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%I #32 Oct 23 2012 17:47:28
%S 1,2,2,2,2,10,8,10,8,4,2,10,4,20,18,10,2,20,58,8,40,2,40,20,10,90,2,
%T 20,10,116,2,8,20,10,2,10,20,26,4,146,8,34,10,40,34,40,2,20,2,160,50,
%U 10,180,2,180,90,58,40,130,16,116,194,50,136,74,34,52,40
%N Squared distance between consecutive primes of the form 4k+1 (see below).
%C Every prime p of the form 4k+1 has a unique solution p = x^2 + y^2. This sequence gives the squared distance between points (x,y) for consecutive primes of this form.
%C The squares mutual distance consecutive points with coordinates x(n) = A002331(n) and y(n) = A002330(n), where x(n)^2 + y(n)^2 = A002313(n) is prime.
%C Theorem: a(n) =/= A082073(n-1) for all n > 1. Generally, it can be shown that there is no pair of primes p = a^2 + b^2 and q = x^2 + y^2 such that (a - x)^2 + (b - y)^2 = |p - q| > 0.
%H T. D. Noe, <a href="/A217503/b217503.txt">Table of n, a(n) for n = 1..1000</a>
%e 5 = 1^2 + 2^2 and 13 = 2^2 + 3^2. The squared distance between the points (1,2) and (2,3) is 2, the second term of this sequence.
%t nn = 200; p = Select[Prime[Range[nn]], Mod[#, 4] == 1 &]; q = {1, 1}; Table[pp = PowersRepresentations[p[[i]], 2, 2][[1]]; d = pp - q; q = pp; d[[1]]^2 + d[[2]]^2, {i, Length[p] - 1}] (* _T. D. Noe_, Oct 19 2012 *)
%Y Cf. A002313, A002330, A002331.
%K nonn
%O 1,2
%A _Thomas Ordowski_, Oct 05 2012
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