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A275011
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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.
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1
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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
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1,1
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COMMENTS
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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
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LINKS
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EXAMPLE
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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.
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MAPLE
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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:
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MATHEMATICA
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primesNoSqDiffs = {2}; p = 3; Do[While[MemberQ[IntegerQ[Sqrt[#]] & /@ (p - s), True], p = NextPrime[p]]; AppendTo[primesNoSqDiffs, p], {60}]; primesNoSqDiffs
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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