

A137364


Prime numbers n such that n = p1^2 + p2^2 + p3^2, a sum of squares of 3 distinct prime numbers.


3



83, 179, 227, 347, 419, 419, 467, 491, 563, 587, 659, 659, 827, 971, 1019, 1019, 1091, 1259, 1427, 1499, 1499, 1667, 1811, 1811, 1907, 1907, 1979, 1979, 2027, 2243, 2267, 2339, 2339, 2531, 2579, 2699, 2819, 2843, 2939, 3347, 3539, 3539, 3659, 3659, 3779
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OFFSET

1,1


COMMENTS

Multiple solutions with different sets {p1,p2,p3} are indicated by repeating the entry for each solution.  R. J. Mathar, Apr 12 2008
All terms are congruent to 5 modulo 6. The smallest of the primes {p1,p2,p3} is always 3.  Zak Seidov, Jun 03 2014


LINKS



EXAMPLE

83 = 3^2 + 5^2 + 7^2;
179 = 3^2 + 7^2 + 11^2;
227 = 3^2 + 7^2 + 13^2.


MATHEMATICA

Array[r, 99]; Array[y, 99]; For[i = 0, i < 10^2, r[i] = y[i] = 0; i++ ]; z = 4^2; n = 0; For[i1 = 1, i1 < z, a = Prime[i1]; a2 = a^2; For[i2 = i1 + 1, i2 < z, b = Prime[i2]; b2 = b^2; For[i3 = i2 + 1, i3 < z, c = Prime[i3]; c2 = c^2; p = a2 + b2 + c2; If[PrimeQ[p], Print[a2, " + ", b2, " + ", c2, " = ", p]; n++; r[n] = p]; i3++ ]; i2++ ]; i1++ ]; Sort[Array[r, 39]]
lst= {}; Do[p = Prime[q]^2 + Prime[r]^2 + Prime[s]^2; If[PrimeQ@p, AppendTo[lst, p]], {q, 26}, {r, q1}, {s, r1}]; Take[Sort@lst, 72] (* Vincenzo Librandi, Jun 15 2013 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



