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A159803
Number of primes p with (2m+1)^2 - 2m <= p < (2m+1)^2.
1
1, 1, 2, 2, 1, 3, 2, 3, 4, 4, 3, 5, 3, 5, 4, 4, 5, 2, 6, 4, 4, 7, 3, 8, 5, 7, 6, 5, 7, 8, 10, 5, 8, 7, 10, 8, 7, 10, 9, 7, 10, 9, 13, 10, 11, 11, 11, 11, 11, 12, 9, 9, 11, 14, 12, 11, 12, 12, 11, 15, 12, 11, 14, 12, 12, 14, 15, 12, 15, 14, 17, 18, 20, 18, 17, 14, 18, 12, 15, 15, 15, 14, 21
OFFSET
1,3
COMMENTS
1) Immediate connection to unsolved problem, is there always a prime between n^2 and (n+1)^2 ("full" interval of two consecutive squares).
2) See sequence A145354 and A157884 for more details to this new improved conjecture.
3) Second ("right") half interval: number of primes p with (2m+1)^2-2m <= p < (2m+1)^2.
4) It is conjectured that a(m) >= 1.
5) No a(m) with m>5 is known, where a(m)=1.
This is a bisection of A094189 and hence related to a conjecture of Oppermann. - T. D. Noe, Apr 22 2009
REFERENCES
L. E. Dickson, History of the Theory of Numbers, Vol, I: Divisibility and Primality, AMS Chelsea Publ., 1999
R. K. Guy, Unsolved Problems in Number Theory (2nd ed.) New York: Springer-Verlag, 1994
P. Ribenboim, The New Book of Prime Number Records. Springer. 1996
EXAMPLE
1) m=1: 7 <= p < 9 => prime 7: a(1)=1.
2) m=2: 21 <= p < 25 => prime 23: a(2)=1.
3) m=3: 43 <= p < 49 => primes 43, 47: a(3)=2.
4) m=30: 3661 <= p < 3721 => primes 3671,3673,3677,3691,3697,3701,3709,3719: a(30)=8.
MAPLE
A159803 := proc(n) local a, p; a := 0 ; for p from 4*n^2+2*n+1 to 4*n^2+4*n do if isprime(p) then a := a+1 ; fi; od: a ; end: seq(A159803(n), n=1..120) ; # R. J. Mathar, Apr 22 2009
CROSSREFS
KEYWORD
nonn
AUTHOR
Ulrich Krug (leuchtfeuer37(AT)gmx.de), Apr 22 2009
EXTENSIONS
More terms from R. J. Mathar, Apr 22 2009
STATUS
approved