[seqfan] Re: Conjecture: a(n) = n only if n = 1 or 9

D. S. McNeil dsm054 at gmail.com
Sat Apr 17 20:50:57 CEST 2021

I think it's true.  Sketch I'm too lazy to formalize (and whenever I break
things apart by cases I feel like I'm missing something obvious):

Say some composite n is a pure prime power p^k.

For k=2, we pick up a factor of p every p in the product (except for p^2
itself), and so A(p^2) = p^(p-1).  The only solution p=3 generates the n=9
solution.  For p > 3, A(p^2) > n.

For k=3, we pick up additional factors of p^2, and if I'm counting
correctly A(p^3) = p^(p^2+p-2), which generates no solutions.  Moreover,
A(p^3) > n for all p.

For k>=4, we have at least prod(p^i,i=1..k-1) as part of the product, and
so p^((k-1)*k/2) | A(n).  Since (k-1)*k/2 > k for k >= 4, this suffices to
show A(n) > n.

Now assume it's not a prime power, and we have n = p_0^k_0 p_1^k_1..
p_m^k_m.  It suffices to show A(p^k*q) > p^k*q for gcd(p, q) = 1.

We'll pick up factors of p^k and q immediately in the expanded product, and
thus p^k * q | A(p^k*q), so all we need is one other component to exceed
p^k * q.

If k>=2, p itself contributes in the terms: [p, p^k, q] giving p^(k+1) * q
| A(n).
If k = 1 and p=2, then q >= 3, n >= 6, and thus [p=2, 2*p=4, q] giving
p^2*q | A(n).
If k = 1 and q=2, then p >= 3, n >= 6, and thus [p, q=2, 2*q=4] giving
p*q^2 | A(n)
If k = 1 and p,q >= 3, then we have [p, 2*p, q] giving p^2 * q| A(n)

And I think that's all the cases.

If I didn't make a silly error in the above it'd be the first time ever,
but I'm pretty sure some argument along these lines works.

Doug #lockdownboredom

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