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Search: id:A098993
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| A098993 |
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Let h be the smallest value for which h, h+1, ..., h+n-1 are all lengths of hypotenuses of Pythagorean triangles. Then a(n)=h. |
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+0
2
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| 5, 25, 39, 50, 218, 403, 403, 403, 403, 1597, 2190, 2820, 6050, 8577, 12423, 27325, 34075, 37088, 37088, 43795, 43795, 43795, 87594, 87594, 87594, 87594, 87594, 169160, 169160, 169160, 169160, 169160, 169160, 1884817, 1884817, 1884817
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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"We can also prove (this is more difficult) that for an arbitrary natural number m there exist m Pythagorean triangles the hypotenuses of which are given by successive natural numbers, n, n+1, n+2, ..., n+m-1." Sierpinski (p. 28). No proof is given in book.
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REFERENCES
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W. Sierpinski, Pythagorean Triangles, Dover Publications, Mineola NY, 2003.
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LINKS
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D. L. Vestal, Review of "Pythagorean Triangles"(Chapter 6) by W. Sierpinski
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EXAMPLE
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a(4)=50 since 50, 51, 52 and 53 is the first occurrence of 4 consecutive integers which are lengths of hypotenuses of Pythagorean triangles.
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MATHEMATICA
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lmt = 5*10^6; hyp = {5}; Do[ mn = m^2 + n^2; hyp = Join[hyp, Table[k*mn, {k, Floor[lmt/mn]}]]; hyp = Union[hyp], {n, 2, 1150}, {m, Min[n - 1, Floor[ Sqrt[ lmt - n^2]]]}]; f[n_] := Block[{k = 1}, While[phk[[k]] + n - 1 != phk[[k + n - 1]], k++ ]; phk[[k]]]; Do[ Print[ f[n]], {n, 33} (from Robert G. Wilson v Nov 10 2004)
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CROSSREFS
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See A099799 for another version.
Sequence in context: A070390 A018724 A070389 this_sequence A099799 A093534 A070388
Adjacent sequences: A098990 A098991 A098992 this_sequence A098994 A098995 A098996
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KEYWORD
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nonn
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AUTHOR
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Charlie Marion (charliemath(AT)optonline.net), Nov 05 2004
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EXTENSIONS
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More terms from Ray Chandler (rayjchandler(AT)sbcglobal.net) and Robert G. Wilson (rgwv(AT)rgwv.com) Nov 10 2004
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