For a positive integer n that is not a power of two, we define t(n) as the greatest odd divisor of n and r(n) as the smallest positive odd divisor of n unequal to 1. Determine all positive integers n that are not a power of two and for which we have n = 3t(n) + 5r(n).
I will rewrite the equation as n = 5r + 3t because r<=t.
One group includes all odd primes times 8. Such numbers have r(n) = t(n) = the prime itself
For these cases, n:r:t are in the ratio 8:1:1
I believe this covers all instances where r(n) = t(n).
If these are excluded then I found only two solutions testing n up to 2000000:
n r(n) t(n)
60, 3, 15
100, 5, 25
Note that n:r:t are in the ratio 20:1:5
We get n = 3(5r) + 5r = 20r
Suppose we want to try some odd number 's', s>5, hoping it will be r(N) for some N, while using the same ratios.
Then we hope that (n,r,t) = (20s, s, 5s) works for some odd s>5.
But it doesn't because 20s includes a factor of 5, so any potential candidate s, then r(20s) is either 3 or 5.
So assuming that N=20s, there are no further solutions.
But I have not proved that there is no other set of ratios that work, other than:
8:1:1 n = 8*π where π is any prime > 2
20:1:5 n = {60, 100}
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Posted by Larry
on 2024-09-10 12:27:18 |
I found ...
