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A005558
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a(n) is the number of n-step walks on square lattice such that 0 <= y <= x at each step.
(Formerly M2598)
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12
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1, 1, 3, 6, 20, 50, 175, 490, 1764, 5292, 19404, 60984, 226512, 736164, 2760615, 9202050, 34763300, 118195220, 449141836, 1551580888, 5924217936, 20734762776, 79483257308, 281248448936, 1081724803600, 3863302870000, 14901311070000, 53644719852000
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OFFSET
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0,3
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COMMENTS
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Number of n-step walks that start at the origin, constrained to stay in the first octant (0 <= y <= x). (Conjectured) - Benjamin Phillabaum, Mar 11 2011, corrected by Robert Israel, Oct 07 2015
For n >= 1, a(n-1) is the number of Dyck Paths with semilength n having floor((n+2)/2) U's in odd numbered positions. Example: (U is in odd numbered position and u is in even numbered position) Dyck path with n=5, floor ((5+2)/2)=3: UuddUuUddd. - Roger Ford, May 27 2017
The ratio of the number of n-step walks on the octant with an equal number of North steps and South steps to the total number of n-step walks on the octant is A005817(n)/a(n). For the reduced ratio, if n is divisible by 4 or n-1 is divisible by 4 the ratio is 1:floor(n/4)+1 and for all other values of n the ratio is 2:floor(n/2)+2. Example n = 4: A005817(4) = 10; EEEE, EEEW, EEWE, EWEE, EWEW, EEWW, ENSE, ENES, ENSW, EENS; a(4) = 20; 10:20 reduces to 1:2. - Roger Ford, Nov 04 2019
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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a(n) = C(n+1, ceiling(n/2))*C(n, floor(n/2)) - C(n+1, ceiling((n-1)/2))*C(n, floor((n-1)/2)). - Paul D. Hanna, Apr 16 2004
G.f.: (1/(4x^2))*((16*x^2-1)*(hypergeom([1/2, 1/2],[1],16*x^2)+2*x*(4*x-1)*hypergeom([3/2, 3/2],[2],16*x^2))-2*x+1). - Mark van Hoeij, Oct 13 2009
E.g.f (conjectured): BesselI(1,2*x)*(BesselI(0,2*x)+BesselI(1,2*x))/x. - Benjamin Phillabaum, Feb 25 2011
Conjecture: (2*n+1)*(n+3)*(n+2)*a(n) - 4*(2*n^2+4*n+3)*a(n-1) - 16*n*(2*n+3)*(n-1)*a(n-2) = 0. - R. J. Mathar, Apr 02 2017
Conjecture: (n+3)*(n+2)*a(n) - 4*(n^2+3*n+1)*a(n-1) + 16*(-n^2+n+1)*a(n-2) + 64*(n-1)*(n-2)*a(n-3) = 0. - R. J. Mathar, Apr 02 2017
a(n) = Sum_{k=0..floor(n/2)} n!/(k!*k!*(floor(n/2)-k)!*(floor((n+1)/2)-k)!*(k+1)) (conjectured). - Roger Ford, Aug 04 2017
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MAPLE
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A:= proc(n, x, y) option remember;
local j, xpyp, xp, yp, res;
xpyp:= [[x-1, y], [x+1, y], [x, y-1], [x, y+1]];
res:= 0;
for j from 1 to 4 do
xp:= xpyp[j, 1];
yp:= xpyp[j, 2];
if xp < 0 or xp > yp or xp + yp > n then next fi;
res:= res + procname(n-1, xp, yp)
od;
return res
end proc:
A(0, 0, 0) := 1:
seq(add(add(A(n, x, y), y = x .. n - x), x = 0 .. floor(n/2)), n = 0 .. 50); # Robert Israel, Oct 07 2015
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MATHEMATICA
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a[n_] := 1/2*Binomial[2*Floor[n/2]+1, Floor[n/2]+1]*CatalanNumber[1/2*(n+Mod[n, 2])]*(Mod[n, 2]+2); Table[a[n]//Abs, {n, 0, 27}] (* Jean-François Alcover, Mar 13 2014 *)
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PROG
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(PARI) a(n)=binomial(n+1, ceil(n/2))*binomial(n, floor(n/2)) - binomial(n+1, ceil((n-1)/2))*binomial(n, floor((n-1)/2))
(Magma) [Binomial(n+1, Ceiling(n/2))*Binomial(n, Floor(n/2)) - Binomial(n+1, Ceiling((n-1)/2))*Binomial(n, Floor((n-1)/2)): n in [0..30]]; // Vincenzo Librandi, Sep 30 2015
(Python)
from sympy import ceiling as c, binomial
def a(n):
return binomial(n + 1, c(n/2))*binomial(n, n//2) - binomial(n + 1, c((n - 1)/2))*binomial(n, (n - 1)//2)
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CROSSREFS
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KEYWORD
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nonn,walk
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AUTHOR
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STATUS
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approved
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