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 1-2 of 2 results.

A098430 a(n) = 4^n*(2*n)!/(n!)^2.

Original entry on oeis.org

1, 8, 96, 1280, 17920, 258048, 3784704, 56229888, 843448320, 12745441280, 193730707456, 2958796259328, 45368209309696, 697972450918400, 10768717814169600, 166556168859156480, 2581620617316925440, 40091049586568724480, 623638549124402380800, 9715632133727531827200
Offset: 0

Views

Author

Paul Barry, Sep 07 2004

Keywords

Comments

a(n) counts walks of 2n steps North, East, South or West that start at the origin and end on the line y=x. For example, a(1)=8 counts EW, EN, NE, NS, WE, WS, SN, SW. If the walk has i East and j North steps, then it must have n-j West and n-i South steps. There are Multinomial[i,j,n-j,n-i] ways to arrange these steps and summing over i and j gives the result. - David Callan, Oct 11 2005
Number of lattice paths from (0,0) to (n,n) using steps (1,0), (0,1), both of two kinds. - Joerg Arndt, Jul 01 2011
Hankel transform is A121913. - Philippe Deléham, Mar 01 2009
Convolving a(n) with itself yields A001025, the powers of 16. Thus the limiting ratio of this sequence is 16. - Bob Selcoe, Jul 16 2014
Number of strings x of length 4n over the alphabet {1, -1} such that the dot product of x with (x reversed) is 0. - Jeffrey Shallit, Mar 06 2017
Number of orthogonal pairs of vectors of length 2n, constructed with any symmetric binary-valued symbol set. - Ross Drewe, May 18 2018
Diagonal of the rational function 1 / (1 - 4*x - y). - Ilya Gutkovskiy, Apr 24 2025

Links

Crossrefs

Cf. A000302, A000984, A174301, A249308.

Programs

  • Haskell
    a098430 n = a000302 n * a000984 n -- Reinhard Zumkeller, Nov 14 2014
  • Magma
    [4^n*Factorial(2*n)/Factorial(n)^2: n in [0..20]]; // Vincenzo Librandi, Jul 05 2011
  • Maple
    A098430 := n -> 4^n*binomial(2*n,n): seq(A098430(n), n=0..30); # Wesley Ivan Hurt, Jul 16 2014
  • Mathematica
    CoefficientList[Series[1/Sqrt[1 - 16 x], {x, 0, 16}], x] (* Robert G. Wilson v, Jun 28 2012 *)
Table[4^n(2n)!/(n!)^2,{n,0,20}] (* Harvey P. Dale, Aug 13 2021 *)
  • PARI
    /* as lattice paths: same as in A092566 but use */
steps=[[1,0], [1,0], [0,1], [0,1]]; /* note the double [1,0] and [0,1] */
/* Joerg Arndt, Jul 01 2011 */
  • Sage
    a = lambda n: 16^n*hypergeometric([-2*n, 1/2], [1], 2)
[simplify(a(n)) for n in range(23)] # Peter Luschny, May 19 2015

Formula

a(n) = 4^n*binomial(2*n, n) = 4^n*A000984(n).
E.g.f.: exp(8*x)*BesselI(0, 8*x).
G.f.: 1/sqrt(1-16*x). - Zerinvary Lajos, Dec 20 2008, corrected R. J. Mathar, May 18 2009
a(n) = (1/Pi)*Integral_{x=-2..2} (2*x)^(2*n)/sqrt((2-x)*(2+x)) dx. - Peter Luschny, Sep 12 2011
D-finite with recurrence: n*a(n) + 8*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Nov 10 2014
a(n) = A249308(2*n). - Reinhard Zumkeller, Nov 14 2014
a(n) = 16^n*hypergeometric([-2*n, 1/2], [1], 2). - Peter Luschny, May 19 2015
a(n) = A174301(2n,n). - Alois P. Heinz, Apr 15 2019
From Amiram Eldar, Jul 21 2020: (Start)
Sum_{n>=0} 1/a(n) = 16/15 + 16*sqrt(15)*arcsin(1/4)/225.
Sum_{n>=0} (-1)^n/a(n) = 16/17 - 16*sqrt(17)*arcsinh(1/4)/289. (End)
a(n) = Sum_{k = 0..2*n} (-1)^k *A000984(k) * A000984(2*n-k). Cf. Sum_{k = 0..2*n} A000984(k) * A000984(2*n-k) = 16^n. - Peter Bala, Aug 23 2025

A174303 A symmetrical triangle: T(n,k) = A008292(n+1, k) * f(n,k), where f(n,k) = 2^k when floor(n/2) >= k, otherwise 2^(n-k).

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 22, 22, 1, 1, 52, 264, 52, 1, 1, 114, 1208, 1208, 114, 1, 1, 240, 4764, 19328, 4764, 240, 1, 1, 494, 17172, 124952, 124952, 17172, 494, 1, 1, 1004, 58432, 705872, 2499040, 705872, 58432, 1004, 1, 1, 2026, 191360, 3641536, 20965664, 20965664, 3641536, 191360, 2026, 1
Offset: 0

Views

Author

Roger L. Bagula, Mar 15 2010

Keywords

Comments

Row sums are: {1, 2, 10, 46, 370, 2646, 29338, 285238, 4029658, ...}.

Examples

			Triangle begins as:
  1;
  1,    1;
  1,    8,     1;
  1,   22,    22,      1;
  1,   52,   264,     52,       1;
  1,  114,  1208,   1208,     114,      1;
  1,  240,  4764,  19328,    4764,    240,     1;
  1,  494, 17172, 124952,  124952,  17172,   494,    1;
  1, 1004, 58432, 705872, 2499040, 705872, 58432, 1004,   1;

Links

Crossrefs

Cf. A008292, A144463, A144470, A174301.

Programs

  • Magma
    Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >; [[Floor(n/2) ge k select 2^k*Eulerian(n+1,k) else 2^(n-k)*Eulerian(n+1,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Apr 15 2019
  • Mathematica
    Eulerian[n_, k_]:= Sum[(-1)^j*Binomial[n+1,j]*(k-j+1)^n, {j,0,k+1}];
Table[Eulerian[n+1,m]*If[Floor[n/2] >= m, 2^m, 2^(n-m)], {n,0,10}, {m,0,n} ]//Flatten (* modified by G. C. Greubel, Apr 15 2019 *)
  • PARI
    {eulerian(n,k) = sum(j=0,k+1, (-1)^j*binomial(n+1,j)*(k-j+1)^n)};
for(n=0,10, for(k=0,n, print1(eulerian(n+1,k)*if(floor(n/2)>=k, 2^k, 2^(n-k)), ", "))) \\ G. C. Greubel, Apr 15 2019
  • Sage
    def Eulerian(n,k): return sum((-1)^j*binomial(n+1,j)*(k-j+1)^n for j in (0..k+1))
def T(n,k):
   if floor(n/2)>=k: return 2^k*Eulerian(n+1,k)
   else: return 2^(n-k)*Eulerian(n+1,k)
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Apr 15 2019

Formula

T(n,k) = Eulerian(n+1, k)*if(floor(n/2) greater than or equal to k then 2^m otherwise 2^(n-k)), where the Eulerian numbers are defined as A008292(n,k).

Extensions

Edited by G. C. Greubel, Apr 15 2019
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.