login
A375351
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).
1
50234415, 1085, 3, 5, 7, 43493, 11, 13, 232087, 17, 19, 579617, 23, 940141, 5208547, 29, 31, 4196617, 3301747, 37, 675790721971, 41, 43, 15940937, 47, 24692861, 4807811, 53, 5461783, 21086917, 59, 61, 22771807, 11287877, 67, 17498441, 71, 73, 22057411, 32126917
OFFSET
1,1
COMMENTS
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).
Conjecture: a(n) exists for any positive integer n.
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.
LINKS
Zhi-Wei Sun, On finite products of (p+4)/(p+2) with p prime, Question 476951 at MathOverflow, August 15, 2024.
EXAMPLE
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).
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).
MATHEMATICA
tab={}; Do[m=2; Label[bb]; If[SquareFreeQ[m]==False, m=m+1; Goto[bb]]; f=FactorInteger[m];
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]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 14 2024
EXTENSIONS
During August 24-25, 2024, Daniel Weber and Deyi Chen found the values of a(21) and a(54), respectively,
STATUS
approved