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Search: id:A106199
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| A106199 |
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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. |
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+0
2
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| 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
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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Denominators form A106200.
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REFERENCES
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E. Deutsch and M. S. Klamkin, Problem 10540, Amer. Math. Monthly 108, (2001), p. 172.
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FORMULA
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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).
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EXAMPLE
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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 fiftyseven times the value 0; then the considered probablity is 12/81=4/27.
0,0,4/27,15/64,36/125,103/324,832/2401
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MAPLE
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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);
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CROSSREFS
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Cf. A106200.
Sequence in context: A113693 A077414 A015653 this_sequence A113289 A033813 A112666
Adjacent sequences: A106196 A106197 A106198 this_sequence A106200 A106201 A106202
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KEYWORD
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frac,nonn
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 24 2005
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