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A024364
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Ordered perimeters of primitive Pythagorean triangles.
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30
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12, 30, 40, 56, 70, 84, 90, 126, 132, 144, 154, 176, 182, 198, 208, 220, 234, 240, 260, 286, 306, 312, 330, 340, 374, 380, 390, 408, 418, 420, 442, 456, 462, 476, 494, 510, 532, 544, 546, 552, 570, 598, 608, 644, 646, 650, 672, 684, 690, 700, 714, 736, 756
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OFFSET
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1,1
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COMMENTS
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Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives perimeters A+B+C.
k is in this sequence iff A070109(k) > 0. This is a subsequence of A010814.
Any term in this sequence can be generated by f(m,k) = 2*m*(m+k), where m and k are positive coprime integers and m > 1, k < m, and m and k are not both odd. For example: f(2,1) = 2*2*(2+1) = 12. - Agola Kisira Odero, Apr 29 2016
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LINKS
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FORMULA
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MAPLE
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isA024364 := proc(an) local r::integer, s::integer ; for r from floor((an/4)^(1/2)) to floor((an/2)^(1/2)) do for s from r-1 to 1 by -2 do if 2*r*(r+s) = an and gcd(r, s) < 2 then RETURN(true) ; fi ; if 2*r*(r+s) < an then break ; fi ; od ; od : RETURN(false) ; end : for n from 2 to 400 do if isA024364(n) then printf("%d, ", n) ; fi ; od ; # R. J. Mathar, Jun 08 2006
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MATHEMATICA
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lst={}; amx=99; Do[For[b=a+1, b<(a^2/2), c=(a^2+b^2)^(1/2); If[c==IntegerPart[c]&&GCD[a, b, c]==1, p=a+b+c; AppendTo[lst, p]]; b=b+2], {a, 3, amx}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 08 2008 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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