A few observations: It’s quite clear that all ##g## for ##n>2## are congruent to 1 modulo 4. In addition I’ve found that all ##g## are relatively prime. It’s also not quite true that the ##p## increase as ##m## increases; for instance, 941 has ##p## of 941, while 969581 has ##p## of 521...
Let $$g(n)$$ be the numerators of the elements of the recursion $$i(n)=i(n-1)+\frac{1}{i(n-1)}$$ when they are expressed in simplest form, with $$i(0)=1$$. Let $$p$$ be the smallest prime factor of $$g(m)$$. Show that $$p>4m-4$$.Homework Equations
Euler's Theorem?
The Attempt at a Solution
OEIS...