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%I #23 Nov 19 2017 18:20:34
%S 1,2,7,7,6,21,11,44,52,78,33,91,28,154,119,187,143,57,266,91,221,364,
%T 418,136,299,483,616,323,130,385,840,897,1020,1155,1071,1235,266,782,
%U 203,986,1638,1190,1653,1683,2046,2387,1463,2002,460,2852,2204,357
%N One-thirtieth the area of the unique Pythagorean triangle whose hypotenuse is A002144(n), the n-th prime of the form 4k+1.
%C Every such prime p has a unique representation as p = r^2 + s^2 with 1 <= r < s. The corresponding right triangle has legs of lengths s^2 - r^2 and 2rs and area rs(s^2 - r^2). For p > 5, this is divisible by 30.
%C Calling A002330(n) and A002331(n) respectively u and v, we have a(n) = u*v*(u-v)*(u+v), for n > 1. - _Lekraj Beedassy_, Mar 12 2002
%C The corresponding Pythagorean triple (A, B, C) with A^2 = B^2 + C^2, (A > B > C) is given by {A002144(n), A002365(n), A002366(n)}, so that a(n) = B*C/(2*30) = A002365(n)*A002366(n)/60. - _Lekraj Beedassy_, Oct 27 2003
%e The 7th prime of the form 4k+1 is 53 = 2^2 + 7^2. So the right triangle has sides 7^2 - 2^2 = 45, 2*2*7 = 28 and 53. Its area is 1/2 * 45 * 28 = 630, so a(7) = 630/30 = 21.
%t a30[p_] := For[r=1, True, r++, If[IntegerQ[s=Sqrt[p-r^2]], Return[r s(s^2-r^2)/30]]]; a30/@Select[Prime/@Range[4, 150], Mod[ #, 4]==1&]
%t areat[p_]:=Module[{c=Flatten[PowersRepresentations[p,2,2]],a,b},a= First[c];b= Last[c];((b^2-a^2)(2a b))/2]; areat[#]/30&/@Select[Prime[ Range[4,200]],IntegerQ[(#-1)/4]&] (* _Harvey P. Dale_, Jun 21 2011 *)
%Y Cf. A008846, A002144.
%K easy,nonn
%O 2,2
%A _Lekraj Beedassy_, Mar 08 2002
%E Edited by _Dean Hickerson_, Mar 14 2002
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