A145010 a(n) = area of Pythagorean triangle with hypotenuse p, where p = A002144(n) = n-th prime == 1 (mod 4).

6, 30, 60, 210, 210, 180, 630, 330, 1320, 1560, 2340, 990, 2730, 840, 4620, 3570, 5610, 4290, 1710, 7980, 2730, 6630, 10920, 12540, 4080, 8970, 14490, 18480, 9690, 3900, 11550, 25200, 26910, 30600, 34650, 32130, 37050, 7980, 23460, 6090, 29580, 49140, 35700

Offset: 1

Author

M. F. Hasler, Feb 24 2009

Keywords

• nonn

Comments

Pythagorean primes, i.e., primes of the form p = 4k+1 = A002144(n), have exactly one representation as sum of two squares: A002144(n) = x^2+y^2 = A002330(n+1)^2+A002331(n+1)^2. The corresponding (primitive) integer-sided right triangle with sides { 2xy, |x^2-y^2| } = { A002365(n), A002366(n) } has area xy|x^2-y^2| = a(n). For n>1 this is a(n) = 30*A068386(n).

Examples

			The following table shows the relationship between several closely related sequences:
Here p = A002144 = primes == 1 (mod 4), p = a^2+b^2 with a < b;
a = A002331, b = A002330, t_1 = ab/2 = A070151;
p^2 = c^2+d^2 with c < d; c = A002366, d = A002365,
t_2 = 2ab = A145046, t_3 = b^2-a^2 = A070079,
with {c,d} = {t_2, t_3}, t_4 = cd/2 = ab(b^2-a^2).
  ---------------------------------
   p  a  b  t_1  c   d t_2 t_3  t_4
  ---------------------------------
   5  1  2   1   3   4   4   3    6
  13  2  3   3   5  12  12   5   30
  17  1  4   2   8  15   8  15   60
  29  2  5   5  20  21  20  21  210
  37  1  6   3  12  35  12  35  210
  41  4  5  10   9  40  40   9  180
  53  2  7   7  28  45  28  45  630
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A002144, A002365, A002366, A144954.

Programs

Mathematica

Reap[For[p = 2, p < 500, p = NextPrime[p], If[Mod[p, 4] == 1, area = x*y/2 /. ToRules[Reduce[0 < x <= y && p^2 == x^2 + y^2, {x, y}, Integers]]; Sow[area]]]][[2, 1]] (* Jean-François Alcover, Feb 04 2015 *)

PARI

forprime(p=1,499, p%4==1 | next; t=[p,lift(-sqrt(Mod(-1,p)))]; while(t[1]^2>p,t=[t[2],t[1]%t[2]]); print1(t[1]*t[2]*(t[1]^2-t[2]^2)","))

PARI

{Q=Qfb(1,0,1);forprime(p=1,499,p%4==1|next;t=qfbsolve(Q,p); print1(t[1]*t[2]*(t[1]^2-t[2]^2)","))} \\ David Broadhurst

Formula

a(n) = A002365(n)*A002366(n)/2.

a(n) = x*y*(x^2-y^2), where x = A002330(n+1), y = A002331(n+1).