A068386 One-thirtieth the area of the unique Pythagorean triangle whose hypotenuse is A002144(n), the n-th prime of the form 4k+1.

1, 2, 7, 7, 6, 21, 11, 44, 52, 78, 33, 91, 28, 154, 119, 187, 143, 57, 266, 91, 221, 364, 418, 136, 299, 483, 616, 323, 130, 385, 840, 897, 1020, 1155, 1071, 1235, 266, 782, 203, 986, 1638, 1190, 1653, 1683, 2046, 2387, 1463, 2002, 460, 2852, 2204, 357

Offset: 2

Author

Lekraj Beedassy, Mar 08 2002

Keywords

• easy
• nonn

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

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

Examples

			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.
		

Crossrefs

Cf. A008846, A002144.

Programs

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 *)

Extensions

Edited by Dean Hickerson, Mar 14 2002