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A075432
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Primes with no squarefree neighbors.
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14
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17, 19, 53, 89, 97, 127, 149, 151, 163, 197, 199, 233, 241, 251, 269, 271, 293, 307, 337, 349, 379, 449, 487, 491, 521, 523, 557, 577, 593, 631, 701, 727, 739, 751, 773, 809, 811, 881, 883, 919, 953, 991, 1013, 1049, 1051, 1061, 1063, 1097, 1151, 1171, 1249
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OFFSET
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1,1
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COMMENTS
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From Ludovicus (luiroto(AT)yahoo.com), Dec 07 2009: (Start)
I propose a shorter name: non-Euclidean primes. That is justified by the Euclid's demonstration of the infinitude of primes. It appears that the proportion of non-Euclidean primes respect to primes tend to the limit 1-2A where A = 0.37395581... is Artin's constant. This table calculated by Jens K. Andersen corroborates it:
10^5: 2421 / 9592 = 0.2523978315
10^6: 19812 / 78498 = 0.2523885958
10^7: 167489 / 664579 = 0.2520227091
10^8: 1452678 / 5761455 = 0.2521373507
10^9: 12817966 / 50847534 = 0.2520862860
10^10: 114713084 / 455052511 = 0.2520875750
10^11: 1038117249 / 4118054813 = 0.2520892256
It comes close to the expected 1-2A. (End)
This sequence is infinite by Dirichlet's theorem, since there are infinitely many primes == 17 or 19 (mod 36) and these have no squarefree neighbors. Ludovicus's conjecture about density is correct. Capsule proof: either p-1 or p+1 is divisible by 4, so it suffices to consider the other number (without loss of generality, p+1). For some fixed bound L, p is not divisible by any prime q < L (with finitely many exceptions) so there are q^2 - q possible residue classes for p. The primes in each are uniformly distributed so the probability that p+1 is divisible by q^2 is 1/(q^2 - q). The product of the complements goes to 2A as L increases without bound, and since 2A is an upper bound the limit is sandwiched between. - Charles R Greathouse IV, Aug 27 2014
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LINKS
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FORMULA
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a(n) ~ Cn log n, where C = 1/(1 - 2A) = 1/(1 - Product_{p>2 prime} (1 - 1/(p^2-p))), where A is the constant in A005596. - Charles R Greathouse IV, Aug 27 2014
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MAPLE
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filter:= n -> isprime(n) and not numtheory:-issqrfree(n+1) and not numtheory:-issqrfree(n-1):
select(filter, [seq(2*i+1, i=1..1000)]); # Robert Israel, Aug 27 2014
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MATHEMATICA
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lst={}; Do[p=Prime[n]; If[ !SquareFreeQ[Floor[p-1]] && !SquareFreeQ[Floor[p+1]], AppendTo[lst, p]], {n, 6!}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 20 2008 *)
Select[Prime[Range[300]], !SquareFreeQ[#-1]&&!SquareFreeQ[#+1]&] (* Harvey P. Dale, Apr 24 2014 *)
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PROG
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(Haskell)
a075432 n = a075432_list !! (n-1)
a075432_list = f [2, 4 ..] where
f (u:vs@(v:ws)) | a008966 v == 1 = f ws
| a008966 u == 1 = f vs
| a010051' (u + 1) == 0 = f vs
| otherwise = (u + 1) : f vs
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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