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Concerning hypotenuses of triangles such that the sum of the two legs is a perfect square.
2

%I #9 Jan 13 2013 02:57:09

%S 9,19,27,33,57,51,51,73,89,83,107,99,139,129,137,123,129,187,187,163,

%T 177,171,209,257,201,233,267,227,251,337,243,321,313,307,297,289,291,

%U 387,411,363,347,393,339,379,369,363,417,401,393,491,499,473,593,449

%N Concerning hypotenuses of triangles such that the sum of the two legs is a perfect square.

%C Last digit is never 5.

%C Frenicle considers numbers N (apparently the set of A058529 or A120681) and their squares N^2. These have representations N=2*b^2-a^2 = d^2-2*c^2 with d=b+c and N^2 = 2*f^2-e^2 = h^2-2*g^2 with h=f+g. For example N=7 with a=1, b=2, c=1, d=3 and N^2=49 with e=1, f=5, g=4, h=9. The current sequence contains the list of h's.

%C Apparently the list of N^2 is A089552, the list of a in A143732, the list of b in A147847, the list of e (in different order) in A152910, the list of f (sorted into a different order) in A020882.

%H M. de Frenicle, <a href="http://gallica.bnf.fr/ark:/12148/bpt6k5493994j/">Methode pour trouver la solutions des problemes par les exclusions</a>, in: Divers ouvrages des mathematiques et de physique par messieurs de l'academie royale des sciences, (1693) pp 1-44.

%e (a,b,c,d,e,f,g,h) = (1,2,1,3,1,5,4,9) with N=7 or (1,3,2,5,7,13,6,19) with N=17 or (3,4,1,5,7,17,10,27) with N=23 or (1,4,3,7,17,25,8,33) with N=31.

%Y Cf. A144407

%K nonn,uned

%O 0,1

%A _Paul Curtz_, Oct 23 2008