EricDesbiaux/SequencesCheckup

[Eric]

Hello,

excuse me
but
i'm novice...and my english is poor

i was thinking that (5*n-1)*binomial(n+2,3)/4 is not the good formula for
A002418...
and that it was 24 instead of 4....
 i add just this "unit numbers always 0,1,9,5,6,4"

you mean that i have to use standard web interface to add the correct
formula?
but
you seems better to do this than me, you expression is cleaner than mine

A010879(a(n)) one of {0,1,9,5,6,4} because
a(n) mod 5 = 0,1,4,0,0, periodically extended with period 5.
[Proof via A002413(n) mod 5 = 1,3,1,0,0 with period 5, plus the
fact that a(n) are the partial sums of A002413.]

in your name,
write it

but if you consider i have to do it
so i will do it

i'm beginner
and not familiar with rules of the database that's why i speack a lots of
for nothing :o)
sorry

Thanks
Best Regards
Eric

-----Message d'origine-----
De : Richard Mathar [mailto:mathar at strw.leidenuniv.nl]
Envoyé : mercredi 19 mars 2008 20:54
À : seqfan at ext.jussieu.fr
Objet : Re: EricDesbiaux/SequencesCheckup



ed> From seqfan-owner at ext.jussieu.fr  Tue Mar 18 22:06:23 2008
ed> From: "Eric" <moongerms at wanadoo.fr>
ed> To: <seqfan at ext.jussieu.fr>
ed> Subject: EricDesbiaux/SequencesCheckup
ed> Date: Tue, 18 Mar 2008 22:05:50 +0100
ed>
ed> A002418
ed> unit numbers always 0,1,9,5,6,4
ed> (5n-1)n(n+1)(n+2)/24

I propose you use the standard web interface to add the formula for A002418
in a fashion like:

A010879(a(n)) one of {0,1,9,5,6,4} because
a(n) mod 5 = 0,1,4,0,0, periodically extended with period 5.
[Proof via A002413(n) mod 5 = 1,3,1,0,0 with period 5, plus the
fact that a(n) are the partial sums of A002413.]

Richard, http://www.strw.leidenuniv.nl/~mathar





Dear Eric,  I am glad that you followed my suggestion
and joined the sequence fans mailing list!

There may be too many messages some days,
there may sometimes be criticisms of people, etc.,
but the only important thing is that there are

and that is the important thing!


Neil