

A024361


Number of primitive Pythagorean triangles with leg n.


12



0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 0, 2, 1, 1, 0, 1, 2, 2, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 2, 2, 1, 0, 1, 2, 2, 0, 1, 2, 1, 0, 2, 2, 1, 0, 2, 2, 2, 0, 1, 4, 1, 0, 2, 1, 2, 0, 1, 2, 2, 0, 1, 2, 1, 0, 2, 2, 2, 0, 1, 2, 1, 0, 1, 4, 2, 0, 2, 2, 1, 0, 2, 2, 2, 0, 2, 2, 1, 0, 2, 2, 1, 0, 1, 2, 4
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OFFSET

1,12


COMMENTS

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives number of times A or B takes value n.
For n > 1, a(n) = 0 for n == 2 (mod 4) (n in A016825).
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 a(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) and u > v. Let s = u + v, t = u  v, then n = s*t, where s and t are unitary divisors of n and s > t, so the number of representations is A034444(n)/2 if n > 1 and 0 if n = 1;
(b) if n is divisible by 4, then a(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) and u > v. So u and v must be unitary divisors of n/2, so the number of representations is A034444(n/2)/2. Since n is divisible by 4, A034444(n/2) = A034444(n) so a(n) = A034444(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 a(n) = 0.
a(n) = 0 if n = 1 or n == 2 (mod 4), otherwise a(n) is a power of 2.
The earliest occurrence of 2^k is 2*A002110(k+1) for k > 0. (End)


LINKS



FORMULA



EXAMPLE

a(12) = 2 because 12 appears twice, in (A,B,C) = (5,12,13) and (12,35,37).


MATHEMATICA

Table[If[n == 1  Mod[n, 4] == 2, 0, 2^(Length[FactorInteger[n]]  1)], {n, 100}]


PROG

(PARI) A024361(n) = if(1==n(2==(n%4)), 0, 2^(omega(n)1)); \\ (after the Mathematica program)  Antti Karttunen, Nov 10 2018


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS

Incorrect comment removed by Ant King, Jan 28 2011


STATUS

approved



