%I #38 Aug 25 2024 18:41:53
%S 50234415,1085,3,5,7,43493,11,13,232087,17,19,579617,23,940141,
%T 5208547,29,31,4196617,3301747,37,675790721971,41,43,15940937,47,
%U 24692861,4807811,53,5461783,21086917,59,61,22771807,11287877,67,17498441,71,73,22057411,32126917
%N Least squarefree integer m > 1 such that the product of all those (p + 4)/(p + 2) (with p a prime divisor of m) equals (2*n + 1)/(2*n - 1).
%C For m > k >= 0, clearly (2*m + 1)/(2*k + 1) is the product of those (2*n + 1)/(2*n - 1) (n = 1..m).
%C Conjecture: a(n) exists for any positive integer n.
%C Perhaps, for any m > k >= 0, we can write (2*m + 1)/(2*k + 1) as a product of distinct numbers of the form (p + 4)/(p + 2) with p an odd prime.
%H Zhi-Wei Sun, <a href="/A375351/b375351.txt">Table of n, a(n) for n = 1..74</a>
%H Zhi-Wei Sun, <a href="https://mathoverflow.net/questions/476951">On finite products of (p+4)/(p+2) with p prime</a>, Question 476951 at MathOverflow, August 15, 2024.
%e a(1) = 50234415 = 3*5*7*11*23*31*61 with (2*1+1)/(2*1-1) = (3+4)/(3+2)*(5+4)/(5+2)*(7+ 4)/(7+2)*(11+4)/(11+2)*(23+4)/(23+2)*(31+4)/(31+2)*(61+4)/(61+2).
%e a(2) = 1085 = 5*7*31 with (2*2+1)/(2*2-1) = (5+4)/(5+2)*(7+4)/(7+2)*(31+4)/(31+2).
%t tab={};Do[m=2;Label[bb];If[SquareFreeQ[m]==False,m=m+1;Goto[bb]];f=FactorInteger[m];
%t If[Product[(f[[i]][[1]]+4)/(f[[i]][[1]]+2),{i,1,Length[f]}]==(2n+1)/(2n-1),tab=Append[tab,m];Goto[aa]];m=m+1;Goto[bb];Label[aa],{n,1,20}];Print[tab]
%Y Cf. A000040, A005117, A005408, A375262.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Aug 14 2024
%E During August 24-25, 2024, Daniel Weber and Deyi Chen found the values of a(21) and a(54), respectively,