OFFSET
1,1
COMMENTS
I conjecture that the sequence is infinite (in fact for any initial term).
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
LINKS
Zak Seidov, Table of n, a(n) for n = 1..1000
EXAMPLE
After 2, the sequence can't continue with 3 because 3 - 2 = 1^2. So instead we take 5, which gives 5 - 2 = 3.
Then 7, for which we verify that 7 - 2 = 5 and 7 - 5 = 2.
And then we can't use 11 because 11 - 2 = 3^2.
MAPLE
A:= <2>;
for n from 2 to 100 do
p:= nextprime(A[n-1]);
while ormap(t -> issqr(p - t), A) do
p:= nextprime(p)
od;
A(n):= p
od:
convert(A, list); # Robert Israel, Nov 20 2016
MATHEMATICA
primesNoSqDiffs = {2}; p = 3; Do[While[MemberQ[IntegerQ[Sqrt[#]] & /@ (p - s), True], p = NextPrime[p]]; AppendTo[primesNoSqDiffs, p], {60}]; primesNoSqDiffs
CROSSREFS
KEYWORD
nonn
AUTHOR
Zak Seidov, Nov 11 2016
STATUS
approved