[seqfan] Re: Intermediate result for ordering problem: 2-D \phi()
Hugo van der Sanden
human at google.com
Sun Nov 30 14:02:06 CET 2008
Yes, but only to zero, and the extension a(0, n) = a(n, 0) = 0 falls
naturally out of the definition.
Hugo
2008/11/30 David Wilson <davidwwilson at comcast.net>
> Yes, your sum is prettier.
>
> Does it actually define the same function though?
>
> Doesn't it force you to extend the function to non-positive arguments?
>
> ----- Original Message -----
> From: <hv at crypt.org>
> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Cc: <njas at research.att.com>
> Sent: Sunday, November 30, 2008 3:13 AM
> Subject: [seqfan] Re: Intermediate result for ordering problem: 2-D \phi()
>
>
> > That's very nice. Though it makes the computation less obvious, I'd
> rather
> > express it as:
> >
> > mn = sum_{k=1}^\inf { a([m/k], [n/k]) }
> >
> > .. in which each component of the sum counts the pairs with gcd equal to
> k
> > (and analogously for higher dimension).
> >
> > Hugo
>
>
>
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