * sqrt(2*Pi*n) * (m+1)^((m+1)*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020

Examples

			n-th order Taylor polynomial of c(x)^(5*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^5 = 1 + 5*x + O(x^2)
  n = 2: c(x)^10 = 1 + 10*x + 65*x^2 + O(x^3)
  n = 3: c(x)^15 = 1 + 15*x + 135*x^2 + 950*x^3 + O(x^4)
  n = 4: c(x)^20 = 1 + 20*x + 230*x^2 + 2000*x^3 + 14625*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 5 = 6, a(2) = 1 + 10 + 65 = 76, a(3) = 1 + 15 + 135 + 950 = 1101 and a(4) = 1 + 20 + 230 + 2000 + 14625 = 16876.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(5*n), n >= 0, in descending powers of x begins
                                                row sums
  n = 0 |     1                                     1
  n = 1 |     5        1                            6
  n = 2 |    65       10      1                    76
  n = 3 |   950      135     15    1             1101
  n = 4 | 14625     2000    230   20    1       16876
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 1566455916243068 - 6 = 2*(13^3)*104701*3404923 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 11627033261887689372357353 - 1101 = (2^2)*(7^4)*19*29* 2197177609353575713 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 1034770243516278817426081673131 - 266881 = 2*3*(5^7)*31* 13305359*5351978496238483 == 0 ( mod 5^6 ).
		

Crossrefs

Programs

  • Maple
    seq(add(5*n/(5*n+k)*binomial(5*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)^(5*n), x, 151):
    seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);
  • Mathematica
    Join[{1}, Table[5*Binomial[7*n-1, n] * HypergeometricPFQ[{1, -6*n, -n}, {1/2 - 7*n/2, 1 - 7*n/2}, 1/4]/6, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

Formula

a(n) = Sum_{k = 0..n} 5*n/(5*n+k)*binomial(5*n+2*k-1,k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^5(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 5*x + 45*x^2 + 500*x^3 + 6200*x^4 + ... = (1/x)*Revert( x/c^5(x) ) is the o.g.f. of A233834.
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 * 7^(7*n + 3/2) / (43 * sqrt(Pi*n) * 2^(6*n + 1) * 3^(6*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 5*n/(5*n+2*k)*binomial(5*n+2*k, k) for n >= 1. - Peter Bala, Apr 20 2024

A233833 a(n) = 3*binomial(7*n+3, n)/(7*n+3).

Original entry on oeis.org

1, 3, 24, 253, 3045, 39627, 543004, 7718340, 112752783, 1682460520, 25533901536, 392912889915, 6116090678334, 96133810101609, 1523687678528400, 24324750346691480, 390786855500604195, 6313161418594235271, 102494297789621214400, 1671366110239940499000
Offset: 0

Views

Author

Tim Fulford, Dec 16 2013

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p=7, r=3.

Crossrefs

Programs

  • Magma
    [3*Binomial(7*n+3, n)/(7*n+3): n in [0..30]];
  • Mathematica
    Table[3 Binomial[7 n + 3, n]/(7 n + 3), {n, 0, 30}]
  • PARI
    a(n)=3*binomial(7*n+3,n)/(7*n+3);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(7/3))^3+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p=7, r=3.
From Ilya Gutkovskiy, Sep 14 2018: (Start)
E.g.f.: 6F6(3/7,4/7,5/7,6/7,8/7,9/7; 2/3,5/6,1,7/6,4/3,3/2; 823543*x/46656).
a(n) ~ 7^(7*n+5/2)/(sqrt(Pi)*3^(6*n+5/2)*4^(3*n+2)*n^(3/2)). (End)

A233907 9*binomial(7*n+9, n)/(7*n+9).

Original entry on oeis.org

1, 9, 99, 1218, 16065, 222138, 3178140, 46656324, 698868216, 10639125640, 164128169205, 2560224004884, 40314178429707, 639948824981928, 10230035192533800, 164541833894991240, 2660919275605834701, 43239781879996449825, 705687913212419321800, 11561996402992103418000, 190100812111989146008641
Offset: 0

Views

Author

Tim Fulford, Dec 17 2013

Keywords

Comments

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

Crossrefs

Programs

  • Magma
    [9*Binomial(7*n+9, n)/(7*n+9): n in [0..30]];
  • Mathematica
    Table[9 Binomial[7 n + 9, n]/(7 n + 9), {n, 0, 30}]
  • PARI
    a(n) = 9*binomial(7*n+9,n)/(7*n+9);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(7/9))^9+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p=7, r=9.
D-finite with recurrence 72*n*(6*n+5)*(3*n+2)*(2*n+3)*(3*n+4)*(6*n+7)*a(n) -7*(7*n+4)*(7*n+8)*(7*n+5)*(7*n+2)*(7*n+6)*(7*n+3)*a(n-1)=0. - R. J. Mathar, Nov 22 2024

A386380 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/6)} a(6*k) * a(n-1-6*k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 15, 24, 34, 45, 57, 70, 154, 253, 368, 500, 650, 819, 1827, 3045, 4495, 6200, 8184, 10472, 23562, 39627, 59052, 82251, 109668, 141778, 320866, 543004, 814506, 1142295, 1533939, 1997688, 4540200, 7718340, 11633440, 16398200, 22137570
Offset: 0

Views

Author

Seiichi Manyama, Jul 20 2025

Keywords

Crossrefs

Programs

  • Maple
    A386380 := proc(n)
        option remember ;
        if n = 0 then
            1;
        else
            add(procname(6*k)*procname(n-1-6*k),k=0..floor((n-1)/6)) ;
        end if;
    end proc:
    seq(A386380(n),n=0..80) ; # R. J. Mathar, Jul 30 2025
  • PARI
    apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
    a(n) = apr(n\6, 7, n%6+1);

Formula

For k=0..5, a(6*n+k) = (k+1) * binomial(7*n+k+1,n)/(7*n+k+1).
G.f. A(x) satisfies A(x) = 1/(1 - x * Product_{k=0..5} A(w^k*x)), where w = exp(Pi*i/3).
Showing 1-10 of 10 results.