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A075247
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Largest possible z-value of an integer solution (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z. The x and y components are in A075245 and A075246.
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14
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12, 6, 20, 42, 210, 42, 90, 240, 1122, 156, 468, 812, 3660, 420, 510, 2070, 9120, 930, 1806, 4422, 19182, 1806, 2100, 8372, 35910, 3192, 9048, 14520, 61752, 5256, 9900, 23562, 99540, 8190, 22940, 36290, 152490, 12210, 6314, 53592, 224202, 17556
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OFFSET
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3,1
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COMMENTS
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LINKS
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EXAMPLE
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For n = 6 we have a(n) = 42 the largest possible z in a solution of 4/n = 2/3 = 1/x + 1/y + 1/z with 0 < x < y < z in the integers. Indeed, from 1/x < 2/3 < 3/x we have 3/2 < x < 9/2. For x = 2 we get 2/y > 2/3 - 1/2 = 1/6 > 1/y <=> 6 < y < 12, and each of these y except y = 11 yields a solution, with z = 42, 24, 18, 15 and 12. There are no other possible z values: x = 3 gives 2/y > 1/3 <=> y < 6 and indeed y = 4 gives a solution with z = 12, no solution for y = 5; finally, x = 4 gives 2/y > 5/12 <=> y < 24/5, impossible with y > x.
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MAPLE
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A075247:= proc () local t, n, a, b, t1, largey, largez; for n from 3 to 100 do t := 4/n; largez := 0; for a from floor(1/t)+1 to floor(3/t) do t1 := t-1/a; for b from max(a, floor(1/t1)+1) to floor(2/t1) do if `and`(type(1/(t1-1/b), integer), a < b, b < 1/(t1-1/b)) then if largez < 1/(t1-1/b) then largez := 1/(t1-1/b) end if end if end do end do; lprint(n, largez) end do end proc; # [program derived from A192787] Patrick J. McNab, Aug 20 2014
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MATHEMATICA
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For[xLst={}; yLst={}; zLst={}; n=3, n<=100, n++, cnt=0; xr=n/4; If[IntegerQ[xr], x=xr+1, x=Ceiling[xr]]; While[yr=1/(4/n-1/x); If[IntegerQ[yr], y=yr+1, y=Ceiling[yr]]; cnt==0&&y>x, While[zr=1/(4/n-1/x-1/y); cnt==0&&zr>y, If[IntegerQ[zr], z=zr; cnt++; AppendTo[xLst, x]; AppendTo[yLst, y]; AppendTo[zLst, z]]; y++ ]; x++ ]]; zLst
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PROG
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(PARI) apply( {A075247(n, c=1, t)=for(x=n\4+1, 3*n\4, for(y=max(1\t=4/n-1/x, x)+1, ceil(2/t)-1, t-1/y >= c && break; numerator(t-1/y)==1 && c=t-1/y)); 1/c}, [3..99]) \\ M. F. Hasler, Jul 02 2022
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CROSSREFS
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KEYWORD
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hard,nice,nonn
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AUTHOR
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STATUS
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approved
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