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 11-14 of 14 results.

A289352 Irregular triangle read by rows: T(n,k) = the number of Dyck paths of semilength n with floor((n+2)/2) up movements in odd numbered positions and k returns to the x axis.

Original entry on oeis.org

1, 0, 1, 1, 2, 1, 2, 3, 6, 8, 6, 10, 15, 15, 10, 50, 60, 45, 20, 105, 140, 126, 84, 35, 490, 560, 420, 224, 70, 1176, 1470, 1260, 840, 420, 126, 5292, 5880, 4410, 2520, 1050, 252, 13860, 16632, 13860, 9240, 4950, 1980, 462, 60984, 66528, 49896, 29568, 13860, 4752, 924
Offset: 1

Views

Author

Roger Ford, Jul 03 2017

Keywords

Examples

			n\k  1    2    3    4    5
1    1
2    0    1
3    1    2
4    1    2    3
5    6    8    6
6   10   15   15   10
7   50   60   45   20
8  105  140  126   84   35
9  490  560  420  224   70
T(4,3)=3: (U = up in odd position, u = up in even position, d = down, _ = return to x axis, floor ((n+2)/2) = 3 up movements in odd position)  Ud_Ud_Uudd_, Uudd_Ud_Ud_, Ud_Uudd_Ud_.
		

Crossrefs

Formula

T(1,1)=1, T(2,1)=0, T(2,2)=1, For n >= 3, T(n,k) = (1/floor((n-1)/2))*C(n-1,floor((n-3)/2))*C(n-1-k,floor((n-3)/2))*k (conjectured).
Row sums of T(n,k) = A005558(a(n-1)).
T(n,1) = A001263(T(n-1,floor(n/2)).
T(n,floor((n+2)/2)) = A001405(a(n-1)).

A297672 Array with four columns read by rows: T(n,k) = number of n step walks in the first octant on a square lattice with last step being right (k=1), left (k=2), up (k=3) or down (k=4).

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 1, 0, 3, 1, 1, 1, 6, 6, 6, 2, 20, 10, 10, 10, 50, 50, 50, 25, 175, 105, 105, 105, 490, 490, 490, 294, 1764, 1176, 1176, 1176, 5292, 5292, 5292, 3528, 19404, 13860, 13860
Offset: 1

Views

Author

Roger Ford, Jan 02 2018

Keywords

Comments

Sum_{k=1..4} T(n,k) = A005558(n).
For n >= 1, the ratio of the numbers of right or up last steps to left or down last steps is floor((n+2)/2): floor(n/2). - Roger Ford, Oct 28 2019

Examples

			k=    1    2    3    4    total
N   right left up   down  walks
1     1    0    0    0    =1
2     1    1    1    0    =3
3     3    1    1    1    =6
4     6    6    6    2    =20
There are 6 walks of 4 steps in the octant with the last step right. T(4,1)=6 RRRR, RRLR, RLRR, RUDR, RURR, RRUR.
		

Crossrefs

Formula

n=1: T(1,1)=1, T(1,2)=0, T(1,3)=0, T(1,4)=0;
n>1: T(n,1) = C(n,floor(n/2))*C(n-1,floor((n-1)/2)) - C(n,floor((n-1)/2))*C(n-1,floor((n-2)/2));
T(n,2) = T(n,3) = C(n-1,floor(n/2)-1)*C(n,floor(n/2)-1)/floor(n/2);
n odd: T(n,4) = T(n,2);
n even: T(n,4) = T(n,2)*((n/2-1)/(n/2+1));
For n > 1, T(n,2) = T(n,3) = A001263(n,floor(n/2)).

A298122 a(n) is the number of n nonintersecting arches above the x-axis that start and/or end with an arch length equal to one and have floor((n+2)/2) arches starting in odd numbered positions.

Original entry on oeis.org

1, 1, 2, 5, 11, 34, 90, 300, 875, 3038, 9408, 33516, 108108, 392040, 1302444, 4785066, 16256955, 60324550, 208579800, 780088452, 2735682092, 10296854984, 36532677272, 138231751840, 495241833996, 1882201158264, 6799413051200, 25939319270000, 94374970110000
Offset: 1

Views

Author

Roger Ford, Jan 12 2018

Keywords

Examples

			Example: For n = 4 the a(4) = 5 solutions are as follows. (The numbers under the arches represent arches starting in an odd-numbered position on the x-axis.)
    /\                                /\
   //\\      /\            /\        //\\        /\
  ///\\\/\, //\\ /\ /\, /\//\\/\, /\///\\\, /\/\//\\.
  1 3   7   1    5  7   1 3   7   1 3 5     1 3 5
		

Crossrefs

Formula

a(1) = a(2) = 1, a(3) = 2; for n > 3, a(n) = 2*(C(n-1, floor((n-1)/2))*C(n-2, floor((n-2)/2)) - (C(n-1, floor((n-2)/2))*C(n-2, floor((n-3)/2)))) - (C(n-3, floor((n-1)/2))*C(n-2, floor((n-1)/2))/(floor((n-1)/2)+1)).

A302185 Number of 3D n-step walks of type acc.

Original entry on oeis.org

1, 2, 7, 24, 98, 400, 1785, 7980, 37674, 178164, 874146, 4294752, 21667932, 109436184, 563910633, 2908233900, 15235550330, 79870553620, 424021948350, 2252356700880, 12088746573540, 64913104882080, 351594254659830, 1905139854213960, 10399223643879420, 56783986550235000
Offset: 0

Views

Author

N. J. A. Sloane, Apr 09 2018

Keywords

Comments

See Dershowitz (2017) for precise definition.

Crossrefs

Programs

  • Maple
    b:= n-> binomial(n, floor(n/2))*binomial(n+1, floor((n+1)/2)):
    C:= n-> binomial(2*n, n)/(n+1):
    a:= n-> add(binomial(n, 2*k)*C(k)*b(n-2*k), k=0..n/2):
    seq(a(n), n=0..25);  # Alois P. Heinz, Dec 06 2024
    # second Maple program:
    a:= proc(n) option remember; `if`(n<4, [1, 2, 7, 24][n+1],
          (8*(14*n^4+85*n^3+190*n^2+188*n+63)*a(n-1)+4*(n-1)*
          (80*n^4+418*n^3+676*n^2+269*n-108)*a(n-2)-96*(n-1)*(n-2)*
          (10*n^2+31*n+27)*a(n-3)-144*(n-1)*(n-2)*(n-3)*(8*n^2+33*n+36)*
           a(n-4))/((n+4)*(n+3)*(n+2)*(8*n^2+17*n+11)))
        end:
    seq(a(n), n=0..25);  # Alois P. Heinz, Dec 06 2024
  • Mathematica
    b[n_] := Binomial[n, Floor[n/2]]*Binomial[n+1, Floor[(n+1)/2]];
    c[n_] := Binomial[2*n, n]/(n+1);
    a[n_] := Sum[Binomial[n, 2*k]*c[k]*b[n - 2*k], {k, 0, n/2}];
    Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 28 2025, after Alois P. Heinz *)
  • Python
    from math import comb as binomial
    def C(n): return (binomial(2*n, n)//(n+1)) # Catalan numbers
    def a(n):
        return sum(binomial(n, k)*C((k+1)//2)*C(k//2)*(2*(k//2)+1)*binomial(n-k, (n-k)//2) for k in range(n+1))
    print([a(n) for n in range(26)]) # Mélika Tebni, Dec 06 2024

Formula

From Mélika Tebni, Dec 06 2024: (Start)
E.g.f.: (BesselI(0, 2*x) + BesselI(1, 2*x))^2*BesselI(1, 2*x) / x.
a(n) = Sum_{k=0..n} binomial(n, k)*A005558(k)*A001405(n-k).
a(2*n+1) = 2*A302182(2*n+1) = A135394(n) / (n+1).
For n > 0, a(A000918(n)) is odd. (End)

Extensions

a(13)-a(25) from Mélika Tebni, Dec 06 2024
Previous Showing 11-14 of 14 results.