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A156679
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Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, gcd (A, B) = 1, A < B<C); sequence gives values of C, sorted to correspond to increasing A (A020884(n)).
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7
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5, 13, 25, 17, 41, 61, 37, 85, 113, 65, 145, 181, 29, 101, 221, 265, 145, 313, 365, 53, 197, 421, 481, 257, 65, 545, 613, 85, 325, 685, 89, 761, 401, 841, 925, 125, 485, 1013, 1105, 73, 577, 1201, 149, 1301, 173, 677, 1405, 1513, 785, 185, 1625, 1741, 109, 229
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OFFSET
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1,1
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COMMENTS
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The ordered sequence of A values is A020884(n) and the ordered sequence of C values is A020882(n) (allowing repetitions) and A008846(n) (excluding repetitions).
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REFERENCES
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Beiler, Albert H.: Recreations In The Theory Of Numbers, Chapter XIV, The Eternal Triangle, Dover Publications Inc., New York, 1964, pp. 104-134.
Sierpinski, W.; Pythagorean Triangles, Dover Publications, Inc., Mineola, New York, 2003.
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LINKS
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EXAMPLE
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As the first four primitive Pythagorean triples (ordered by increasing A) are (3,4,5), (5,12,13), (7,24,25) and (8,15,17), then a(1)=5, a(2)=13, a(3)=25 and a(4)=17.
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MATHEMATICA
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PrimitivePythagoreanTriplets[n_]:=Module[{t={{3, 4, 5}}, i=4, j=5}, While[i<n, If[GCD[i, j]==1, h=Sqrt[i^2+j^2]; If[IntegerQ[h] && j<n, AppendTo[t, {i, j, h}]]; ]; If[j<n, j+=2, i++; j=i+1]]; t]; k=38; data1=PrimitivePythagoreanTriplets[2k^2+2k+1]; data2=Select[data1, #[[1]]<=2k+1 &]; #[[3]] &/@data2
With[{nn=61}, Take[Sqrt[#[[1]]^2+#[[2]]^2]&/@Union[Sort/@({Times@@#, (Last[ #]^2- First[#]^2)/2}&/@(Select[Subsets[Range[1, nn, 2], {2}], GCD@@# == 1&]))], nn]] (* Harvey P. Dale, May 10 2020 *)
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PROG
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(Haskell)
a156679 n = a156679_list !! (n-1)
a156679_list = f 1 1 where
f u v | v > uu `div` 2 = f (u + 1) (u + 2)
| gcd u v > 1 || w == 0 = f u (v + 2)
| otherwise = w : f u (v + 2)
where uu = u ^ 2; w = a037213 (uu + v ^ 2)
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CROSSREFS
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KEYWORD
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easy,nice,nonn
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AUTHOR
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STATUS
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approved
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