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%I #14 Feb 22 2020 20:54:24
%S 0,0,1,1,1,1,1,1,1,2,1,1,1,2,1,1,1,2,1,1,1,2,1,2,1,2,1,1,1,3,1,1,2,2,
%T 1,1,1,2,2,1,1,3,1,2,1,2,1,2,1,2,2,2,1,2,1,1,2,2,1,3,1,2,1,1,2,3,1,2,
%U 2,3,1,1,1,2,2,2,1,3,1,2,1,2,1,3,2,2,2,1,1,3,1,2,2,2,2,2,1,2,1,2,1,4,1
%N Number of unitary divisors of n that are smaller than sqrt((sqrt(2) - 1)*n).
%C Related to A024359: (Start)
%C 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:
%C (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));
%C (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 (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.
%C 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)
%C For k >= 2, the earliest occurrence of k is at n = A132404(k)/2 if A132404(k) is even (and thus being a multiple of 4). Conjecture: this is always the case.
%H Ray Chandler, <a href="/A307706/b307706.txt">Table of n, a(n) for n = 1..10000</a>
%F A024359(n) = a(n) for odd n; A024359(n) = a(n/2) for n divisible by 4.
%e 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.
%o (PARI) a(n) = my(i=0); for(k=1, sqrt((sqrt(2)-1)*n), if(!(n%k) && gcd(k,n/k)==1, i++)); i
%Y Cf. A024359, A034444, A132404.
%K nonn
%O 1,10
%A _Jianing Song_, Apr 23 2019
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