cp's OEIS Frontend

A063669 Hypotenuses of reciprocal Pythagorean triangles: number of solutions to 1/(12n)^2 = 1/b^2 + 1/c^2 [with b >= c > 0]; also number of values of A020885 (with repetitions) which divide n.

%I A063669 #8 Dec 12 2021 22:54:11
%S A063669 1,1,1,1,2,1,1,1,1,3,1,1,1,2,2,1,1,1,1,3,1,1,1,1,2,1,1,2,1,4,1,1,1,1,
%T A063669 4,1,1,1,1,3,1,2,1,1,2,1,1,1,1,3,1,1,1,1,3,2,1,1,1,4,1,1,1,1,2,1,1,1,
%U A063669 1,6,1,1,1,1,2,1,1,1,1,3,1,1,1,3,2,1,1,1,1,4,2,1,1,1,2,1,1,2,1,3,1,1,1,1,5
%N A063669 Hypotenuses of reciprocal Pythagorean triangles: number of solutions to 1/(12n)^2 = 1/b^2 + 1/c^2 [with b >= c > 0]; also number of values of A020885 (with repetitions) which divide n.
%C A063669 Primitive reciprocal Pythagorean triangles 1/a^2 = 1/b^2 + 1/c^2 have a=fg, b=ef, c=eg where e^2 = f^2 + g^2; i.e., e,f,g represent the sides of primitive Pythagorean triangles. But the product of the two legs of primitive Pythagorean triangles are multiples of 12 and so the reciprocal of hypotenuses of reciprocal Pythagorean triangles are always multiples of 12 (A008594).
%e A063669 a(1)=1 since 1/(12*1)^2 = 1/12^2 = 1/15^2 + 1/20^2;
%e A063669 a(70)=6 since 1/(12*70)^2 = 1/840^2 = 1/875^2 + 1/3000^2 = 1/888^2 + 1/2590^2 = 1/910^2 + 1/2184^2 = 1/952^2 + 1/1785^2 = 1/1050^2 + 1/1400^2 = 1/1160^2 + 1/1218^2.
%e A063669 Looking at A020885, 1 is divisible by 1, while 70 is divisible by 1, 5, 10, 14, 35 and again 35.
%Y A063669 Cf. A046080, A063664, A063665, A063014.
%K A063669 nonn
%O A063669 1,5
%A A063669 _Henry Bottomley_, Jul 28 2001