A217503 Squared distance between consecutive primes of the form 4k+1 (see below).

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, 20, 10, 116, 2, 8, 20, 10, 2, 10, 20, 26, 4, 146, 8, 34, 10, 40, 34, 40, 2, 20, 2, 160, 50, 10, 180, 2, 180, 90, 58, 40, 130, 16, 116, 194, 50, 136, 74, 34, 52, 40

Offset: 1

Author

Thomas Ordowski, Oct 05 2012

Keywords

• nonn

Comments

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.

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.

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A002313, A002330, A002331.

Programs

Mathematica

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 *)