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%I #25 Jul 03 2022 09:17:21
%S 1,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5,6,6,6,6,7,7,7,7,8,8,8,8,9,9,9,9,10,
%T 10,10,10,11,11,11,11,12,12,12,12,13,14,13,13,14,14,14,14,15,15,15,15,
%U 16,16,16,16,17,17,17,17,18,18,18,18,19,20,19,19,20,20,20,20,21,21,21
%N x-value of the solution (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and having the largest z-value. The y and z components are in A075246 and A075247.
%C See A073101 for more details.
%C a(n) = floor(n/4) + 1, at least up to n = 2000, except for some n = 8k+1 (k = 6, 9, 11, 14, 20, 21, 24, 29, 30, 35, 39, 41, 44, 45, 50, ...), where a(n) is one larger than a(n-1) and a(n+1). - _M. F. Hasler_, Jul 02 2022
%H M. F. Hasler, <a href="/A075245/b075245.txt">Table of n, a(n) for n = 3..2000</a>
%F Conjecture: a(n) = floor(n/4) + d, with d = 1 except for some n = 8k+1 (k = 6, 9, 11, 14, 20, 21, 24, 29, 30, 35, 39, ...) where d = 2 . - _M. F. Hasler_, Jul 02 2022
%e For n = 3, we have a(3) = 1 = x in 4/3 = 1/x + 1/y + 1/z with y = 4 and z = 12 which is the largest possible z: Indeed, x < y < z gives 4/3 < 3/x, so only x = 1 and 2 are possible, and then with y < z, 2/y > 4/3 - 1/x is impossible for x = 2 < y < 12/5 and for x = 1 < y < 6 only y = 4 gives a solution.
%p A075245:= proc () local t, n, a, b, t1, largex, largez; for n from 3 to 100 do t := 4/n; largez := 0; largex := 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); largex := a end if end if end do end do; lprint(n, largex) end do end proc; # [program derived from A192787] _Patrick J. McNab_, Aug 20 2014
%t 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++ ]]; xLst
%o (PARI) apply( {A075245(n,c=1,a)=for(x=n\4+1, 3*n\4, my(t=4/n-1/x); for(y=max(1\t,x)+1, ceil(2/t)-1, t-1/y >= c && break; numerator(t-1/y)==1 && [c,a]=[t-1/y,x])); a}, [3..99]) \\ _M. F. Hasler_, Jul 02 2022
%Y Cf. A073101, A075246, A075247, A192787.
%K hard,nice,nonn
%O 3,2
%A _T. D. Noe_, Sep 10 2002
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