|
Search: id:A068386
|
|
|
| A068386 |
|
One-thirtieth the area of the unique Pythagorean triangle whose hypotenuse is A002144(n), the n-th prime of the form 4k+1. |
|
+0
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
(list; graph; listen)
|
|
|
OFFSET
|
2,2
|
|
|
COMMENT
|
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 (blekraj(AT)yahoo.com), 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 (blekraj(AT)yahoo.com), Oct 27 2003
|
|
EXAMPLE
|
The 7-th 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&]
|
|
CROSSREFS
|
Cf. A008846, A002144.
Sequence in context: A057105 A016536 A063503 this_sequence A021040 A003061 A087385
Adjacent sequences: A068383 A068384 A068385 this_sequence A068387 A068388 A068389
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Lekraj Beedassy (blekraj(AT)yahoo.com), Mar 08 2002
|
|
EXTENSIONS
|
Edited by Dean Hickerson (dean(AT)math.ucdavis.edu), Mar 14 2002
|
|
|
Search completed in 0.002 seconds
|
