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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, 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 (list; graph; refs; listen; history; text; internal format)
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
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
Sequence in context: A107454 A371212 A250261 * A306489 A319734 A211005
KEYWORD
nonn
AUTHOR
Henry Bottomley, Jul 28 2001
STATUS
approved

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Last modified March 28 20:05 EDT 2024. Contains 371254 sequences. (Running on oeis4.)