[seqfan] Re: A187824: a(n) is the largest m such that n is congruent to -1, 0 or 1 mod k for all k from 1 to m.
Vladimir Shevelev
shevelev at bgu.ac.il
Mon Dec 31 11:54:38 CET 2012
The sequence is, indeed, interesting.
Let n==0(mod 20). Then always a(n-2)=a(n+2)=3, while a(n)=5,5,6,5,5,8,5,5,6,5,5,6,5,5,7,5,5,6,5,5,7,...
with records
a(20)=5,a(60)=6,a(120)=8,a(720)=10,a(2520)=12,a(9360)=13,...
If n==0(mod 5), but is not multiple of 20, then always a(n-2)=a(n+2)=4,
while a(n)=6,3,5,6,3,7,5,3,9,6,3,5,7,3,6,5,3,6,7,3,5,... and it seems that it is bounded by 9. Is'nt it?
Happy New 2013 Year!
Regards,
Vladimir
----- Original Message -----
From: Neil Sloane <njasloane at gmail.com>
Date: Sunday, December 30, 2012 6:55
Subject: [seqfan] A187824: a(n) is the largest m such that n is congruent to -1, 0 or 1 mod k for all k from 1 to m.
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> A187824 was submitted recently, and seems quite interesting.
> There are several questions: is it unbounded? (cf. A056697, A220891)
> If so does it take every value? (cf. A220890)
> The answer to both is probably Yes.
> Neil
>
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Shevelev Vladimir
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