[seqfan] Re: Conjecture: a(n) = n only if n = 1 or 9
Alonso Del Arte
alonso.delarte at gmail.com
Sun Apr 18 20:05:21 CEST 2021
Thanks, D. S., I think it checks out. I appreciate your taking the time to
sketch this out. In all honesty, the most I did was scan the B-file.
Al
On Sat, Apr 17, 2021 at 2:51 PM D. S. McNeil <dsm054 at gmail.com> wrote:
> 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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>
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Alonso del Arte
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