[seqfan] Re: both n and n^2 use just digits (2,3,5,7}

David Wilson davidwwilson at comcast.net
Fri Sep 16 06:33:19 CEST 2016

5, 235, and 72335 are the only such numbers <= 31 digits.
I suspect there is a probabilistic argument that the number of such numbers is finite, and the probability of the existence of such a number with >= 32 digits is extremely small.
The only methods I know for categorically solving this question involve showing that for sufficient d, if a number n consists of d good digits, then n^2 either cannot start with d good digits or end with d good digits. These arguments do not work for the set of digits {2,3,5,7}.
So the answer to your question is almost certainly yes, but I doubt we'll ever prove it.

> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Zak
> Seidov via SeqFan
> Sent: Monday, August 15, 2016 7:35 AM
> To: Sequence Fanatics Discussion list
> Cc: Zak Seidov
> Subject: [seqfan] both n and n^2 use just digits (2,3,5,7}
>  Are {5,235,72335}
> the only numbers n
> such that  both n and n^2 use just digits (2,3,5,7}?
> Cf. A191486, A275971.
> zak
> --
> Seqfan Mailing list - http://list.seqfan.eu/

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