A106200 a(n)=denominator 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.

1, 1, 27, 64, 125, 324, 2401, 512, 6561, 2500, 14641, 324, 28561, 2401, 50625, 8192, 83521, 8748, 130321, 1250, 194481, 14641, 279841, 82944, 390625, 114244, 531441, 153664, 707281, 202500, 923521, 262144, 1185921, 334084, 1500625, 209952

Offset

1,3

Links

Table of n, a(n) for n=1..36.

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

Formula

a(n)=denominator 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)=27 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(denom(simplify(p(n))), n=1..45);

Crossrefs

Cf. A106199 (numerators).

Sequence in context: A088248 A319389 A340700 * A303972 A361147 A099865

Adjacent sequences: A106197 A106198 A106199 * A106201 A106202 A106203

Keyword

frac,nonn

Author

Emeric Deutsch, Apr 24 2005

Status

approved