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 11 results. Next

A003168 Number of blobs with 2n+1 edges.

Original entry on oeis.org

1, 1, 4, 21, 126, 818, 5594, 39693, 289510, 2157150, 16348960, 125642146, 976789620, 7668465964, 60708178054, 484093913917, 3884724864390, 31348290348086, 254225828706248, 2070856216759478, 16936016649259364
Offset: 0

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Author

Keywords

Comments

a(n) is the number of ways to dissect a convex (2n+2)-gon with non-crossing diagonals so that no (2m+1)-gons (m>0) appear. - Len Smiley
a(n) is the number of plane trees with 2n+1 leaves and all non-leaves having an odd number > 1 of children. - Jordan Tirrell, Jun 09 2017
a(n) is the number of noncrossing cacti with n+1 nodes. See A361242. - Andrew Howroyd, Mar 07 2023

Examples

			a(2)=4 because we may place exactly one diagonal in 3 ways (forming 2 quadrilaterals), or not place any (leaving 1 hexagon).
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A049124 (no 2m-gons).
Row sums of A102537, A243662. Column 2 of A336573.

Programs

  • Haskell
    import Data.List (transpose)
    a003168 0 = 1
    a003168 n = sum (zipWith (*)
       (tail $ a007318_tabl !! n)
       ((transpose $ take (3*n+1) a007318_tabl) !! (2*n+1)))
       `div` fromIntegral n
    -- Reinhard Zumkeller, Oct 27 2013
  • Maple
    Order := 40; solve(series((A-2*A^3)/(1-A^2),A)=x,A);
    A003168 := n -> `if`(n=0,1,A100327(n)/2): seq(A003168(n),n=0..20); # Peter Luschny, Jun 10 2017
  • Mathematica
    a[0] = 1; a[n_] = (2^(-n-1)*(3n)!* Hypergeometric2F1[-1-2n, -2n, -3n, -1])/((2n+1)* n!*(2n)!); Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jul 25 2011, after Vladimir Kruchinin *)
  • PARI
    a(n)=if(n<0,0,polcoeff(serreverse((x-2*x^3)/(1-x^2)+O(x^(2*n+2))),2*n+1))
    
  • PARI
    {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=(1+x*A)/(1-x*A)^2); sum(k=0,n,polcoeff(A^(n-k),k))} \\ Paul D. Hanna, Nov 17 2004
    
  • PARI
    seq(n) = Vec( 1 + serreverse(x/((1+2*x)*(1+x)^2) + O(x*x^n)) ) \\ Andrew Howroyd, Mar 07 2023
    

Formula

a(n) = Sum_{k=1..n} binomial(n, k)*binomial(2*n+k, k-1)/n.
G.f.: A(x) = Sum_{n>=0} a(n)*x^(2*n+1) satisfies (A-2*A^3)/(1-A^2)=x. - Len Smiley.
D-finite with recurrence 4*n*(2*n + 1)*(17*n - 22)*a(n) = (1207*n^3 - 2769*n^2 + 1850*n - 360)*a(n - 1) - 2*(17*n - 5)*(n - 2)*(2*n - 3)*a(n - 2). - Vladeta Jovovic, Jul 16 2004
G.f.: A(x) = 1/(1-G003169(x)) where G003169(x) is the g.f. of A003169. - Paul D. Hanna, Nov 17 2004
a(n) = JacobiP(n-1,1,n+1,3)/n for n > 0. - Mark van Hoeij, Jun 02 2010
a(n) = (1/(2*n+1))*Sum_{j=0..n} (-1)^j*2^(n-j)*binomial(2*n+1,j)*binomial(3*n-j,2*n). - Vladimir Kruchinin, Dec 24 2010
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) = upper left term in M^n, M = the production matrix:
1, 1
3, 3, 1
5, 5, 3, 1
7, 7, 5, 3, 1
9, 9, 7, 5, 3, 1
... (End)
a(n) ~ sqrt(14+66/sqrt(17)) * (71+17*sqrt(17))^n / (sqrt(Pi) * n^(3/2) * 2^(4*n+4)). - Vaclav Kotesovec, Jul 01 2015
From Peter Bala, Oct 05 2015: (Start)
a(n) = (1/n) * Sum_{i = 0..n} 2^(n-i-1)*binomial(2*n,i)* binomial(n,i+1).
O.g.f. = 1 + series reversion( x/((1 + 2*x)*(1 + x)^2) ).
Logarithmically differentiating the modified g.f. 1 + 4*x + 21*x^2 + 126*x^3 + 818*x^4 + ... gives the o.g.f. for A114496, apart from the initial term. (End)
G.f.: A(x) satisfies A = 1 + x*A^3/(1-x*A^2). - Jordan Tirrell, Jun 09 2017
a(n) = A100327(n)/2 for n>=1. - Peter Luschny, Jun 10 2017

A156894 a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*n+k-1,k).

Original entry on oeis.org

1, 3, 19, 138, 1059, 8378, 67582, 552576, 4563235, 37972290, 317894394, 2674398268, 22590697614, 191475925332, 1627653567916, 13870754053388, 118464647799075, 1013709715774130, 8689197042438274, 74594573994750972, 641252293546113434, 5519339268476249676, 47558930664216470628
Offset: 0

Views

Author

Paul Barry, Feb 17 2009

Keywords

Crossrefs

Programs

  • Magma
    A156894:= func< n | (&+[ Binomial(n,k)*Binomial(2*n+k-1,k): k in [0..n]]) >;
    [A156894(n): n in [0..30]]; // G. C. Greubel, Jan 06 2022
    
  • Maple
    a := n -> hypergeom([-n, 2*n], [1], -1);
    seq(round(evalf(a(n),32)), n=0..19); # Peter Luschny, Aug 02 2014
  • Mathematica
    Table[Sum[Binomial[n,k]Binomial[2n+k-1,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Nov 12 2014 *)
  • PARI
    a(n) = if (n < 1, 1, sum(k=0, n, binomial(n,k)*binomial(2*n+k-1,k)));
    vector(50, n, a(n-1)) \\ Altug Alkan, Oct 05 2015
    
  • Sage
    [round( hypergeometric([-n, 2*n], [1], -1) ) for n in (0..30)] # G. C. Greubel, Jan 06 2022

Formula

a(n) = [x^n] ((1+x)/(1-x)^2)^n.
a(n) = (4*(n+1)*(2*n+1)*A003169(n+1) - (5*n+1)*(2*n-1)*A003169(n))/(17*n + 5) for n>0. - Mark van Hoeij, Jul 14 2010
a(n) = Hypergeometric2F1([-n, 2*n], [1], -1). - Peter Luschny, Aug 02 2014
Conjecture: 64*n*(2*n-1)*a(n) -16*(89*n^2 -134*n +63)*a(n-1) +4*(661*n^2 -2619*n +2576)*a(n-2) -3*(119*n^2 -713*n +1092)*a(n-3) +6*(2*n-7)*(n-4)*a(n-4) = 0. - R. J. Mathar, Feb 05 2015
Conjecture: 16*n*(782*n +5365)*(2*n-1)*a(n) +8*(3128*n^3 -362053*n^2 +593930*n -290328)*a(n-1) -3*(726869*n^3 -5105981*n^2 +11667946*n -8715544)*a(n-2) +158*(2*n-5)*(n-3)*(391*n -764)*a(n-3) = 0. - R. J. Mathar, Feb 05 2015
Conjecture: 4*n*(2*n-1)*(17*n^2 -52*n +39)*a(n) -(1207*n^4 -4899*n^3 +6692*n^2 -3504*n +576)*a(n-1) +2*(n-2)*(2*n-3)*(17*n^2 -18*n +4)*a(n-2) = 0. - R. J. Mathar, Feb 05 2015 [the Maple command sumrecursion (binomial(n,k) * binomial(2*n+k-1,k), k, a(n)) verifies this recurrence. - Peter Bala, Oct 05 2015 ]
a(n) ~ sqrt(578 + 306*sqrt(17)) * (71 + 17*sqrt(17))^n / (17 * sqrt(Pi*n) * 2^(4*n+2)). - Vaclav Kotesovec, Feb 05 2015
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 14*x^2 + 79*x^3 + ... is the o.g.f. of A003169 (taken with offset 0). - Peter Bala, Oct 05 2015
From Peter Bala, Mar 20 2020: (Start)
a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A119259.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)

A339710 a(n) = Sum_{k=0..n} binomial(n, k)*binomial(2*n + k, k)*2^k.

Original entry on oeis.org

1, 7, 81, 1051, 14353, 201807, 2891409, 41976627, 615371169, 9089130967, 135048608401, 2016306678987, 30224723308081, 454603719479839, 6857319231939537, 103694587800440931, 1571449259865571137, 23860205774602899111, 362897293035114695121, 5527773456878667951483
Offset: 0

Views

Author

Yifan Zhang, Dec 13 2020

Keywords

References

  • Frits Beukers, Some Congruences for Apery Numbers, Mathematisch Instituut, University of Leiden, 1983, pages 1-2.

Crossrefs

Cf. A000079 (Sum(binomial(n, k))), A000984 (Sum(binomial(n, k)^2)), A026375 (Sum(binomial(n, k)*binomial(2*k, k))), A001850 (Sum(binomial(n, k)*binomial(n+k, k))), A005809 (Sum(binomial(n, k)*binomial(2*n, k))).

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]*Binomial[2n+k,k]*2^k,{k,0,n}],{n,0,20}] (* or *)
    Table[Hypergeometric2F1[-n,1+2 n,1,-2],{n,0,20}] (* Stefano Spezia, Dec 17 2020 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*binomial(2*n + k, k)*2^k); \\ Michel Marcus, Feb 12 2021

Formula

a(n) = 2F1([-n, 1 + 2*n], [1], -2), where 2F1 is the hypergeometric function. - Stefano Spezia, Dec 17 2020
From Vaclav Kotesovec, May 11 2021: (Start)
Recurrence: 3*n*(2*n - 1)*(26*n - 35)*a(n) = (2444*n^3 - 5734*n^2 + 3830*n - 729)*a(n-1) - (n-1)*(2*n - 3)*(26*n - 9)*a(n-2).
a(n) ~ sqrt(3/8 + 11/(8*sqrt(13))) * ((47 + 13*sqrt(13))/6)^n / sqrt(Pi*n). (End)

Extensions

More terms from Stefano Spezia, Dec 17 2020

A156886 a(n) = Sum_{k=0..n} C(n,k)*C(3*n+k,k).

Original entry on oeis.org

1, 5, 43, 416, 4239, 44485, 475780, 5156548, 56437231, 622361423, 6904185523, 76964141600, 861408728964, 9673849095708, 108954068684616, 1230185577016156, 13920106205444335, 157814104889538739
Offset: 0

Views

Author

Paul Barry, Feb 17 2009

Keywords

Comments

a(n)=[x^n](1+5x+9x^2+7x^3+2x^4)^n. The coefficients (1,5,9,7,2) are the 5th row of A029635.

Crossrefs

Programs

  • Maple
    A156886 := proc(n)
        add(binomial(n,k)*binomial(3*n+k,k), k = 0..n);
    end proc:
    seq(A156886(n), n = 0..20); # Peter Bala, Feb 11 2018
  • Mathematica
    a[n_] := Sum[ Binomial[n, k] Binomial[3n + k, k], {k, 0, n}]; Array[a, 21, 0] (* Robert G. Wilson v, Feb 11 2018 *)

Formula

From Peter Bala, Feb 11 2018: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n-k)*C(n,k)*C(3*n+k,n)*2^k.
a(n) = Sum_{k = 0..n} C(n,k)*C(3*n,k)*2^(n-k),
12*n*(3*n-1)*(3*n-2)*(238*n^2 - 663*n + 457)*a(n) = 2*(150416*n^5 - 644640*n^4 + 1020351*n^3 - 734334*n^2 + 237007*n - 26880)*a(n-1) - (3*n-3)*(3*n-4)*(3*n-5)*(238*n^2 - 187*n + 32)*a(n-2). (End)
a(n) = P_n(0,2*n,3) where P_n(a,b,x) is the n-th Jacobi polynomial with parameters a and b. - Robert Israel, Feb 11 2018
a(n) ~ sqrt(1/3 + 11/(12*sqrt(7))) * ((316 + 119*sqrt(7))/54)^n / sqrt(Pi*n). - Vaclav Kotesovec, Jan 09 2023

A156887 a(n) = Sum_{k=0..n} C(n,k)*C(4*n+k,k).

Original entry on oeis.org

1, 6, 64, 768, 9708, 126386, 1676956, 22548168, 306167324, 4188703512, 57649462164, 797294161824, 11071026740964, 154250752864812, 2155368246401224, 30192512693210888, 423859798484668188, 5961793387214958792, 83998039356129372448, 1185277027372535468544
Offset: 0

Views

Author

Paul Barry, Feb 17 2009

Keywords

Comments

a(n)=[x^n] (1+6x+14x^2+16x^3+9x^4+2x^5)^n. The coefficients (1,6,14,16,9,2) are the 6th row of A029635.

Crossrefs

Programs

  • Maple
    A156887 := proc(n)
        add(binomial(n,k)*binomial(4*n+k,k),k=0..n) ;
    end proc: # R. J. Mathar, Feb 25 2015
  • Mathematica
    Table[Sum[Binomial[n,k]Binomial[4n+k,k],{k,0,n}],{n,0,30}] (* Harvey P. Dale, Jul 24 2018 *)
  • PARI
    {a(n) = sum(k=0, n, binomial(n, k)*binomial(4*n+k, k))} \\ Seiichi Manyama, Feb 02 2019

Formula

Conjecture: 576*n*(32901928701*n-65877527665)*(4*n-3)*(2*n-1)*(4*n-1)*a(n) +(-8795436181229177*n^5 +35251410418024655*n^4 -47934714902592853*n^3 +29414167990853161*n^2 -9060238526902314*n +1466702211905280)*a(n-1) +8*(10299715469615*n^5 -136961193094719*n^4 +872530072905392*n^3 -2699499511785411*n^2 +3902106377543903*n -2123717948975100)*a(n-2) -64*(2*n-5)*(4*n-9)*(n-2)*(27741827*n-2925269736)*(4*n-11)*a(n-3)=0. - R. J. Mathar, Feb 25 2015
From Peter Bala, Feb 11 2018: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n-k)*C(n,k)*C(4*n+k,n)*2^k.
a(n) = Sum_{k = 0..n} C(n,k)*C(4*n,k)*2^(n-k).
8*(4*n)*(4*n-1)*(4*n-2)**(4*n-3)*(4633*n^3-19662*n^2+27593*n-12804)*a(n) = (137604733*n^7-859190528*n^6+2179882848*n^5-2890753162*n^4+2144669963*n^3-880916550*n^2+182941416*n-14515200)*a(n-1) - (4*n-4)*(4*n-5)*(4*n-6)*(4*n-7)*(4633*n^3-5763*n^2+2168*n-240)*a(n-2). A proof of Mathar's conjectured third-order recurrence above follows easily using this second-order recurrence. (End)
a(n) ~ sqrt(5 + 33/sqrt(41)) * ((29701 + 4633*sqrt(41)))^n / (sqrt(Pi*n) * 2^(12*n + 2)). - Vaclav Kotesovec, Jan 09 2023

A114497 Numerators of Apéry-style convergents to 4/11 log 2.

Original entry on oeis.org

0, 1, 865, 12643, 13619843, 323746091, 115021083581, 2224431220019, 161734891776923, 57221149255770431, 7283680944060726343, 655528909167704911, 8662089991175424531107, 33378033361711480198829
Offset: 0

Views

Author

Eric Rowland, Dec 01 2005

Keywords

Comments

Denominators are given by A114498.

Crossrefs

A114498 Denominators of Apéry-style convergents to 4/11 log 2.

Original entry on oeis.org

1, 4, 3432, 50160, 54035520, 1284433920, 456335953920, 8825233697280, 641668847339520, 227019837729484800, 28897358537888096256, 2600752842642898944, 34366074253507451879424, 132424388815310703820800
Offset: 0

Views

Author

Eric Rowland, Dec 01 2005

Keywords

Comments

Numerators are given by A114497.

Crossrefs

A359646 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(5*n+k,k).

Original entry on oeis.org

1, 7, 89, 1273, 19181, 297662, 4707971, 75459496, 1221388525, 19919031781, 326797222834, 5387618403526, 89178832899887, 1481143718244912, 24671054686539336, 411966653603163008, 6894167059382069485, 115593504497163747167, 1941434442814233362939, 32656575110841643234631
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 09 2023

Keywords

Comments

In general, for m>0, Sum_{k=0..n} binomial(n,k) * binomial(m*n+k,k) ~ (m+c) / sqrt(2*Pi*c*m * (m*(2-c)+c)*n) * d^n, where d = (m+c)^(m+c) / ((1-c)^(1-c) * c^(2*c) * m^m) and c = (sqrt(m^2 + 6*m + 1) + 1 - m)/4.
Equivalently, d = (3 + m + sqrt(1 + m*(6 + m))) * (1 + 3*m + sqrt(1 + m*(6 + m)))^m / (2^(2*m + 1) * m^m).

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n, k]*Binomial[5*n+k, k], {k, 0, n}], {n, 0, 20}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k) * binomial(5*n+k,k)) \\ Andrew Howroyd, Jan 09 2023

Formula

a(n) ~ sqrt(3/10 + 23/(20*sqrt(14))) * ((108007 + 28854*sqrt(14))/12500)^n / sqrt(Pi*n).

A268543 The diagonal of 1/(1 - (y + z + x z + x w + x y w)).

Original entry on oeis.org

1, 8, 156, 3800, 102340, 2919168, 86427264, 2626557648, 81380484900, 2559296511200, 81443222791216, 2616761264496288, 84749038859067856, 2763262653898544000, 90615128199047200800, 2986287891921565639200, 98841887070519004625700
Offset: 0

Views

Author

N. J. A. Sloane, Feb 29 2016

Keywords

Comments

From Gheorghe Coserea, Jul 03 2016: (Start)
Also diagonal of rational function R(x,y,z) = 1/(1 - x - y - z - x*y).
Annihilating differential operator: x*(2*x+3)*(16*x^2-71*x+2)*Dx^2 + 2*(32*x^3+x^2-213*x+3)*Dx + 8*x^2+48*x-48.
(End)

Crossrefs

Programs

  • Maple
    A268543 := proc(n)
        1/(1-y-z-x*z-x*w-x*y*w) ;
        coeftayl(%,x=0,n) ;
        coeftayl(%,y=0,n) ;
        coeftayl(%,z=0,n) ;
        coeftayl(%,w=0,n) ;
    end proc:
    seq(A268543(n),n=0..40) ; # R. J. Mathar, Mar 11 2016
    #alternative program
    with(combinat):
    seq(binomial(2*n,n)*add(binomial(n,k)*binomial(2*n+k,k), k = 0..n), n = 0..20); # Peter Bala, Jan 27 2018
  • Mathematica
    CoefficientList[Series[HypergeometricPFQ[{1/12, 5/12}, {1}, 1728*x^3*(2 - 71*x + 16*x^2)/(1 - 32*x + 16*x^2)^3]*(1 - 32*x + 16*x^2)^(-1/4), {x, 0, 20}], x] (* Vaclav Kotesovec, Jul 05 2016 *)
  • PARI
    my(x='x, y='y, z='z, w='w);
    R = 1/(1 - x - y - z - x*y);
    diag(n, expr, var) = {
      my(a = vector(n));
      for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
      for (k = 1, n, a[k] = expr;
           for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
      return(a);
    };
    diag(10, R, [x,y,z])
    
  • PARI
    \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
    read("hypergeom.gpi");
    N = 20; x = 'x + O('x^N);
    Vec(hypergeom_sym([1/12,5/12],[1],1728*x^3*(16*x^2-71*x+2)/(16*x^2-32*x+1)^3, N)/(16*x^2-32*x+1)^(1/4))  \\ Gheorghe Coserea, Jul 03 2016

Formula

Conjecture: 2*n^2*(17*n-23)*a(n) +(-1207*n^3+2840*n^2-1897*n+360)*a(n-1) + 4*(17*n-6)*(-3+2*n)^2*a(n-2) = 0. - R. J. Mathar, Mar 11 2016
G.f.: hypergeom([1/12, 5/12], [1], 1728*x^3*(2-71*x+16*x^2)/(1-32*x+16*x^2)^3)*(1-32*x+16*x^2)^(-1/4). - Gheorghe Coserea, Jul 01 2016
0 = x*(2*x+3)*(16*x^2-71*x+2)*y'' + 2*(32*x^3+x^2-213*x+3)*y' + (8*x^2+48*x-48)*y, where y is the g.f. - Gheorghe Coserea, Jul 03 2016
a(n) ~ sqrt(3 + 13/sqrt(17)) * (71+17*sqrt(17))^n / (Pi * n * 2^(2*n + 3/2)). - Vaclav Kotesovec, Jul 05 2016
From Peter Bala, Jan 27 2018: (Start)
a(n) = binomial(2*n,n)*Sum_{k = 0..n} binomial(n,k)* binomial(2*n+k,k) (apply Eger, Theorem 3 to the set of column vectors S = {[1,0,0], [0,1,0], [0,0,1], [1,1,0]}). Using this binomial sum, Maple confirms the above recurrence of Mathar.
a(n) = A000984(n)*A114496(n). (End)

A306280 a(n) = Sum_{k=0..n} binomial(n,k)*binomial(n^2+k,k).

Original entry on oeis.org

1, 3, 26, 416, 9708, 297662, 11306572, 512307336, 26968496504, 1617489748394, 108885682104744, 8129721925098468, 666736347200187804, 59582961423951290184, 5762936296492591067968, 599807329803134064385488, 66843498592187788579795440
Offset: 0

Views

Author

Seiichi Manyama, Feb 02 2019

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[Binomial[n,k] * Binomial[n^2+k,k], {k, 0, n}]; Array[a, 20, 0] (* Amiram Eldar, Feb 03 2019 *)
  • PARI
    {a(n) = sum(k=0, n, binomial(n, k)*binomial(n^2+k, k))}

Formula

From Vaclav Kotesovec, Feb 08 2019: (Start)
a(n) ~ exp(1) * A135860(n).
a(n) ~ exp(n + 3/2) * n^(n - 1/2) / sqrt(2*Pi). (End)
Showing 1-10 of 11 results. Next