A106199 a(n)=numerator of the probability that (x-y)/(x+y)+(y-z)/(y+z)+(z-u)/(z+u)+ (u-x)/(u+x)>0, assuming that each random quadruple of integers (x,y,z,u), with a<=x,y,z,u<=n, is equally likely.

0, 0, 4, 15, 36, 103, 832, 187, 2500, 981, 5904, 133, 11944, 1018, 21716, 3551, 36568, 3859, 57936, 559, 87464, 6619, 127156, 37835, 178916, 52517, 245116, 71083, 328180, 94187, 430688, 122529, 555508, 156813, 705696, 98899, 884432, 123199

Offset

1,3

Links

Table of n, a(n) for n=1..38.

E. Deutsch and M. S. Klamkin, Counting the Solutions: Problem 10540, Amer. Math. Monthly 108, (2001), p. 172.

Formula

a(n)=numerator of [n(n-1)(n^2-n-1)+4sum(floor(n/(k^2))*phi(k), k=2...floor(sqrt(n)))-2sum(floor(n/k)^2*phi(k), k=2... n)]/(2n^4).

Example

a(3)=4 because at the 81 quadruples (x,y,z,u) (1<=x,y,z,u<=3) the function
(x-y)/(x+y)+(y-z)/(y+z)+(z-u)/(z+u)+(u-x)/(u+x) assumes twelve times the value 1/30, twelve times the value -1/30 and fifty-seven times the value 0; then the considered probability is 12/81=4/27.
0,0,4/27,15/64,36/125,103/324,832/2401

Maple

with(numtheory): a:=n*(n-1)*(n^2-n-1): b:=4*sum(floor(n/k^2)*phi(k), k=2..floor(sqrt(n))): c:=2*sum((floor(n/k))^2*phi(k), k=2..n): p:=proc(n) (a+b-c)/2/n^4 end: seq(numer(simplify(p(n))), n=1..45);

Crossrefs

Cf. A106200 (denominators).

Sequence in context: A304487 A350689 A015653 * A113289 A212974 A033813

Adjacent sequences: A106196 A106197 A106198 * A106200 A106201 A106202

Keyword

frac,nonn

Author

Emeric Deutsch, Apr 24 2005

Status

approved