A075246 y-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 x and z components are in A075245 and A075247.
4, 3, 4, 7, 15, 7, 10, 16, 34, 13, 18, 29, 61, 21, 30, 46, 96, 31, 43, 67, 139, 43, 60, 92, 190, 57, 78, 121, 249, 73, 100, 154, 316, 91, 124, 191, 391, 111, 154, 232, 474, 133, 181, 277, 565, 157, 99, 326, 664, 183, 248, 379, 771, 211, 286, 436, 886, 241, 326
Offset: 3
Links
- M. F. Hasler, Table of n, a(n) for n = 3..2000
Crossrefs
Programs
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Maple
A075246:=proc() local t, n,a,b,t1,largey,largez; for n from 3 to 100 do t:=4/n; largez:=0; largey:=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),alargez then largez:=(1/(t1 - 1/b)); largey:=b; end if end if end do; end do; lprint(n, largey) end do; end proc; # [program derived from A192787] Patrick J. McNab, Aug 20 2014
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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++ ]]; yLst
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PARI
apply( {A075246(n, c=1, a, 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, a]=[t-1/y, y])); a}, [3..99]) \\ M. F. Hasler, Jul 03 2022
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