

A275011


a(1) = 2; for n > 1, a(n) is the least prime p > a(n1) such that none of pa(1), ..., pa(n1) is a square.


1



2, 5, 7, 13, 19, 31, 37, 59, 61, 79, 89, 103, 109, 127, 193, 199, 211, 239, 241, 251, 281, 283, 307, 313, 353, 367, 373, 379, 397, 421, 439, 463, 487, 547, 557, 571, 577, 601, 619, 643, 661, 673, 727, 733, 739, 751, 757, 809, 823, 829, 853, 941
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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[n1]);
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

Sequence in context: A019359 A038959 A069351 * A094743 A175075 A262392
Adjacent sequences: A275008 A275009 A275010 * A275012 A275013 A275014


KEYWORD

nonn


AUTHOR

Zak Seidov, Nov 11 2016


STATUS

approved



