

A075432


Primes with no squarefree neighbors.


14



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

1,1


COMMENTS

Complement of A075430 in A000040.
From Ludovicus (luiroto(AT)yahoo.com), Dec 07 2009: (Start)
I propose a shorter name: nonEuclidean primes. That is justified by the Euclid's demonstration of the infinitude of primes. It appears that the proportion of nonEuclidean primes respect to primes tend to the limit 12A 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 12A. (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 p1 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


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Pieter Moree, Artin's primitive root conjecture a survey , arXiv:math/0412262 [math.NT], 20042012.
Carlos Rivera, Conjecture 65. NonEuclidean primes, The Prime Puzzles and Problems Connection.


FORMULA

a(n) ~ Cn log n, where C = 1/(1  2A) = 1/(1  Product_{p>2 prime} (1  1/(p^2p))), where A is the constant in A005596.  Charles R Greathouse IV, Aug 27 2014


MAPLE

filter:= n > isprime(n) and not numtheory:issqrfree(n+1) and not numtheory:issqrfree(n1):
select(filter, [seq(2*i+1, i=1..1000)]); # Robert Israel, Aug 27 2014


MATHEMATICA

lst={}; Do[p=Prime[n]; If[ !SquareFreeQ[Floor[p1]] && !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 *)


PROG

(Haskell)
a075432 n = a075432_list !! (n1)
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
 Reinhard Zumkeller, May 04 2013
(PARI) is(n)=!issquarefree(if(n%4==1, n+1, n1)) && isprime(n) \\ Charles R Greathouse IV, Aug 27 2014


CROSSREFS

Cf. A039787, A049097, A005117, A000040, A008966, A010051.
Sequence in context: A243437 A144709 A132239 * A232882 A232878 A226681
Adjacent sequences: A075429 A075430 A075431 * A075433 A075434 A075435


KEYWORD

nonn


AUTHOR

Reinhard Zumkeller, Sep 15 2002


EXTENSIONS

More terms (that were already in the bfile) from Jeppe Stig Nielsen, Apr 23 2020


STATUS

approved



