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A263206
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Number of primes p with n^2 < prime(p) < (n+2)^2.
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1
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2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 3, 2, 2, 3, 3, 3, 2, 3, 5, 3, 1, 3, 3, 3, 3, 3, 5, 4, 4, 3, 1, 3, 5, 5, 4, 4, 5, 4, 1, 4, 4, 2, 5, 5, 3, 4, 6, 5, 4, 4, 4, 5, 5, 5, 4, 3, 4, 4, 5, 5, 5, 6, 5, 5, 6, 5, 4, 4, 5, 6, 6, 4, 4, 7, 5, 5, 7, 4, 4, 5, 5, 6, 6, 5, 6, 7, 6, 7, 7, 5, 5, 5, 5, 7, 7, 4
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OFFSET
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1,1
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COMMENTS
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Conjecture: a(n) > 0 for all n > 0. In other words, for each n = 1,2,3,... the interval (n^2, (n+2)^2) contains a prime with prime subscript.
We also guess that a(n) = 1 only for n = 25, 35, 43.
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LINKS
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Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
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EXAMPLE
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a(1) = 2 since 1^2 < prime(2) = 3 < prime(3) = 5 < (1+2)^2 with 2 and 3 both prime.
a(25) = 1 since 25^2 = 625 < prime(127) = 709 < (25+2)^2 = 729 with 127 prime.
a(35) = 1 since 35^2 = 1225 < prime(211) = 1297 < (35+2)^2 = 1369 with 211 prime.
a(43) = 1 since 43^2 = 1849 < prime(293) = 1913 < (43+2)^2 = 2025 with 293 prime.
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MATHEMATICA
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Do[r=0; Do[If[PrimeQ[k], r=r+1], {k, PrimePi[n^2]+1, PrimePi[(n+2)^2-1]}]; Print[n, " ", r]; Continue, {n, 1, 100}]
Table[Count[PrimePi/@Select[Range[n^2+1, (n+2)^2-1], PrimeQ], _?PrimeQ], {n, 100}] (* Harvey P. Dale, May 26 2020 *)
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CROSSREFS
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Cf. A000040, A000290, A006450, A263204.
Sequence in context: A337301 A076925 A100961 * A230501 A287272 A300287
Adjacent sequences: A263203 A263204 A263205 * A263207 A263208 A263209
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KEYWORD
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nonn
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AUTHOR
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Zhi-Wei Sun, Oct 12 2015
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STATUS
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approved
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