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A063669
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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.
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1
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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; internal format)
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OFFSET
| 1,5
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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).
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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.
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CROSSREFS
| Cf. A046080, A063664, A063665, A063014.
Sequence in context: A137773 A175010 A107454 * A162154 A134505 A076933
Adjacent sequences: A063666 A063667 A063668 * A063670 A063671 A063672
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KEYWORD
| nonn
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AUTHOR
| Henry Bottomley (se16(AT)btinternet.com), Jul 28 2001
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