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%I #12 May 18 2020 04:32:41
%S 1,1,1,1,2,2,3,4,1,3,4,2,4,4,4,5,6,5,3,6,5,7,6,6,6,5,7,6,7,8,7,8,11,8,
%T 8,8,7,11,4,11,9,10,9,11,8,10,13,9,14,12,11,12,12,11,15,12,9,13,15,8,
%U 14,13,16,12,14,11,11,15,9,16,16,11,14,14,13,13,13
%N Number of primes q with (2n-1)^2+1 <= q < (2n)^2-(2n-1).
%C Immediate connection to unsolved problem, is there always a prime between n^2 and (n+1)^2 ("full" interval of two consecutive squares).
%C See sequence A145354 and A157884 for more details to this new improved conjecture.
%C First ("left") half interval: number of primes q (2m-1)^2+1 <= q < (2m)^2-(2m-1).
%C It is conjectured that a(n) >= 1 for n >= 1.
%C No a(m) with m>9 is known, where a(m)=1.
%C This is a bisection of A089610 and hence related to a conjecture of Oppermann. [_T. D. Noe_, Apr 22 2009]
%D L. E. Dickson, History of the Theory of Numbers, Vol, I: Divisibility and Primality, AMS Chelsea Publ., 1999
%D R. K. Guy, Unsolved Problems in Number Theory (2nd ed.) New York: Springer-Verlag, 1994
%D P. Ribenboim, The New Book of Prime Number Records. Springer. 1996
%e n=1: 2 <= q < 3 => prime 2: a(1)=1;
%e n=5: 82 <= q < 91 => primes 83,89: a(5)=2;
%e n=9: 290 <= q < 307 => prime 293: a(9)=1;
%e n=30: 3482 <= q < 3541 => prime 3491,3499,3511,3517,3527,3529,3533,3539: a(30)=8.
%o (PARI) a(n) = if (n==1, 1, primepi((2*n)^2-(2*n-1)-1) - primepi((2*n-1)^2+1)); \\ _Michel Marcus_, May 18 2020
%Y Cf. A014085, A089610, A145354, A157884.
%K nonn
%O 1,5
%A Ulrich Krug (leuchtfeuer37(AT)gmx.de), Apr 22 2009
%E More terms from _Michel Marcus_, May 18 2020
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