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 11-18 of 18 results.

A233669 a(n) = 7*binomial(5*n+7, n)/(5*n+7).

Original entry on oeis.org

1, 7, 56, 490, 4550, 44051, 439824, 4496388, 46834095, 495260150, 5303177880, 57385471962, 626548297648, 6893781417320, 76362138282400, 850867975145160, 9530515916642385, 107249427630005661, 1211964598880990640, 13747501038498835300
Offset: 0

Views

Author

Tim Fulford, Dec 14 2013

Keywords

Comments

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

Crossrefs

Programs

  • Magma
    [7*Binomial(5*n+7,n)/(5*n+7): n in [0..30]];
  • Mathematica
    Table[7 Binomial[5 n + 7, n]/(5 n + 7), {n, 0, 30}]
  • PARI
    a(n) = 7*binomial(5*n+7,n)/(5*n+7);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(5/7))^7+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, here p=5, r=7.
From Ilya Gutkovskiy, Sep 14 2018: (Start)
E.g.f.: 4F4(7/5,8/5,9/5,11/5; 1,9/4,5/2,11/4; 3125*x/256).
a(n) ~ 7*5^(5*n+13/2)/(sqrt(Pi)*2^(8*n+31/2)*n^(3/2)). (End)

A233736 a(n) = 8*binomial(5*n + 8, n)/(5*n + 8).

Original entry on oeis.org

1, 8, 68, 616, 5850, 57536, 581196, 5995184, 62891499, 668922800, 7197169980, 78195588168, 856708896784, 9454328800896, 104997940138300, 1172624772468960, 13161188646791865, 148375147999406328, 1679436658449372744, 19078164706488179600
Offset: 0

Views

Author

Tim Fulford, Dec 15 2013

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p=5, r=8.

Crossrefs

Programs

  • Magma
    [8*Binomial(5*n+8,n)/(5*n+8): n in [0..30]]; // Vincenzo Librandi, Dec 16 2013
  • Mathematica
    Table[8 Binomial[5 n + 8, n]/(5 n + 8), {n, 0, 40}] (* Vincenzo Librandi, Dec 16 2013 *)
  • PARI
    a(n) = 8*binomial(5*n+8,n)/(5*n+8);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(5/8))^8+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, where p=5, r=8.
From Ilya Gutkovskiy, Sep 14 2018: (Start)
E.g.f.: 5F5(8/5,9/5,2,11/5,12/5; 1,9/4,5/2,11/4,3; 3125*x/256).
a(n) ~ 5^(5*n+15/2)/(sqrt(Pi)*2^(8*n+29/2)*n^(3/2)). (End)

A233737 a(n) = 9*binomial(5*n+9, n)/(5*n+9).

Original entry on oeis.org

1, 9, 81, 759, 7371, 73656, 752913, 7838298, 82832706, 886322710, 9583986555, 104568156819, 1149793519368, 12728471356944, 141747219186705, 1586867219265060, 17848735288114995, 201607141031660871, 2285899896222757346, 26008027474874327190, 296840444852078282610, 3397721117411729991960
Offset: 0

Views

Author

Tim Fulford, Dec 15 2013

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p=5, r=9.

Crossrefs

Programs

  • Magma
    [9*Binomial(5*n+9,n)/(5*n+9): n in [0..30]];
  • Mathematica
    Table[9 Binomial[5 n + 9, n]/(5 n + 9), {n, 0, 30}]
  • PARI
    a(n) = 9*binomial(5*n+9,n)/(5*n+9);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(5/9))^9+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, here p=5, r=9.
From Ilya Gutkovskiy, Sep 14 2018: (Start)
E.g.f.: 5F5(9/5,2,11/5,12/5,13/5; 1,5/2,11/4,3,13/4; 3125*x/256).
a(n) ~ 9*5^(5*n+17/2)/(sqrt(Pi)*2^(8*n+39/2)*n^(3/2)). (End)

A143338 G.f. A(x) satisfies A(x) = 1 + x*A(x)^3*A(-x).

Original entry on oeis.org

1, 1, 2, 8, 26, 127, 478, 2536, 10250, 56900, 239880, 1370272, 5940054, 34607146, 153018932, 904441648, 4058644842, 24254529036, 110096276440, 663665021280, 3040205250984, 18455364854839, 85176971647470, 520059936017128
Offset: 0

Views

Author

Paul D. Hanna, Aug 09 2008

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 26*x^4 + 127*x^5 + 478*x^6 +...
Compare bisections of A(x)^2, A(x)^2*A(-x), and A(x)^4*A(-x)^2:
A(x)^2 = 1 + 2*x + 5*x^2 + 20*x^3 + 72*x^4 + 338*x^5 + 1378*x^6 + 6952*x^7 +...
A(x)^2*A(-x) = 1 + x + 5*x^2 + 11*x^3 + 72*x^4 + 191*x^5 + 1378*x^6 + 3979*x^7 +...
A(x)^4*A(-x)^2 = 1 + 2*x + 11*x^2 + 32*x^3 + 191*x^4 + 636*x^5 + 3979*x^6 +...
Related expansions:
A(x)^3 = 1 + 3*x + 9*x^2 + 37*x^3 + 144*x^4 + 669*x^5 + 2882*x^6 + 14229*x^7 +...
A(x)^3*A(-x) = 1 + 2*x + 8*x^2 + 26*x^3 + 127*x^4 + 478*x^5 + 2536*x^6 +...
A(x)^3*A(-x)^2 = 1 + x + 8*x^2 + 14*x^3 + 127*x^4 + 264*x^5 + 2536*x^6 +...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+x+O(x^21));for(i=0,n,A=1+x*A^3*subst(A,x,-x));polcoeff(A,n)}

Formula

a(0) = 1; a(n) = Sum_{i, j, k, l>=0 and i+j+k+l=n-1} (-1)^i * a(i) * a(j) * a(k) * a(l). - Seiichi Manyama, Jul 08 2025

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

Views

Author

N. J. A. Sloane, Jul 12 2011

Keywords

Comments

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

Crossrefs

Column k=11 of A369929 and k=12 of A370062.
Cf. A143048.

Programs

  • PARI
    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

Formula

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

Extensions

a(11) onwards from Andrew Howroyd, Jan 26 2024
a(0)=1 prepended by Andrew Howroyd, Feb 08 2024

A369473 Number of chiral pairs of polyominoes composed of n hexagonal cells of the hyperbolic regular tiling with Schläfli symbol {6,oo}.

Original entry on oeis.org

7, 50, 448, 3810, 34200, 314655, 2982040, 28897440, 285577500, 2868769045, 29227672960, 301429078080, 3141983233130, 33059729519325, 350763428176480, 3749420512083472, 40348040467611800, 436827334389425980
Offset: 4

Views

Author

Robert A. Russell, Jan 23 2024

Keywords

Comments

A stereographic projection of the {6,oo} tiling on the Poincaré disk can be obtained via the Christensson link. Each member of a chiral pair is a reflection but not a rotation of the other.

Crossrefs

Polyominoes: A221184(n-1) (oriented), A004127 (unoriented), A143546 (achiral), A369471 {5,oo}.

Programs

  • Mathematica
    p=6; Table[(Binomial[(p-1)n, n]/(((p-2)n+1)((p-2)n+2))-If[OddQ[n], If[OddQ[p], Binomial[(p-1)n/2, (n-1)/2]/n, (p+1)Binomial[((p-1)n-1)/2, (n-1)/2]/((p-2)n+2)-Binomial[((p-1)n+1)/2, (n-1)/2]/((p-1)n+1)], Binomial[(p-1)n/2, n/2]/((p-2)n+2)]+DivisorSum[GCD[p, n-1], EulerPhi[#]Binomial[((p-1)n+1)/#, (n-1)/#]/((p-1)n+1)&, #>1&])/2, {n, 4, 30}]

Formula

a(n) = A221184(n-1) - A004127(n) = (A221184(n-1) - A143546(n)) / 2 = A004127(n) - A143546(n).

A192894 Number of symmetric 13-ary factorizations of the n-cycle (1,2...n).

Original entry on oeis.org

1, 1, 1, 7, 13, 112, 247, 2310, 5525, 53998, 135408, 1360289, 3518515, 36017352, 95223414, 988172368, 2655417765, 27844071255, 75769712590, 801012669457, 2201663313200, 23428926096576, 64924369564353, 694644371065372, 1938034271677595, 20829931845958872, 58448142042957576
Offset: 0

Views

Author

N. J. A. Sloane, Jul 12 2011

Keywords

Comments

The six sequences displayed in Table 1 of the Bousquet-Lamathe reference are A047749, A143546, A143547, A143554, A192893, A192894. From this one should be able to guess a g.f.

Crossrefs

Column k=13 of A369929 and k=14 of A370062.
Cf. A143049.

Formula

From Seiichi Manyama, Jul 07 2025: (Start)
G.f. A(x) satisfies A(x) = 1/( 1 - x*(A(x)*A(-x))^6 ).
G.f. A(x) satisfies A(x)*A(-x) = (A(x) + A(-x))/2 = G(x^2), where G(x) = 1 + x*G(x)^13.
a(0) = 1; a(n) = Sum_{x_1, x_2, ..., x_7>=0 and x_1+2*(x_2+x_3+...+x_7)=n-1} a(x_1) * Product_{k=2..7} a(2*x_k). (End)
a(0) = 1; a(n) = Sum_{x_1, x_2, ..., x_13>=0 and x_1+x_2+...+x_13=n-1} (-1)^(x_1+x_2+x_3+x_4+x_5+x_6) * Product_{k=1..13} a(x_k). - Seiichi Manyama, Jul 09 2025

Extensions

a(11) onwards from Andrew Howroyd, Jan 26 2024
a(0)=1 prepended by Seiichi Manyama, Jul 07 2025

A385687 E.g.f. A(x) satisfies A(x) = exp( x*((A(x) + A(-x))/2)^2 ).

Original entry on oeis.org

1, 1, 1, 7, 25, 341, 2161, 44115, 404209, 11010025, 132273601, 4508793983, 67085545033, 2747071330173, 48765277295281, 2331905267846731, 48106649137922017, 2631174441142423505, 61862217319644572161, 3809106344377237185399, 100542158725584301036921
Offset: 0

Views

Author

Seiichi Manyama, Jul 06 2025

Keywords

Crossrefs

Programs

  • Mathematica
    terms = 21;  A[] = 1; Do[A[x] = Exp[x*((A[x] + A[-x])/2)^2] + O[x]^terms // Normal, terms]; CoefficientList[A[x], x]Range[0,terms-1]! (* Stefano Spezia, Jul 07 2025 *)

Formula

E.g.f. A(x) satisfies A(-x) = 1/A(x).
a(0) = 1; a(n) = (n-1)! * Sum_{i, j, k>=0 and i+2*j+2*k=n-1} (n-i) * a(i) * a(2*j) * a(2*k)/(i! * (2*j)! * (2*k)!).
Previous Showing 11-18 of 18 results.