

A068386


Onethirtieth the area of the unique Pythagorean triangle whose hypotenuse is A002144(n), the nth prime of the form 4k+1.


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
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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*(uv)*(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


LINKS

Table of n, a(n) for n=2..53.


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[pr^2]], Return[r s(s^2r^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^2a^2)(2a b))/2]; areat[#]/30&/@Select[Prime[ Range[4, 200]], IntegerQ[(#1)/4]&] (* Harvey P. Dale, Jun 21 2011 *)


CROSSREFS

Cf. A008846, A002144.
Sequence in context: A063503 A244976 A248675 * A021040 A246553 A330479
Adjacent sequences: A068383 A068384 A068385 * A068387 A068388 A068389


KEYWORD

easy,nonn


AUTHOR

Lekraj Beedassy, Mar 08 2002


EXTENSIONS

Edited by Dean Hickerson, Mar 14 2002


STATUS

approved



