OFFSET
1,10
COMMENTS
Related to A024359: (Start)
Note that all the primitive Pythagorean triangles are given by A = min{2*u*v, u^2 - v^2}, B = max{2*u*v, u^2 - v^2}, C = u^2 + v^2, where u, v are coprime positive integers, u > v and u - v is odd. As a result:
(a) if n is odd, then A024359(n) is the number of representations of n to the form n = u^2 - v^2, where u, v are coprime positive integers (note that this guarantees that u - v is odd), u > v and u^2 - v^2 < 2*u*v. Let s = u + v, t = u - v, then n = s*t, where s and t are unitary divisors of n, s > t and s*t < (s^2 - t^2)/2, so t is in the range (0, sqrt((sqrt(2) - 1)*n));
(b) if n is divisible by 4, then A024359(n) is the number of representations of n to the form n = 2*u*v, where u, v are coprime positive integers (note that this also guarantees that u - v is odd because n/2 is even), u > v and 2*u*v < u^2 - v^2. So u and v must be unitary divisors of n/2, and v is in the range (0, sqrt((sqrt(2) - 1)*(n/2))).
(c) if n == 2 (mod 4), then n/2 is odd, so n = 2*u*v implies that u and v are both odd, which is not acceptable, so A024359(n) = 0.
Similarly, let b(n) be the number of unitary divisors of n in the range (sqrt((sqrt(2) - 1)*n), sqrt(n)) (= A034444(n)/2 - a(n) for n > 1), then the number of times B takes value n is b(n) for odd n > 1, b(n/2) if n is divisible by 4 and 0 if n = 1 or n == 2 (mod 4). (End)
LINKS
Ray Chandler, Table of n, a(n) for n = 1..10000
EXAMPLE
The unitary divisors of 210 that are smaller than sqrt((sqrt(2) - 1)*210) = 9.3265... are 1, 2, 3, 5, 6 and 7, so a(210) = 6. Correspondingly, A024359(420) = 6.
PROG
(PARI) a(n) = my(i=0); for(k=1, sqrt((sqrt(2)-1)*n), if(!(n%k) && gcd(k, n/k)==1, i++)); i
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Apr 23 2019
STATUS
approved