[seqfan] Re: changes to A020882

Frank Adams-Watters franktaw at netscape.net
Sun Mar 15 22:52:08 CET 2015

Start with the comment of M. F. Hasler. This comment is IMO more complex than it needs to be; an equivalent formulation is that this is the numbers a^2+b^2 where gcd(a, b) = 1 and a and b are not both odd. (It should be noted that a and b need to be strictly positive integers, BTW, and we don't count both a,b and b,a.)

The number of points (a,b) in the circle with radius m is asymptotically pi * m. Restricting to a, b > 0 and (wlog) a > b gives us 1/8 of that: pi/8 * m. The probability that a and b are relatively prime is 6/pi^2. Such pairs can be even, odd; odd, even; or odd, odd; so the number allowed by our condition is 2/3 or that, or 4/pi^2. Multiplying by the total number of points, pi/8 * m, gives us n ~ 1 / (2 * pi) * m. But m here is approximately a(n), so a(n) ~ 2 * pi * n, as claimed.

Franklin T. Adams-Watters

-----Original Message-----
From: David Wilson <davidwwilson at comcast.net>
To: 'Sequence Fanatics Discussion list' <seqfan at list.seqfan.eu>
Sent: Sun, Mar 15, 2015 3:35 pm
Subject: [seqfan] Re: changes to A020882

I think A020882(n) ~ 2*pi*n, can anyone prove that?

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