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A145010
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a(n) = area of Pythagorean triangle with hypotenuse p, where p = A002144(n) = n-th prime == 1 (mod 4).
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3
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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
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OFFSET
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1,1
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COMMENTS
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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).
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LINKS
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FORMULA
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EXAMPLE
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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;
with {c,d} = {t_2, t_3}, t_4 = cd/2 = ab(b^2-a^2).
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p a b t_1 c d t_2 t_3 t_4
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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
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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