Search: id:A046079
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%I A046079
%S A046079 0,0,1,1,1,1,1,2,2,1,1,4,1,1,4,3,1,2,1,4,4,1,1,7,2,1,3,4,1,4,1,4,4,1,4,
%T A046079 7,1,1,4,7,1,4,1,4,7,1,1,10,2,2,4,4,1,3,4,7,4,1,1,13,1,1,7,5,4,4,1,4,4,
%U A046079 4,1,12,1,1,7,4,4,4,1,10,4,1,1,13,4,1,4,7,1,7,4,4,4,1,4,13,1,2,7
%N A046079 Number of Pythagorean triangles with leg n.
%C A046079 Number of ways in which n can be the leg (other than the hypotenuse) 
               of a primitive or nonprimitive right triangle.
%C A046079 Number of ways that 2/n can be written as a sum of exactly two distinct 
               unit fractions. For every solution to 2/n = 1/x + 1/y, x < y, the 
               Pythagorean triple is (n, y-x, x+y-n). - T. D. Noe (noe(AT)sspectra.com), 
               Sep 11 2002
%D A046079 A. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 
               116-117, 1966.
%H A046079 H. Becker, <a href="http://home.foni.net/~heinzbecker/pythagoras.html">
               Pythagorean triples triplets in JavaScript</a>
%H A046079 Ron Knott, <a href="http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/
               pythag.html">Pythagorean Triples and Online Calculators</a>
%H A046079 F. Richman, <a href="http://www.math.fau.edu/Richman/mla/pythag3s.htm">
               Pythagorean Triples</a>
%H A046079 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               PythagoreanTriple.html">Link to a section of The World of Mathematics.</
               a>
%H A046079 Project Euler, <a href="http://projecteuler.net/index.php?section=problems&id=176">
               Problem 176: Rectangular triangles that share a cathetus.</a>
%F A046079 For odd n, a(n) = A018892(n) - 1.
%F A046079 Let n = (2^a0)*(p1^a1)*...*(pk^ak). Then a(n) = [(2*a0 - 1)*(2*a1 + 1)*(2*a2 
               + 1)*(2*a3 + 1)*...*(2*ak + 1) - 1]/2. Note that if there is no a0 
               term, i.e. if n is odd, then the first term is simply omitted. - 
               Temple Keller (temple.keller(AT)gmail.com), Jan 05 2008
%F A046079 For odd n, a(n) = (tau(n^2) - 1) / 2; for even n, a(n) = (tau((n / 2)^2) 
               - 1) / 2. - Amber Hu (hupo001(AT)gmail.com), Jan 23 2008
%t A046079 f[n_] := (DivisorSigma[0, If[OddQ[n], n, n / 2]^2] - 1) / 2; Table[f[i], 
               {i, 100}] - Amber Hu (hupo001(AT)gmail.com), Jan 23 2008
%Y A046079 Cf. A046080, A046081, A001227, A018892, A024361, A024362, A024363.
%Y A046079 Sequence in context: A157654 A078692 A033151 this_sequence A165509 A100996 
               A090048
%Y A046079 Adjacent sequences: A046076 A046077 A046078 this_sequence A046080 A046081 
               A046082
%K A046079 nonn
%O A046079 1,8
%A A046079 Eric Weisstein (eric(AT)weisstein.com)

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