[seqfan] Re: Intermediate result for ordering problem: 2-D \phi()
David Wilson
davidwwilson at comcast.net
Sun Nov 30 11:48:09 CET 2008
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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