OFFSET
3,1
COMMENTS
See A073101 for more details.
LINKS
M. F. Hasler, Table of n, a(n) for n = 3..2000
EXAMPLE
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.
MAPLE
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
MATHEMATICA
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
PROG
(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
CROSSREFS
KEYWORD
hard,nice,nonn
AUTHOR
T. D. Noe, Sep 10 2002
STATUS
approved