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.

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, 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, 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

Offset

1,5

Comments

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).

Links

Table of n, a(n) for n=1..105.

Example

a(1)=1 since 1/(12*1)^2 = 1/12^2 = 1/15^2 + 1/20^2;
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.
Looking at A020885, 1 is divisible by 1, while 70 is divisible by 1, 5, 10, 14, 35 and again 35.

Crossrefs

Cf. A046080, A063664, A063665, A063014.

Sequence in context: A107454 A371212 A250261 * A306489 A319734 A211005

Adjacent sequences: A063666 A063667 A063668 * A063670 A063671 A063672

Keyword

nonn

Author

Henry Bottomley, Jul 28 2001

Status

approved