[seqfan] Re: 1, 10793, 161393,
israel at math.ubc.ca
israel at math.ubc.ca
Mon Dec 10 22:25:57 CET 2018
Also consider integers t such that there exists a with 1 < a < t-1 such
that a^(t-a) + (t-a)^a == 0 mod t (Of course for any t, a=1 and a=t-1 would
work).
9, 11, 19, 27, 29, 31, 33, 35, 38, 43, 45, 49, 57, 58, 59, 61, 62, 63, 67,
77, 79, 81, 83, 86, 87, 89, 91, 93, 95, 97, 98, 99, 103, 107, 111, 113,
116, 117, 121, 122, 124, 127, 129, 131, 133, 134, 135, 139, 141, 143, 145,
146, 147, 149, 153, 154, 161, 163, 171, 177, 178, 179, 182, 183, 185, 187,
188, 189, 190, 195, 196, 199 ... not in OEIS.
Cheers,
Robert
On Dec 10 2018, W. Edwin Clark wrote:
>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
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>--
>Seqfan Mailing list - http://list.seqfan.eu/
>
>
More information about the SeqFan
mailing list