[seqfan] Re: Which palindromes are fourth powers?

[Vladimir Shevelev]

For my simple case, we shoul consider the following cases for n:
1) 1+10^k+10^m, 0<k<m,
2) 1+2*10^r, r>0,
3) 2+10^s, s>0,
4) 3*10^t, t>=0.
For the two last cases n^4 trivially is not a palindrome.  Others are 
considered directly.
For example, for r>=2, in the second case we have
(1+2*10^r)^4=1+8*10^r+4*10^(2*r)+
2*10^(2*r+1)+2*10^(3*r)+3*10^(3*r+1)+
6*10^(4*r)+10^(4*r+1)
which cannot be a palindrome.
If r=1, we have
1+8*10+...9*10^4+10^5
which also is not a palindrome.
The first case is a little longer.

Best regards,
Vladimir

________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Neil Sloane [njasloane at gmail.com]
Sent: 24 October 2015 01:58
To: Sequence Fanatics Discussion list
Subject: [seqfan] Which palindromes are fourth powers?

Dear Sequence Fans, A056810 gives numbers n such that n^4 is a palindrome.
The initial terms were calculated by Bob Wilson, back in the year 2000.
They are
0, 1, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001
The question is, is every term of the form 10^k+1?
It would be nice to have some more terms - or some theorems.
There is a claim that:

The sequence contains no term with digit sum 3. - Vladimir Shevelev, May 23
2011

but I don't know if this claim (which seems very plausible) has been
checked.

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com

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