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 39 results. Next

A213684 Logarithmic derivative of A001002.

Original entry on oeis.org

1, 5, 22, 105, 511, 2534, 12720, 64449, 328900, 1688115, 8705060, 45064110, 234054198, 1219053680, 6364813192, 33302104593, 174570695175, 916628799380, 4820160541350, 25381091113455, 133808636072595, 706211862466500, 3730964595817680, 19729042153581150
Offset: 1

Views

Author

Paul D. Hanna, Jun 22 2012

Keywords

Comments

A001002(n) is the number of dissections of a convex (n+2)-gon into triangles and quadrilaterals by nonintersecting diagonals.
The g.f. of A001002 satisfies: G(x) = 1 + x*G(x)^2 + x^2*G(x)^3.
Central terms in A155161: a(n) = A155161(2*n,n). - Reinhard Zumkeller, Apr 17 2013
a(n) is the 2n-th term of the n-fold self-convolution of the Fibonacci numbers. - Alois P. Heinz, Feb 07 2021

Examples

			L.g.f.: L(x) = x + 5*x^2/2 + 22*x^3/3 + 105*x^4/4 + 511*x^5/5 +...
such that
L(x) = x*(1+x) + d/dx x^3*(1+x)^2/2! + d^2/dx^2 x^5*(1+x)^3/3! + d^3/dx^3 x^7*(1+x)^4/4! +...
The g.f. of A001002 begins:
exp(L(x)) = 1 + x + 3*x^2 + 10*x^3 + 38*x^4 + 154*x^5 + 654*x^6 +...
		

Crossrefs

Programs

  • Haskell
    a213684 n = a155161 (2*n) n  -- Reinhard Zumkeller, Apr 17 2013
  • Maple
    with(orthopoly): seq(add(i, i in [seq((-1)^iquo(n-k,2)*coeff(G(n,n,x/2), x, k), k=0..n)]), n=1..24); # Peter Luschny, Jan 26 2018
  • Mathematica
    Table[n*Sum[Binomial[k+n-1,n]*Binomial[k,n-k]/k,{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Oct 20 2012 *)
  • PARI
    {a(n)=n*sum(r=1,n,binomial(r+n-1,n)*binomial(r,n-r)/r)}
    for(n=1, 30, print1(a(n), ", "))
    
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n)=local(A=1); A=(sum(m=1, n+1, Dx(m-1, x^(2*m-1)*(1+x)^m/m!)+x*O(x^n))); n*polcoeff(A, n)}
    for(n=1, 30, print1(a(n), ", "))
    

Formula

a(n) = n * Sum_{r=1..n} binomial(r+n-1,n) * binomial(r,n-r) / r.
L.g.f.: Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1)*(1+x)^n/n! = Sum_{n>=1} a(n)*x^n/n.
Recurrence: 75*(n-1)*n*a(n) = 5*(n-1)*(59*n-12)*a(n-1) + (559*n^2-1503*n+1100)* a(n-2) + 21*(3*n-8)*(3*n-7)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 3^(3*n)/(2*5^(n-1/2)*sqrt(6*Pi*n)). - Vaclav Kotesovec, Oct 20 2012
a(n) = A037027(2*n-1,n-1). - Vladimir Kruchinin, Feb 28 2013
a(n) = Sum_{k=0..n} (-1)^floor((n-k)/2) [x^k] G(n,n,x/2), where G(n,a,x) denotes the n-th Gegenbauer polynomial; row sums of A298610. - Peter Luschny, Jan 26 2018
a(n) = [x^n] (1/(1-x-x^2))^n. - Alois P. Heinz, Feb 07 2021
From Peter Bala, Mar 11 2025: (Start)
a(n) = Sum_{k = 0..n} n/(2*n-k) * binomial(2*n-k, k)*binomial(2*n-2*k, n).
a(n) = (1/2)*binomial(2*n, n)*hypergeom([-n/2, (-n+1)/2], [-2*n+1], -4). Cf. A350383.
Second-order recurrence: 5*n*(n-1)*(8*n-13)*a(n) = 2*(n-1)*(88*n^2-187*n+75)*a(n-1) + 3*(8*n-5)*(3*n-4)*(3*n-5)*a(n-2) with a(1) = 1 and a(2) = 5. (End)

A052706 A simple context-free grammar: convolution square of A001002.

Original entry on oeis.org

0, 0, 1, 2, 7, 26, 105, 444, 1944, 8734, 40040, 186550, 880750, 4204508, 20260498, 98419392, 481442805, 2369551218, 11725590555, 58303117680, 291151523355, 1459590130350, 7342906908645, 37058911816680, 187579329483780, 952006706210196, 4843566974043900
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Crossrefs

Cf. A001002.

Programs

  • Maple
    spec := [S,{C = Union(S,B,Z),B = Prod(S,C),S = Prod(C,C)},unlabeled]: seq(combstruct[count](spec,size = n), n = 0..20);
  • Mathematica
    Flatten[{0,0,Table[2*Sum[Binomial[k,n-k-2]*Binomial[n+k-1,n-1],{k,0,n-2}]/n,{n,2,20}]}] (* Vaclav Kotesovec, Oct 09 2012 *)
  • Maxima
    a(n):=if n<2 then 0 else (2*sum(binomial(k,n-k-2)*binomial(n+k-1,n-1),k,0,n-2))/n; /* Vladimir Kruchinin, May 19 2012 */
    
  • PARI
    a(n) = if(n>1,2*sum(k=0,n-2,binomial(k,n-k-2)*binomial(n+k-1,n-1))/n,0) \\ Jason Yuen, Aug 12 2024

Formula

G.f.: RootOf(-_Z+_Z^2+_Z^3+x)^2.
Recurrence: {a(1) = 0, a(2) = 1, a(3) = 2, (6-27*n+27*n^2)*a(n)+(6+65*n+49*n^2)*a(n+1)+(67*n+66+17*n^2)*a(n+2)+(-5*n^2-25*n-30)*a(n+3)}.
a(n) = 2*(Sum_{k=0..n-2} binomial(k,n-k-2)*binomial(n+k-1,n-1))/n, n>1, a(0)=a(1)=0. - Vladimir Kruchinin, May 19 2012
a(n) ~ 3^(3*n-5/2)/(sqrt(2*Pi)*5^(n-1/2)*n^(3/2)). - Vaclav Kotesovec, Oct 09 2012

A052703 A simple context-free grammar: convolution cube of A001002.

Original entry on oeis.org

0, 0, 0, 1, 3, 12, 49, 210, 927, 4191, 19305, 90285, 427570, 2046324, 9881862, 48090824, 235619133, 1161257580, 5753365015, 28638093270, 143148398085, 718242481770, 3616135914375, 18263111515740, 92500790125770, 469737499557222, 2391192703656054, 12199557377107450
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Crossrefs

Programs

  • Maple
    spec := [S,{C=Prod(B,B),B=Union(S,C,Z),S=Prod(B,C)},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • PARI
    my(N=30, x='x+O('x^N)); concat([0, 0, 0], Vec(serreverse(x-x^2-x^3)-serreverse(x-x^2-x^3)^2-x)) \\ Seiichi Manyama, Nov 22 2024
    
  • PARI
    a(n) = 3*sum(k=0, n-3, binomial(n+k, k)*binomial(k, n-3-k)/(n+k)); \\ Seiichi Manyama, Nov 22 2024

Formula

G.f.: RootOf(-_Z+_Z^2+_Z^3+x)-RootOf(-_Z+_Z^2+_Z^3+x)^2-x
Recurrence: {a(1)=0, a(2)=0, a(3)=1, a(4)=3, (30-135*n+135*n^2)*a(n)+(-130-107*n+29*n^2)*a(n+1)+(-281*n-198-91*n^2)*a(n+2)+(15*n^2+75*n+90)*a(n+3)}
From Seiichi Manyama, Nov 22 2024: (Start)
G.f.: (x*B(x))^3 where B(x) is the g.f. of A001002.
a(n) = 3 * Sum_{k=0..n-3} binomial(n+k,k) * binomial(k,n-3-k)/(n+k). (End)
a(n) ~ 3^(3*n - 5/2) / (sqrt(Pi) * 2^(3/2) * n^(3/2) * 5^(n - 1/2)). - Vaclav Kotesovec, Nov 22 2024

Extensions

More terms from Seiichi Manyama, Nov 21 2024

A259690 Erroneous version of A001002.

Original entry on oeis.org

1, 1, 3, 10, 38, 154, 645, 2853, 12844, 58985
Offset: 0

Views

Author

Keywords

A111006 Another version of Fibonacci-Pascal triangle A037027.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 0, 2, 3, 0, 0, 1, 5, 5, 0, 0, 0, 3, 10, 8, 0, 0, 0, 1, 9, 20, 13, 0, 0, 0, 0, 4, 22, 38, 21, 0, 0, 0, 0, 1, 14, 51, 71, 34, 0, 0, 0, 0, 0, 5, 40, 111, 130, 55, 0, 0, 0, 0, 0, 1, 20, 105, 233, 235, 89, 0, 0, 0, 0, 0, 0, 6, 65, 256, 474, 420, 144
Offset: 0

Views

Author

Philippe Deléham, Oct 02 2005

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 1, -1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
Row sums are the Jacobsthal numbers A001045(n+1) and column sums form Pell numbers A000129.
Maximal column entries: A038149 = {1, 1, 2, 5, 10, 22, ...}.
T(n,k) gives a convolved Fibonacci sequence (A001629, A001872, ...).
Triangle read by rows: T(n,n-k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)). - Philippe Deléham, Feb 17 2014
Diagonal sums are A013979(n). - Philippe Deléham, Feb 17 2014
T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and 1 X 2 tiles. - Emeric Deutsch, Aug 14 2014

Examples

			Triangle begins:
  1;
  0, 1;
  0, 1, 2;
  0, 0, 2, 3;
  0, 0, 1, 5,  5;
  0, 0, 0, 3, 10,  8;
  0, 0, 0, 1,  9, 20, 13;
  0, 0, 0, 0,  4, 22, 38,  21;
  0, 0, 0, 0,  1, 14, 51,  71,  34;
  0, 0, 0, 0,  0,  5, 40, 111, 130,  55;
  0, 0, 0, 0,  0,  1, 20, 105, 233, 235,  89;
  0, 0, 0, 0,  0,  0,  6,  65, 256, 474, 420, 144;
		

Crossrefs

Cf. A000045, A000129, A001045, A037027, A038112, A038149, A084938, A128100 (reversed version).
Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A114197, A162741, A228074.

Programs

  • Haskell
    a111006 n k = a111006_tabl !! n !! k
    a111006_row n = a111006_tabl !! n
    a111006_tabl =  map fst $ iterate (\(us, vs) ->
       (vs, zipWith (+) (zipWith (+) ([0] ++ us ++ [0]) ([0,0] ++ us))
                        ([0] ++ vs))) ([1], [0,1])
    -- Reinhard Zumkeller, Aug 15 2013

Formula

T(0, 0) = 1, T(n, k) = 0 for k < 0 or for n < k, T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-2, k-2).
T(n, k) = A037027(k, n-k). T(n, n) = A000045(n+1). T(3n, 2n) = (n+1)*A001002(n+1) = A038112(n).
G.f.: 1/(1-yx(1-x)-x^2*y^2). - Paul Barry, Oct 04 2005
Sum_{k=0..n} x^k*T(n,k) = (-1)^n*A053524(n+1), (-1)^n*A083858(n+1), (-1)^n*A002605(n), A033999(n), A000007(n), A001045(n+1), A083099(n) for x = -4, -3, -2, -1, 0, 1, 2 respectively. - Philippe Deléham, Dec 02 2006
Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1) for x = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 respectively. - Philippe Deléham, Feb 17 2014

A236339 Association types in 2-dimensional algebra.

Original entry on oeis.org

1, 2, 8, 39, 212, 1232, 7492, 47082, 303336, 1992826, 13299624, 89912992, 614474252, 4238138216, 29463047072, 206234876287, 1452319244772, 10281935334928, 73138728191724, 522475643860940, 3746698673538480, 26961197787989220, 194626504416928080
Offset: 1

Views

Author

Murray R. Bremner, Jan 22 2014

Keywords

Comments

A kind of two-dimensional Catalan number.
This sequence has two equivalent descriptions:
(1) This sequence enumerates the number of decompositions of the unit square into n rectangles obtained by the following algorithm.
(a) Start with the unit square.
(b) Perform the following operation n-1 times:
Choose a rectangle in the current decomposition.
Bisect this rectangle into two rectangles horizontally or vertically.
Different sequences of bisections can produce the same decomposition.
(2) Consider the universal algebra with two nonassociative binary products *1 and *2 related only by the interchange law from 2-category theory:
( a *1 b ) *2 ( c *1 d ) = ( a *2 c ) *1 ( b *2 d )
This sequence enumerates the number of distinct monomials of degree n.

References

  • J.-L. Loday and B. Vallette, Algebraic Operads, Grundlehren 346, Springer, 2012, section 13.10.4, page 544 (for the interchange law)
  • S. Mac Lane, Categories for the Working Mathematician, second edition, Springer, 1978, equation (5), page 43 (also for the interchange law).

Crossrefs

Cf. A000108 (Catalan numbers), A236342.
Column k=2 of A237018.

Programs

  • Maple
    c := table():
    c[1] := 1:
    printf( "\n" ):
    for n from 2 to 50 do
    c[n] := 0:
    for ij in combinat[composition](n,2) do
        c[n] := c[n] + 2*c[ij[1]]*c[ij[2]]
    od:
    for ijkl in combinat[composition](n,4) do
        c[n] := c[n] - c[ijkl[1]]*c[ijkl[2]]*c[ijkl[3]]*c[ijkl[4]]
    od:
       printf( "%2d      %d \n", n, c[n] )
    od:
    # second Maple program:
    a:= proc(n) option remember; `if`(n<3, n, (
          8*(2*n-5)*(148*n-243)*(4*n-13)*(4*n-11)*a(n-3)
          +16*(n-2)*(4736*n^3-31456*n^2+68444*n-48609)*a(n-2)
          -32*(n-1)*(n-2)*(148*n^2-613*n+594)*a(n-1)) /
          (5*n*(n-1)*(n-2)*(148*n-391)))
        end:
    seq(a(n), n=0..25);  # Alois P. Heinz, Jan 22 2014
  • Mathematica
    max = 30; c[1] = 1; c[2] = 2; g = Sum[c[k]*x^k, {k, 1, max}]; eq = Take[Thread[CoefficientList[g^4 - 2*g^2 + g - x, x] == 0], max+1]; sol = Solve[eq] // First; Array[c, max] /. sol (* Jean-François Alcover, Jan 27 2014 *)
    Rest[CoefficientList[InverseSeries[Series[x^4-2*x^2+x, {x, 0, 20}], x],x]] (* Vaclav Kotesovec, Feb 16 2014 *)

Formula

Recurrence relation:
C(1) = 1,
C(n) = 2 Sum_{i,j} C(i)C(j) - Sum_{i,j,k,l} C(i)C(j)C(k)C(l).
The first sum is over all 2-compositions of n into positive integers (i+j=n), and the second sum is over all 4-compositions of n into positive integers (i+j+k+l=n).
Generating function G(x) = Sum_{n>=1} C(n) x^n satisfies a quartic polynomial equation: G(x)^4 - 2*G(x)^2 + G(x) - x = 0.
a(n) ~ (1/r)^(n-1/2) / (2 * sqrt(2*Pi*(1-3*s^2)) * n^(3/2)), where s = 0.2695944364054445582... is the root of the equation 4*s*(1-s^2) = 1, and r = s*(1-2*s+s^3) = 0.1295146671633141285... - Vaclav Kotesovec, Feb 16 2014
From Seiichi Manyama, Jan 10 2023: (Start)
G.f.: Series_Reversion( x * (1-x) * (1-x-x^2) ).
a(n+1) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(n+k,k) * binomial(3*n-k+1,n-2*k). (End)

Extensions

a(17)-a(23) from Alois P. Heinz, Jan 22 2014

A368961 Expansion of (1/x) * Series_Reversion( x * (1-x-x^2)^2 ).

Original entry on oeis.org

1, 2, 9, 48, 286, 1820, 12116, 83334, 587537, 4223582, 30840355, 228111390, 1705509981, 12868775056, 97867753424, 749401318160, 5772939358590, 44708058004740, 347879528717526, 2718400037837988, 21323471768334120, 167844335760482220, 1325332432687278960
Offset: 0

Views

Author

Seiichi Manyama, Jan 10 2024

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := (1/(n+1)) * Sum[Binomial[2*n+k+1,k] * Binomial[3*n-k+1,n-2*k],{k,0,Floor[n/2]}]; Array[a,23,0] (* Stefano Spezia, Aug 11 2025 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x-x^2)^2)/x)
    
  • PARI
    a(n, s=2, t=2, u=0) = sum(k=0, n\s, binomial(t*(n+1)+k-1, k)*binomial((t+u+1)*(n+1)-(s-1)*k-2, n-s*k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(2*n+k+1,k) * binomial(3*n-k+1,n-2*k).

A214692 G.f. A(x) satisfies: x = Sum_{n>=1} 1/A(x)^(4*n) * Product_{k=1..n} (1 - 1/A(x)^(2*k-1)).

Original entry on oeis.org

1, 1, 2, 11, 71, 515, 3997, 32488, 273009, 2352724, 20678966, 184660333, 1670619561, 15279692008, 141048655988, 1312429249996, 12296515232446, 115909188223053, 1098444610424929, 10459429664510189, 100021237512559055, 960168745226226195, 9249466125601138425
Offset: 0

Views

Author

Paul D. Hanna, Jul 26 2012

Keywords

Comments

Compare the g.f. to the identity:
G(x) = Sum_{n>=0} 1/G(x)^(2*n) * Product_{k=1..n} (1 - 1/G(x)^(2*k-1))
which holds for all power series G(x) such that G(0)=1.

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 11*x^3 + 71*x^4 + 515*x^5 + 3997*x^6 +...
The g.f. satisfies:
x = (A(x)-1)/A(x)^5 + (A(x)-1)*(A(x)^3-1)/A(x)^12 + (A(x)-1)*(A(x)^3-1)*(A(x)^5-1)/A(x)^21 + (A(x)-1)*(A(x)^3-1)*(A(x)^5-1)*(A(x)^7-1)/A(x)^32 +
(A(x)-1)*(A(x)^3-1)*(A(x)^5-1)*(A(x)^7-1)*(A(x)^9-1)/A(x)^45 +...
Related expansions.
A(x)^2 = 1 + 2*x + 5*x^2 + 26*x^3 + 168*x^4 + 1216*x^5 + 9429*x^6 +...
A(x)^3 = 1 + 3*x + 9*x^2 + 46*x^3 + 297*x^4 + 2148*x^5 + 16649*x^6 +...
A(x)^4 = 1 + 4*x + 14*x^2 + 72*x^3 + 465*x^4 + 3364*x^5 + 26078*x^6 +...
where 1+x = A(x)^2 + A(x)^3 - A(x)^4.
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[1+InverseSeries[Series[x - 2*x^2 - 3*x^3 - x^4, {x, 0, 20}], x],x] (* Vaclav Kotesovec, Nov 29 2014 *)
  • PARI
    {a(n)=if(n<0, 0, polcoeff(1 + serreverse(x - 2*x^2 - 3*x^3 - x^4 +x^2*O(x^n)), n))}
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    {a(n)=local(A=[1, 1]); for(i=1, n, A=concat(A, 0); A[#A]=-polcoeff(sum(m=1, #A, 1/Ser(A)^(4*m)*prod(k=1, m, 1-1/Ser(A)^(2*k-1))), #A-1)); A[n+1]}
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    /* From 1+x = A(x)^2 + A(x)^3 - A(x)^4: */
    {a(n)=local(A=[1,1]);for(i=1,n,A=concat(A,0);A[#A]=-Vec(Ser(A)^2+Ser(A)^3-Ser(A)^4)[#A]);A[n+1]}
    for(n=0,25,print1(a(n) ,", "))

Formula

G.f. A(x) satisfies:
(1) 1+x = A(y) where y = x - 2*x^2 - 3*x^3 - x^4, which is the g.f. of row 2 in triangle A214690.
(2) x = Sum_{n>=1} 1/A(x)^(n*(n+4)) * Product_{k=1..n} (A(x)^(2*k-1) - 1).
(3) 1+x = A(x)^2 + A(x)^3 - A(x)^4. - Paul D. Hanna, Nov 15 2014
a(n) ~ sqrt(145/sqrt(41)-21) * ((213+41*sqrt(41))/46)^n / (16 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 29 2014

A104978 Triangle read by rows, where the g.f. satisfies A(x, y) = 1 + x*A(x, y)^2 + x*y*A(x, y)^3.

Original entry on oeis.org

1, 1, 1, 2, 5, 3, 5, 21, 28, 12, 14, 84, 180, 165, 55, 42, 330, 990, 1430, 1001, 273, 132, 1287, 5005, 10010, 10920, 6188, 1428, 429, 5005, 24024, 61880, 92820, 81396, 38760, 7752, 1430, 19448, 111384, 352716, 678300, 813960, 596904, 245157, 43263, 4862, 75582, 503880, 1899240, 4476780, 6864396, 6864396, 4326300, 1562275, 246675
Offset: 0

Views

Author

Paul D. Hanna, Mar 30 2005

Keywords

Examples

			The triangle T(n, k) begins:
  [0]    1;
  [1]    1,     1;
  [2]    2,     5,      3;
  [3]    5,    21,     28,     12;
  [4]   14,    84,    180,    165,     55;
  [5]   42,   330,    990,   1430,   1001,    273;
  [6]  132,  1287,   5005,  10010,  10920,   6188,   1428;
  [7]  429,  5005,  24024,  61880,  92820,  81396,  38760,   7752;
  [8] 1430, 19448, 111384, 352716, 678300, 813960, 596904, 245157, 43263;
  ...
The array A(n, k) begins:
  [0]   1,    1,      3,      12,       55,       273,       1428, ...  [A001764]
  [1]   1,    5,     28,     165,     1001,      6188,      38760, ...  [A025174]
  [2]   2,   21,    180,    1430,    10920,     81396,     596904, ...  [A383450]
  [3]   5,   84,    990,   10010,    92820,    813960,    6864396, ...  [A383451]
  [4]  14,  330,   5005,   61880,   678300,   6864396,   65615550, ...
  [5]  42, 1287,  24024,  352716,  4476780,  51482970,  551170620, ...
  [6] 132, 5005, 111384, 1899240, 27457584, 354323970, 4206302100, ...
  [A000108]  |  [A074922][A383452]
         [A002054]
		

Crossrefs

Columns of array: A000108, A002054, A074922, A383452.
Rows of array: A001764, A025174, A383450, A383451.
Cf. A001002 (antidiagonal sums), A001764 (semidiagonal sums), A027307 (row sums), A104979, A383439 (central terms).

Programs

  • Magma
    [Binomial(2*n+k, n+2*k)*Binomial(n+2*k, k)/(n+k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, Jun 08 2021
    
  • Maple
    From Peter Luschny, May 04 2025:  (Start)
    T := (n, k) -> (k + 2*n)!/(k!*(n - k)!*(n + k + 1)!):
    seq(print(seq(T(n, k), k = 0..n)), n = 0..10);
    # Alternatively the array:
    A := (n, k) -> (3*k + 2*n)!/(k!*n!*(n + 2*k + 1)!);
    for n from 0 to 8 do seq(A(n, k), k = 0..7) od;  (End)
  • Mathematica
    T[n_, k_]:= Binomial[2n+k, n+2k]*Binomial[n+2k, k]/(n+k+1);
    Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Jan 27 2019 *)
  • PARI
    T(n,k) = local(A=1+x+x*y+x*O(x^n)+y*O(y^k)); for(i=1,n,A=1+x*A^2+x*y*A^3); polcoeff(polcoeff(A,n,x),k,y)
    for(n=0, 10, for(k=0, n, print1(T(n,k),", ")); print(""))
    
  • PARI
    Dy(n, F)=local(D=F); for(i=1, n, D=deriv(D,y)); D
    T(n,k)=local(A=1); A=1+sum(m=1, n+1, x^m/y^(m+1) * Dy(m-1, (y^2+y^3)^m/m!)) +x*O(x^n)+y*O(y^k); polcoeff(polcoeff(A, n,x),k,y)
    for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Jun 22 2012
    
  • PARI
    x='x; y='y; z='z; Fxyz = 1 - z + x*z^2 + x*y*z^3;
    seq(N) = {
      my(z0 = 1 + O((x*y)^N), z1 = 0);
      for (k = 1, N^2,
        z1 = z0 - subst(Fxyz, z, z0)/subst(deriv(Fxyz, z), z, z0);
        if (z0 == z1, break()); z0 = z1);
      vector(N, n, Vecrev(polcoeff(z0, n-1, 'x)));
    };
    concat(seq(9)) \\ Gheorghe Coserea, Nov 30 2016
    
  • Sage
    flatten([[binomial(2*n+k, n+2*k)*binomial(n+2*k, k)/(n+k+1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 08 2021

Formula

T(n, k) = binomial(2*n+k, n+2*k)*binomial(n+2*k, k)/(n+k+1).
G.f.: A(x, y) = Sum_{n>=0} x^n/y^(n+1) * d^(n-1)/dy^(n-1) (y^2 + y^3)^n / n!. - Paul D. Hanna, Jun 22 2012
G.f. of row n: 1/y^(n+1) * d^(n-1)/dy^(n-1) (y^2+y^3)^n / n!. - Paul D. Hanna, Jun 22 2012
A(n, k) = T(n + k, k) = (3*k + 2*n)! / (k!*n!*(n + 2*k + 1)!). - Peter Luschny, May 04 2025

A038112 a(n) = T(2n,n), where T(n,k) is in A037027.

Original entry on oeis.org

1, 2, 9, 40, 190, 924, 4578, 22968, 116325, 593450, 3045185, 15699840, 81260816, 421993040, 2197653240, 11472991008, 60023749566, 314621200260, 1651883008050, 8685998428800, 45734484854520, 241098942106440, 1272406536645660
Offset: 0

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Author

Keywords

Comments

Number of lattice paths from (0,0) to (n,n) using steps (0,1), (1,0), (2,0). - Joerg Arndt, Jun 30 2011
Diagonal of rational function 1/(1 - (x + y + y^2)). - Gheorghe Coserea, Aug 06 2018

Examples

			G.f.: A(x) = 1 + 2*x + 9*x^2 + 40*x^3 + 190*x^4 + 924*x^5 + 4578*x^6 + ...
		

Programs

  • GAP
    List([0..25],n->Sum([0..n],k->Binomial(n+k,k)*Binomial(k,n-k))); # Muniru A Asiru, Aug 06 2018
  • Maple
    a:=n->sum(binomial(2*j,n)*(binomial(n+j,2*j)),j=0..n): seq(a(n), n=0..21); # Zerinvary Lajos, Aug 22 2006
    series(RootOf((x+1)*(27*x-5)*A^3+4*A+1,A),x=0,30); # Mark van Hoeij, May 01 2013
  • Mathematica
    Table[Sum[Binomial[n+k,k]Binomial[k,n-k],{k,0,n}],{n,0,30}] (* Harvey P. Dale, Sep 30 2012 *)
    Table[Binomial[2 n, n] Hypergeometric2F1[1/2 - n/2, -n/2, -2 n, -4], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 19 2016 *)
  • PARI
    {a(n) = if( n<0, 0, sum(k=0, n\2, (2*n-k)!/ (k! * (n-2*k)!)) / n!)}; /* Michael Somos, Sep 29 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; n * polcoeff(serreverse( x - x^2 - x^3 + x * O(x^n)), n))}; /* Michael Somos, Sep 29 2003 */
    
  • PARI
    /* same as in A092566 but use */
    steps=[[0,1], [1,0], [2,0]]; /* Joerg Arndt, Jun 30 2011 */
    
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D} \\ = d^n/dx^n F
    {a(n)=local(A=x); A=1+sum(m=1, n, Dx(m, x^(2*m)*(1+x+x*O(x^n))^m/m!)); polcoeff(A, n)} \\ Paul D. Hanna, Aug 04 2012
    

Formula

a(n) = Sum_{k=0..n} C(n+k,k)*C(k,n-k). - Paul Barry, May 13 2006
a(n) = Sum_{j=0..n} binomial(2*j, n)*binomial(n+j, 2*j). - Zerinvary Lajos, Aug 22 2006
a(n) = [x^n] (1/(1-x-x^2))^(n+1). - Paul Barry, Mar 23 2011
a(n) = (n+1)*A001002(n+1).
G.f.: Sum_{n>=0} d^n/dx^n x^(2*n)*(1+x)^n/n!. - Paul D. Hanna, Aug 04 2012
Recurrence: 5*(n-1)*n*a(n) = 11*(n-1)*(2*n-1)*a(n-1) + 3*(3*n-4)*(3*n-2)*a(n-2). - Vaclav Kotesovec, Oct 08 2012
a(n) ~ 3^(3*n+3/2)/(2^(3/2)*5^(n+1/2)*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 08 2012
G.f.: A(x) where (x+1)*(27*x-5)*A(x)^3 + 4*A(x) + 1 = 0. - Mark van Hoeij, May 01 2013
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