Minimal positive 5-tuple (b1,b2,b3,b4,b5) such that n = b1/b5 + b2/b1 + b3/b2 + b4/b3 + b5/b4.

1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 4, 2, 1, 3, 4, 2, 6, 1, 5, 4, 2, 5, 1, 2, 3, 1, 6, 1, 8, 6, 8, 6, 1, 9, 6, 4, 6, 1, 6, 1, 6, 2, 1, 1, 10, 4, 10, 1, 1, 12, 2, 3, 1, 6, 1, 9, 6, 1, 16, 8, 4, 2, 1, 6, 4, 1, 12, 1, 18, 6, 3, 2, 1, 8, 6, 1, 12, 1, 18, 3, 2, 6

5

By minimal, we mean that max(b1,b2,b3,b4,b5) is as small as possible. If there are multiple solutions having the same maximum, we take the first one. These numbers are different from those in Lee’s paper; we find the minimal 5-tuple and we compute it for all n from 5 to 100.

T. D. Noe, Plot of rows 5..100

T. D. Noe, Table of rows 5..100

Jungin Lee, On a sum of positive rational numbers whose product is 1, arXiv 1603.08420 (Feb 26 2016)

(Mma) Table[mx = 2*n; best = {}; While[soln = {}; Do[If[b1/b5 + b2/b1 + b3/b2 + b4/b3 + b5/b4 == n, soln = {b1, b2, b3, b4, b5}; Break[]], {b1, mx}, {b2, mx}, {b3, mx}, {b4, mx}, {b5, mx}]; soln != {}, mx = Max[soln] - 1; best = soln]; best, {n, 5, 22}]

nonn,tabl

T. D. Noe, Apr 07 2016