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Search: id:A156681
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| A156681 |
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Consider all Pythagorean triangles A^2 + B^2 = C^2 with A<B<C; sequence gives values of B, sorted to correspond to increasing A (A009004(n)). |
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3
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| 4, 12, 8, 24, 15, 12, 40, 24, 60, 16, 35, 84, 48, 20, 36, 112, 30, 63, 144, 24, 80, 180, 21, 48, 99, 28, 72, 220, 120, 264, 32, 45, 70, 143, 60, 312, 168, 36, 120, 364, 45, 96, 195, 420, 40, 72, 224, 480, 60, 126, 255, 44, 56, 180, 544, 288, 84, 120, 612, 48, 77, 105
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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The ordered sequence of B values is A009012(n) (allowing repetitions) and A009023(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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Ron Knott, Right-angled Triangles and Pythagoras' Theorem
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FORMULA
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Sqrt(A156682(n)^2-A009004(n)^2)
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EXAMPLE
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As the first four Pythagorean triples (ordered by increasing A) are (3,4,5), (5,12,13), (6,8,10) and (7,24,25), then a(1)=4, a(2)=12, a(3)=8 and a(4)=24.
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MATHEMATICA
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PythagoreanTriplets[n_]:=Module[{t={{3, 4, 5}}, i=4, j=5}, While[i<n, h=Sqrt[i^2+j^2]; If[IntegerQ[h] && j<n, AppendTo[t, {i, j, h}]]; If[j<n, j++, i++; j=i+1]]; t]; k=20; data1=PythagoreanTriplets[2k^2+2k+1]; data2=Select[data1, #[[1]]<=2k+1 &]; #[[2]] &/@data2
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CROSSREFS
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A156682, A009004, A009012, A009023
Sequence in context: A133517 A145046 A084415 this_sequence A063608 A074258 A120213
Adjacent sequences: A156678 A156679 A156680 this_sequence A156682 A156683 A156684
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KEYWORD
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easy,nice,nonn
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AUTHOR
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Ant King (mathstutoring(AT)ntlworld.com), Feb 17 2009
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