[seqfan] Re: 1, 10793, 161393,
W. Edwin Clark
wclark at mail.usf.edu
Mon Dec 10 21:13:34 CET 2018
What's special about 3*n?
For example, replace 3*n by k and the following values of k satisfy
2^(k-2) + (k-2)^2 == 0 mod k:
3, 4, 10, 130, 2146, 4810, 12874, 14170, 32379, 33674, 45994, 116194,
185770, 186826, 271210, 426010, 484179, 524290, 601354, 612770, 755794,
996034
which I don't find in the OEIS.
Note that 2 + (k - 2) = k suggesting:
More generally there are numerous pairs a,b such that (a^b + b^a) == 0 mod
(a+b), but I don't see how to make a sequence out of such pairs. Have they
been
studied somewhere? One possiblity for a sequence is a(n) = number of x in
{1,...,n-1} such that x^(n-x) + (n-x)^x == 0 mod n which begins:
1, 2, 3, 2, 3, 2, 5, 4, 7, 4, 7, 2, 3, 2, 9, 2, 7, 6, 7, 8, 3, 2, 15, 6, 9,
10, 11, 4, 17, 4, 17, 4, 9, 8, 15, 2, 5, 10, 15 --- also not in OEIS
On Sun, Dec 9, 2018 at 9:52 AM юрий герасимов <2stepan at rambler.ru> wrote:
> Dear SeqFans, that are the other solutions to (2^(3*n-2)+(3*n-2)^2)==0 mod
> (3n)?
> Thanks you. юрий герасимов
>
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