[seqfan] Re: A100083
David Wilson
davidwwilson at comcast.net
Sat Jun 15 04:39:17 CEST 2013
Let
a_k = {n : !n == k (mod n)}
where !n = A003422(n).
Then
a_1 = (1, 3, 9, 11, 33, 99, ...) = A057245
a_2 = (1, 2, 4, 8, 31, 62, 124, 248, ...) = A100083
a_3 = (1, 467, ...),
a_4 = (1, 2, 3, 5, 6, 10, 15, 30, 41, 82, 123, ...)
a_5 = (1, 37, ...),
a_6 = (1, 2, 4, 7, 14, 28, ...)
a_7 = (1, 3, 277, 831, 7717, 23151, ...)
a_8 = (1, 2, ...)
a_9 = (1, 5, 19, 95, 197, 985, 3743, 18715, ...)
a_10 = (1, 2, 3, 4, 5, 6, 8, 9, 12, 13, 16, 18, 24, ...)
I'm guessing that all a_k share the same nice properties we observed for a_2
= A100083:
If n in a_k and m | n, then m in a_k
If n, m in a_k and (n, m) = 1, then nm in a_k.
> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of
> Giovanni Resta
> Sent: Wednesday, June 12, 2013 5:10 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: FW: A100083
>
> On 06/12/2013 03:22 AM, Robert G. Wilson v wrote:
> > Et al,
> >
> > YES a(16) does indeed equal 92504. I am now looking for a(17).
>
>
> Charles Greathouse and I already found a(16) yesterday, thanks for re-
> checking.
>
> About a(17), currently I can only say that a(17) > 1.2*10^7.
>
> Giovanni
>
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