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%I #25 Nov 30 2016 06:01:34
%S 2,5,7,13,19,31,37,59,61,79,89,103,109,127,193,199,211,239,241,251,
%T 281,283,307,313,353,367,373,379,397,421,439,463,487,547,557,571,577,
%U 601,619,643,661,673,727,733,739,751,757,809,823,829,853,941
%N a(1) = 2; for n > 1, a(n) is the least prime p > a(n-1) such that none of p-a(1), ..., p-a(n-1) is a square.
%C I conjecture that the sequence is infinite (in fact for any initial term).
%C Theorem: The sequence is infinite. Given primes a(1)...a(n), take k(1) ... k(n) such that k(i) is not a square mod a(i). If x == k(i) mod a(i) for i=1..n, then x - a(i) is not a square. By the Chinese Remainder Theorem and Dirichlet's theorem on primes in arithmetic progressions, there is such an x > a(n) that is prime. - _Robert Israel_, Nov 20 2016
%H Zak Seidov, <a href="/A275011/b275011.txt">Table of n, a(n) for n = 1..1000</a>
%e After 2, the sequence can't continue with 3 because 3 - 2 = 1^2. So instead we take 5, which gives 5 - 2 = 3.
%e Then 7, for which we verify that 7 - 2 = 5 and 7 - 5 = 2.
%e And then we can't use 11 because 11 - 2 = 3^2.
%p A:= <2>;
%p for n from 2 to 100 do
%p p:= nextprime(A[n-1]);
%p while ormap(t -> issqr(p - t), A) do
%p p:= nextprime(p)
%p od;
%p A(n):= p
%p od:
%p convert(A,list); # _Robert Israel_, Nov 20 2016
%t primesNoSqDiffs = {2}; p = 3; Do[While[MemberQ[IntegerQ[Sqrt[#]] & /@ (p - s), True], p = NextPrime[p]]; AppendTo[primesNoSqDiffs, p], {60}]; primesNoSqDiffs
%K nonn
%O 1,1
%A _Zak Seidov_, Nov 11 2016
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