[seqfan] Re: Close sequences again

Christopher Gribble chris.eveswell at virgin.net
Tue Oct 20 11:50:39 CEST 2009

```Michael,

Looks like my plain text table got garbled, so I'll try again:

n								1
prime(n)							2
floor(j^2/prime(n)), j=1				0
sum(floor(j^2/prime(n))), j=1:prime(n)-1)	(A)	0
floor((prime(n)-2)(prime(n)-1)/3)		(B)	0
A-B								0

n								2
prime(n)							3
floor(j^2/prime(n)), j=1				0
floor(j^2/prime(n)), j=2				1
sum(floor(j^2/prime(n))), j=1:prime(n)-1)	(A)	1
floor((prime(n)-2)(prime(n)-1)/3)		(B)	0
A-B								1

Chris

-----Original Message-----
From: seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu]
On Behalf Of Michael Porter
Sent: 20 October 2009 09:12
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Close sequences again

> (Sum of the quadratic non-residues of prime(n) - Sum of the quadratic

> residues of prime(n)) / prime(n).  I discovered that this was equivalent
to

> (sum ( floor  (j^2 / prime(n))), j = 1:prime(n) - 1) - floor ((prime(n) -

> 2)(prime(n) - 1) / 3) and that this could be extended to the naturals.

That seems like a non-trivial result, so I'm glad to see you have already
added it to OEIS (A165951).  But for a(1) and a(2), prime(n)=2 and 3, the
expression evaluates to -1/2 and 1/3, right?

> A166128
> A166129
> A166130
> A166468

A166468 is the one we were discussing earlier, and if I read it correctly,
A166128 is just prime(n)-1 with a(1) set to 0, so a similar argument
applies.  I can't seem to get to A166129 and A166130.
- Michael

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