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).
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
Keywords
Examples
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).
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..74
- Zhi-Wei Sun, On finite products of (p+4)/(p+2) with p prime, Question 476951 at MathOverflow, August 15, 2024.
Programs
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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]
Extensions
During August 24-25, 2024, Daniel Weber and Deyi Chen found the values of a(21) and a(54), respectively,
Comments