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A162872
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Primes p such that p-1 and p+1 each contain at least one squared prime in their prime factorization.
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5
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19, 149, 197, 199, 293, 307, 349, 491, 523, 557, 577, 739, 773, 883, 1013, 1051, 1061, 1151, 1171, 1277, 1451, 1493, 1531, 1549, 1601, 1637, 1667, 1693, 1709, 1733, 1747, 1861, 1949, 2069, 2141, 2179, 2251, 2351, 2357, 2467, 2549, 2683, 2789, 2843, 2851
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OFFSET
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1,1
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COMMENTS
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The selection criterion is that p-1 and p+1 are in the subsequence 4=2^2, 9=3^2, 12=2^2*3, 18=2*3^2, ... of nonsquarefree numbers (A013929) that actually display at least one square in their standard prime factorization.
So at least one of the e_i in p-1=product p_i^e_i, and at least one of the e_j in p+1=product p_j^e_j must equal 2. This is more stringent than being nonsquarefree, and the sequence becomes a subsequence of A075432.
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LINKS
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FORMULA
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EXAMPLE
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19 is in the sequence because 19 - 1 = 2*3^2 contains 3^2 and because 19 + 1 = 2^2*5 contains 2^2 in the factorization.
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MAPLE
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isA162872 := proc(n)
if isprime(n) then
isA038109(n-1) and isA038109(n+1) ;
else
false;
end if;
end proc:
n := 1:
for c from 1 to 50000 do
if isA162872(c) then
printf("%d %d\n", n, c) ;
n := n+1 ;
N:= 10^5: # to get all terms < N, where N is even
V:= Vector(N/2):
for i from 1 do
p:= ithprime(i);
if p^2 > N+1 then break fi;
if p = 2 then inds:= 2*[seq(i, i=1..floor(N/4), 2)]
else inds:= p^2*select(t -> t mod p <> 0, [$1..floor(N/2/p^2)])
fi;
V[inds]:= 1;
od:
select(t -> V[(t-1)/2] = 1 and V[(t+1)/2] = 1 and isprime(t), [seq(t, t=3..N, 2)]); # Robert Israel, Dec 08 2015
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MATHEMATICA
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f[n_]:=Module[{a=m=0}, Do[If[FactorInteger[n][[m, 2]]==2, a=1], {m, Length[FactorInteger[n]]}]; a]; lst={}; Do[p=Prime[n]; If[f[p-1]==1&&f[p+1]==1, AppendTo[lst, p]], {n, 7!}]; lst
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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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Role of squarefree numbers clarified by R. J. Mathar, Jul 31 2007
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STATUS
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approved
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