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 1-5 of 5 results.

A141902 Duplicate of A069466.

Original entry on oeis.org

1, 2, 2, 6, 24, 6, 20, 180, 180, 20, 70, 1120, 2520, 1120, 70, 252, 6300, 25200, 25200, 6300, 252, 924, 33264, 207900, 369600, 207900, 33264, 924, 3432, 168168, 1513512, 4204200, 4204200, 1513512, 168168, 3432, 12870, 823680, 10090080, 40360320
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Sep 13 2008

Keywords

A002894 a(n) = binomial(2n, n)^2.

Original entry on oeis.org

1, 4, 36, 400, 4900, 63504, 853776, 11778624, 165636900, 2363904400, 34134779536, 497634306624, 7312459672336, 108172480360000, 1609341595560000, 24061445010950400, 361297635242552100, 5445717990022688400, 82358080713306090000, 1249287673091590440000
Offset: 0

Views

Author

Keywords

Comments

a(n) is the number of monotonic paths (only moving N and E) in the lattice [0..2n] X [0..2n] that contain the points (0,0), (n,n) and (2n,2n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Expansion of K(k) / (Pi/2) in powers of m/16 = (k/4)^2, where K(k) is the complete elliptic integral of the first kind evaluated at k. - Michael Somos, Mar 04 2003
Square lattice walks that start and end at origin after 2n steps. - Gareth McCaughan and Michael Somos, Jun 12 2004
If A is a random matrix in USp(4) (4 X 4 complex matrices that are unitary and symplectic) then a(n)=E[(tr(A^k))^{2n}] for any k > 4. - Andrew V. Sutherland, Apr 01 2008
From R. H. Hardin, Feb 03 2016 and R. J. Mathar, Feb 18 2016: (Start)
Also, number of 2 X (2n) arrays of permutations of 2n copies of 0 or 1 with row sums equal.
For example, some solutions for n=3:
0 1 0 1 0 1 0 1 0 1 1 0 0 0 1 0 1 1 1 1 1 0 0 0
1 0 0 0 1 1 1 1 0 1 0 0 0 0 0 1 1 1 0 0 1 1 0 1
There is a simple combinatorial argument to show that this is a(n): We have 2n copies of 0's and 1's and need equal row sums. Therefore there must be n 1's in each of the two rows. Otherwise there are no constraints, so there are C(2n,n) ways of placing the 1's in the first row and independently C(2n,n) ways of placing the 1's in the second. The product is clearly C(2n,n)^2. (End)
Also the even part of the bisection of A241530. One half of the odd part is given in A000894. - Wolfdieter Lang, Sep 06 2016
From Peter Bala, Jan 26 2018: (Start)
Let S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} be a set of four column vectors. Then a(n) equals the number of 3 X k arrays whose columns belong to the set S and whose row sums are all equal to n (apply Eger, Theorem 3). An example is given below. Equivalently, a(n) equals the number of lattice paths from (0,0,0) to (n,n,n) using steps (1,0,0), (0,1,0), (1,0,1) and (0,1,1).
The o.g.f. for the sequence equals the diagonal of the rational function 1/(1 - (x + y + x*z + y*z)).
Row sums of A069466. (End)
Also, the constant term in the expansion of (x + 1/x + y + 1/y)^(2n). - Christopher J. Smyth, Sep 26 2018
Number of ways to place 2n^2 nonattacking pawns on a 2n x 2n board. - Tricia Muldoon Brown, Dec 12 2018
For n>0, a(n) is the number of Littlewood polynomials of degree 4n-1 that have a closed Lill path. A polynomial p(x) has a closed Lill path if and only if p(x) is divisible by x^(2)+1. - Raul Prisacariu, Aug 28 2024

Examples

			G.f. = 1 + 4*x + 36*x^2 + 400*x^3 + 4900*x^4 + 63504*x^5 + 853776*x^6 + ... - _Michael Somos_, Aug 06 2014
From _Peter Bala_, Jan 26 2018: (Start)
a(2) = 36: The thirty six 3 x k arrays with columns belonging to the set of column vectors S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} and having all row sums equal to 2 are the 6 distinct arrays obtained by permuting the columns of
  /1 1 0 0\
  |0 0 1 1|,
  \0 0 1 1/
the 6 distinct arrays obtained by permuting the columns of
  /0 0 1 1\
  |1 1 0 0|
  \0 0 1 1/
and the 24 arrays obtained by permuting the columns of
  /1 0 1 0\
  |0 1 0 1|. (End)
  \0 0 1 1/
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 591,828.
  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 8.
  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
  • Leonard Lipshitz and A. van der Poorten. "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990): 339-358.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row sums of A069466.
Row 2 of A268367 (even terms).
Equals 4*A060150.
Cf. A000984, A000515, A010370, A054474 (INVERTi transform), A172390, A000897, A002897, A006480, A008977, A186420, A188662, A000894, A241530, A002898 (walks hex lattice).

Programs

  • Magma
    [Binomial(2*n, n)^2: n in [0..20]]; // Vincenzo Librandi, Aug 07 2014
  • Maple
    A002894 := n-> binomial(2*n,n)^2.
  • Mathematica
    CoefficientList[Series[Hypergeometric2F1[1/2, 1/2, 1, 16x], {x, 0, 20}], x]
    Table[Binomial[2n,n]^2,{n,0,20}] (* Harvey P. Dale, Jul 06 2011 *)
    a[ n_] := SeriesCoefficient[ EllipticK[16 x] / (Pi/2), {x, 0, n}]; (* Michael Somos, Aug 06 2014 *)
    a[n_] := 16^n HypergeometricPFQ[{1/2, -2 n, 2 n + 1}, {1, 1}, 1];
    Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 14 2018 *)
  • PARI
    {a(n) = binomial(2*n, n)^2};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( polcoeff( polcoeff( 1 / (1 - x * (y + z + 1/y + 1/z)) + x * O(x^(2*n)), 2*n), 0), 0))}; /* Michael Somos, Jun 12 2004 */
    
  • Sage
    [binomial(2*n, n)**2 for n in range(17)]  # Zerinvary Lajos, Apr 21 2009
    

Formula

D-finite with recurrence: (n+1)^2*a(n+1) = 4*(2*n + 1)^2*a(n). - Matthijs Coster, Apr 28 2004
a(n) ~ Pi^(-1)*n^(-1)*2^(4*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
G.f.: F(1/2, 1/2; 1; 16*x) = 1 / AGM(1, (1 - 16*x)^(1/2)) = K(4*sqrt(x)) / (Pi/2), where AGM(x, y) is the arithmetic-geometric mean of Gauss and Legendre. - Michael Somos, Mar 04 2003
G.f.: 2*EllipticK(4*sqrt(x))/Pi, using Maple's convention for elliptic integrals.
E.g.f.: Sum_{n>=0} a(n)*x^(2*n)/(2*n)! = BesselI(0, 2x)^2.
a(n) = A000984(n)^2. - Jonathan Vos Post, Jun 17 2007
E.g.f.: (BesselI(0, 2*x))^2 = 1+2*x^2/(U(0)-2*x^2); U(k) = 2*x^2*(2*k+1)+(k+1)^3-2*x^2*(2*k+3)*(k+1)^3/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
In generally, for (BesselI(b, 2x))^2=((x^(2*b))/(GAMMA(b+1))^2)*(1+(2*x^2)*(2*b+1)/(Q(0)-(2*x^2)*(2*b+1)); Q(k)=(2*x^2)*(2*k+2*b+1)+(k+1)*(k+b+1)*(k+2*b+1)-(2*x^2)*(k+1)*(k+b+1)*(k+2*b+1)*(2*k+2*b+3)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - 4*(2*k+1)^2*x*(1+4*x)^2/(4*(2*k+1)^2*x*(1+4*x)^2 + (k+1)^2*(1+4*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 01 2013
0 = +a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for all n in Z. - Michael Somos, Aug 06 2014
Integral representation as the n-th moment of a positive function W(x) on (0,16), in Maple notation, W(x) = EllipticK(sqrt(1-x/16))/(2*Pi^2*sqrt(x)); a(n) = Integral_{x=0..16} x^n*W(x) dx, n>=0. The function W(x) is singular at x=0 and W(16) = 1/(16*Pi). This representation is unique since W(x) is the solution of the Hausdorff moment problem. - Stanley Smith and Karol A. Penson, Jun 19 2015
a(n) ~ 16^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)^2/((4*n+1)* Pi). - Peter Luschny, Oct 14 2015
a(n) = binomial(2*n,n)*binomial(2*n,n) = ( [x^n](1 + x)^(2*n) ) *( [x^n](1 + x)^(2*n) ) = [x^n](F(x)^(4*n)), where F(x) = 1 + x + x^2 + 4*x^3 + 20*x^4 + 120*x^5 + 798*x^6 + 5697*x^7 + ... appears to have integer coefficients. For similar results see A000897, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) = Sum_{k = 0..n} binomial(2*n + k,k)*binomial(n,k)^2. Cf. A005258(n) = Sum_{k = 0..n} binomial(n + k,k)*binomial(n,k)^2. - Peter Bala, Jul 27 2016
a(n) = A241530(2*n), n >= 0. - Wolfdieter Lang, Sep 06 2016
E.g.f.: 2F2(1/2,1/2; 1,1; 16*x). - Ilya Gutkovskiy, Jan 23 2018
a(n) = 16^n*hypergeom([1/2, -2*n, 2*n + 1], [1, 1], 1). - Peter Luschny, Mar 14 2018
The right-hand side of the binomial coefficient identity Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k)*(-4)^(n-k) = a(n). - Peter Bala, Mar 16 2018
a(n) = [x^n] (1 - x)^(2*n) * P(2*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A245086(n) = [x^n] (1 - x)^(2*n) * P(n,(1 + x)/(1 - x)). - Peter Bala, Mar 23 2022
a(n) = Sum_{k=0..n} multinomial(2n [k k (n-k) (n-k)]), which is another way to count random walks on Z^2, with steps of (0,+-1) or (+-1,0), that return to the point of origin after 2n steps (not necessarily for the first time), as is C(2n,n)^2. - Shel Kaphan, Jan 12 2023
0 = a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for n>=0. - Michael Somos, May 30 2023
From Peter Bala, Sep 12 2023: (Start)
Right-hand side of the binomial coefficient identities
1) Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k,n)*C(2*n+k,n) = a(n).
2) 2*Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k-1,n)*C(2*n+k-1,n) = a(n) for n >= 1.
3) (4/3)*Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k,n)*C(2*n+k-1,n) = a(n) for n >= 1. (End)

Extensions

Edited by N. J. A. Sloane, Feb 18 2016

A282252 Exponential Riordan array [Bessel_I(0,2*x)^2, x].

Original entry on oeis.org

1, 0, 1, 4, 0, 1, 0, 12, 0, 1, 36, 0, 24, 0, 1, 0, 180, 0, 40, 0, 1, 400, 0, 540, 0, 60, 0, 1, 0, 2800, 0, 1260, 0, 84, 0, 1, 4900, 0, 11200, 0, 2520, 0, 112, 0, 1, 0, 44100, 0, 33600, 0, 4536, 0, 144, 0, 1, 63504, 0, 220500, 0, 84000, 0, 7560, 0, 180, 0, 1
Offset: 0

Views

Author

Peter Bala, Feb 12 2017

Keywords

Comments

Bessel_I(0,2*x) = Sum_{n >= 0} binomial(2*n,n)*x^(2*n)/(2*n)! is a modified Bessel function of the first kind.
Consider the infinite 2-dimensional square lattice Z x Z with an oriented self-loop at each vertex. Then the triangle entry T(n,k) equals the number of walks of length n from the origin to itself having k loops. An example is given below.
See A069466 for walks an infinite 2-dimensional square lattice without self-loops.
This is the square of triangle A109187, whose entries give the number of walks of length n from a vertex to itself having k loops on a 1-dimensional integer lattice with an oriented self-loop at each vertex.
A109187 is the exponential Riordan array [Bessel_I(0,2*x), x]. Note that Bessel_I(0,2*x)^2 = (Sum_{n >= 0} binomial(2*n,n)* x^(2*n)/(2*n)!)^2 = Sum_{n >= 0} binomial(2*n,n)^2*x^(2*n) /(2*n)!.

Examples

			The triangle begins
    1;
    0,   1;
    4,   0,   1;
    0,  12,   0,   1;
   36,   0,  24,   0,   1;
    0, 180,   0,  40,   0,   1;
  400,   0, 540,   0,  60,   0,   1;
  ...
T(3,1) = 12: on the square lattice, let L, R, U, D denote a left step, right step, up step and down step respectively. The 12 walks of length 3 containing a single loop are
    loop L R, loop R L, loop U D, loop D U,
    L loop R, R loop L, U loop D, D loop U,
    L R loop, R L loop, U D loop, D U loop.
The infinitesimal generator of this array has integer entries and begins
      0;
      0,    0;
      4,    0,    0;
      0,   12,    0,    0;
    -12,    0,   24,    0,    0;
      0,  -60,    0,   40,    0,    0;
    160,    0, -180,    0,   60,    0,    0;
      0, 1120,    0, -420,    0,   84,    0,    0;
  -4620,    0, 4480,    0, -840,    0,  112,    0,    0;
  ...
It is the generalized exponential Riordan array [ 2*log(Bessel_I(0,2*x)), x ].
		

Crossrefs

A201805 gives row sums. Cf. A069466, A109187.

Programs

  • Maple
    T := (n, k) -> (1/2)*binomial(n, k)*binomial(n-k, floor((1/2)*n-(1/2)*k))^2*(1+(-1)^(n-k)):
    seq(seq(T(n, k), k = 0..n), n = 0..9);
  • Mathematica
    Table[Binomial[n, k] Binomial[n - k, Floor[(n - k)/2]]^2*(1 + (-1)^(n - k))/2, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 12 2017 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(binomial(n,k)*binomial(n-k,floor((n-k)/2))^2*(1 + (-1)^(n-k))/2, ", "))) \\ G. C. Greubel, Aug 16 2017

Formula

T(n,k) = binomial(n,k)*binomial(n-k,floor((n-k)/2))^2*(1 + (-1)^(n-k))/2.
T(n,n-2*k) = n/(n - 2*k)*T(n-1,n-2*k-1).
T(n,k) = the coefficient of t^k in the expansion of (t + X + 1/X + Y + 1/Y)^n.
T(n,k) = 1/Pi^2 * Integral_{y = 0..Pi} Integral_{x = 0..Pi} ( t + 2*cos(x) + 2*cos(y) )^n dx dy.
E.g.f.: exp(x*t)*Bessel_I(0,2*x)^2 = 1 + t*x + (4 + t^2)*x^2/2! + (12*t + t^3)*x^3/3! + (36 + 24*t^2 + t^4)*x^4/4! + ....
The n-th row polynomial R(n,t) = Sum_{k = 0..floor(n/2)} binomial(n,2*k)*binomial(2*k,k)^2 * t^(n-2*k).
Recurrence: n^2*R(n,t) = t*(3*n^2 - 3*n + 1)*R(n-1,t) + (16 - 3*t^2)*(n - 1)^2*R(n-2,t) + t*(t^2 - 16)*(n - 1)*(n - 2)*R(n-3,t) with R(0,t) = 1, R(1,t) = t and R(2,t) = 4 + t^2.
d/dt(R(n,t)) = n*R(n-1,t).
The zeros of the row polynomials appear to lie on the imaginary axis in the complex plane. Also, the zeros of R(n,t) and R(n+1,t) appear to interlace on the imaginary axis.

A320824 T(n, k) = (m*n)!/(k!*(n-k)!)^m with m = 3; triangle read by rows, 0 <= k <= n.

Original entry on oeis.org

1, 6, 6, 90, 720, 90, 1680, 45360, 45360, 1680, 34650, 2217600, 7484400, 2217600, 34650, 756756, 94594500, 756756000, 756756000, 94594500, 756756, 17153136, 3705077376, 57891834000, 137225088000, 57891834000, 3705077376, 17153136
Offset: 0

Views

Author

Peter Luschny, Oct 21 2018

Keywords

Examples

			Triangle starts:
[0]      1;
[1]      6,        6;
[2]     90,      720,        90;
[3]   1680,    45360,     45360,      1680;
[4]  34650,  2217600,   7484400,   2217600,    34650;
[5] 756756, 94594500, 756756000, 756756000, 94594500, 756756;
		

Crossrefs

Cf. A007318 (Pascal, m=1), A069466 (m=2), this sequence (m=3).

Programs

  • GAP
    Flat(List([0..6],n->List([0..n],k->Factorial(3*n)/(Factorial(k)*Factorial(n-k))^3))); # Muniru A Asiru, Oct 27 2018
  • Magma
    [[(Factorial(3*n)/(Factorial(n))^3)*Binomial(n, k)^3: k in [0..n]]: n in [0..15]]; // G. C. Greubel, Oct 27 2018
    
  • Maple
    T := (n, k, m) -> (m*n)!/(k!*(n-k)!)^m:
    seq(seq(T(n, k, 3), k=0..n), n=0..7);
  • Mathematica
    Table[((3*n)!/(n!)^3)*Binomial[n, k]^3, {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 27 2018 *)
  • PARI
    t(n, k) = (3*n)!/(k!*(n-k)!)^3
    trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
    /* Print initial 6 rows of triangle as follows: */
    trianglerows(6) \\ Felix Fröhlich, Oct 21 2018
    

Formula

T(n, k) = ((3*n)!/(n!)^3) * binomial(n, k)^3 = A006480(n)*A181543(n, k).

A068218 Triangle of numbers of square lattice walks that start and end at origin after 2k steps and contain exactly r steps to the east, not touching origin at intermediate stages.

Original entry on oeis.org

1, 2, 2, 2, 16, 2, 4, 84, 84, 4, 10, 400, 1056, 400, 10, 28, 1820, 9184, 9184, 1820, 28, 84, 8064, 66276, 126720, 66276, 8064, 84, 264, 35112, 426888, 1329768, 1329768, 426888, 35112, 264, 858, 151008, 2546544, 11737440, 19123776, 11737440
Offset: 0

Views

Author

Martin Wohlgemuth, Mar 24 2002

Keywords

Comments

The given recurrences do not provide a means to calculate T(2r,r). But T(2r,r) is computable by the formula relating T(k,r) to A069466(k,r).

Examples

			T(3,1)=84 because there are 84 distinct lattice walks of length 2*3=6 starting and ending at the origin and containing exactly 1 step to the east and not touching origin at intermediate steps. Let E, W, S, N denote the 4 possible directions, then NNEWSS and NWSSNE are examples of such walks.
		

Crossrefs

T(k, 0) = A002420(k) = A069466(k)/(2k-1).
Cf. A054474 (row sums).

Programs

  • Mathematica
    A069466[k_, r_] := Binomial[2 k, k]*Binomial[k, r]^2; t[k_, r_] := t[k, r] = A069466[k, r] - Sum[Sum[t[i, j]*A069466[k - i, r - j], {j, 0, r}], {i, 1, k - 1}]; Table[t[k, r], {k, 0, 8}, {r, 0, k}] // Flatten (* Jean-François Alcover, Nov 21 2012, from formula *)

Formula

T(k, r) = 2*(2k-3)/(k-2r) * ( T(k-1, r) - T(k-1, r-1) ), for k > 2r. T(1, 0)=2, T(1, 1)=2 Sum[T(k, r), r=0, ..., k] = A054474(k) T(k, r)=A069466(k, r) - Sum[ Sum[ T(i, j)*A069466(k-i, r-j), j=0...r], i=1, k-1]
Showing 1-5 of 5 results.