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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, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
LINKS
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
KEYWORD
frac,nonn
AUTHOR
Emeric Deutsch, Apr 24 2005
STATUS
approved

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Last modified March 28 14:38 EDT 2024. Contains 371254 sequences. (Running on oeis4.)