The OEIS will be on holiday for the rest of the year!

[cino hilliard]

Hi, Lßbos ElemÚr <Labos at ana.sote.hu>
> > Labos Elemer said:
> >
> > >From seqfan-owner at ext.jussieu.fr  Fri Jan  6 02:46:48 2006
> > >Date: Fri, 6 Jan 2006 08:37:51 +0100
> > >Subject: RE: The OEIS will be on holiday for the rest of the year!
> > >
> > >> back to business, yes - but please only submit really important 
>sequences
> > >>
> > >> don't sit down and try to make up new sequences
> > >>
> > >> that is strongly discouraged
> > >>
> > >> Neil
> > >
> > >Does it mean for the rest of 2006 or that of 2005 ?
> > >
> > >Labos
> > >
> >
> >
> > Me: for the rest of time!
> >
> > If a series comes up in your work, send it in.
> >
> > But don't sit down and try to make up new sequences.
> >
> > That is strongly discouraged.
> >
> > Neil
well this is kinder than what I got

>>"back to business, yes - but please only submit really important sequences

>>"don't sit down and try to make up new sequences

>>"that is strongly discouraged

>>"Neil

Apparently,
> > If a series comes up in your work, send it in.
is now OK.and realy important sequences is now relaxed in the sense if it is 
in your work it
is important to you? Doing math for fun is my work since I am retired. I 
have a degree in math(40 years ago) but I am an amateur. I guess if I am 
working on some concept such as properties of the
year 2006. 2006 = 2*17*59 and study the patterns in 17^n + 2 and come up 
with the sequence

0,1,105,369
n such that 17^n+2 is prime.
2006 = 2*17*59
17^n+2 is composite for all n of the form 4k+3 and and 6k+5
Proof
If n = 4k+3 then 5 divides 17^n + 2 (15+2)^(4k+3)+ 2 = 5H + 2^(4k+3)+ 2
= 5H + 2(4^(2k+1)+1)) == 0 mod 5 If n = 6k+5 then 7 divides 17^n + 2
(14+3)^(6k+5)+ 2 = 7H + 3^(6k+5)+ 2 = 5H + (7-4)^(6k+5) + 2 = 7H + 4^
(6k+5)-2 = 7H+2^(12k+10)+2 = 7H + 2*(2^(12k+9)-1)=7H + 2*(8^(4k+3)-1)
=7H+2*7I == 0 mod 7.
(PARI) forstep(n=0,1000,2,y=17^n+2;if(ispseudoprime(y),print1(n",")))
simillar sequence on 59^n +2
3,61,3483,205381,12117363,714924301,42180533643,2488651484821,
146830437604323,8662995818654941,511116753300641403,
30155888444737842661,1779197418239532716883,104972647676132430295981,
6193386212891813387462763,365409786560616989860302901,
21559177407076402401757871043,1271991467017507741703714391421,
75047496554032956760519149093723,4427802296687944448870629796529541
Numbers of the form 59^n + 2.
The year 2006 = 2*17*59. Here are some properties of 59 and 2.
n      divisor of 59^n + 2
2k             3
6k+5           7
10k+3          11
12k+5          13
8n+7           17
Proofs are similar to the case for 17^n+2.
59^1 + 2 = 61 which is prime.
(PARI) for
These are important to me. Is that the criteria? If I think they are really  
important I should
submit?

>Funy. Moreover sad.
Me too.

>Something should happen that you reached to this position or
>you are fatigued.
>

>After several months without submission I just wanted to
>submit again but....
>Let me tell you that I am completely discouraged to submit
>seuences  and I bet I am not alone.
You are not.

>So discouraging is  effective.
Always
>

>PS: I am sad.
Me too

Cino