

A307706


Number of unitary divisors of n that are smaller than sqrt((sqrt(2)  1)*n).


0



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, 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, 2, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2
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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)
For k >= 2, the earliest occurence 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.


LINKS

Table of n, a(n) for n=1..87.


FORMULA

A024359(n) = a(n) for odd n; A024359(n) = a(n/2) for n divisible by 4.


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

Cf. A024359, A034444, A132404.
Sequence in context: A129252 A327936 A022929 * A161102 A276329 A161101
Adjacent sequences: A307703 A307704 A307705 * A307707 A307708 A307709


KEYWORD

nonn


AUTHOR

Jianing Song, Apr 23 2019


STATUS

approved



