OFFSET
2,2
COMMENTS
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.
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
EXAMPLE
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.
MATHEMATICA
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&]
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 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Lekraj Beedassy, Mar 08 2002
EXTENSIONS
Edited by Dean Hickerson, Mar 14 2002
STATUS
approved