

A157884


For each natural number m exist TWO primes Q=Q(m) and P=P(m) with: (2m1)^2 + 1 <= Q <= (2m)^2  (2m1) <= P <= (2m)**2 (*) notes: If more than one prime, we take the smallest In some intervals one prime only: 2, 3, 11, 13, 29, 31, ..., 97


6



2, 3, 11, 13, 29, 31, 53, 59, 83, 97, 127, 137, 173, 191, 227, 241, 293, 307, 367, 383, 443, 463, 541, 557, 631, 653, 733, 757, 853, 877, 967, 997, 1091, 1123, 1229, 1277, 1373, 1409, 1523, 1567, 1693, 1723, 1861, 1901, 2027, 2081, 2213, 2267, 2411, 2459
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OFFSET

1,1


COMMENTS

Second part of numerical results to that Classic more than 200 years UNSOLVED famous problem: There is always a prime p in the interval of two consecutive square numbers: n^2 <= p <= (n+1)^2


REFERENCES

Dickson, History of the theory of numbers


LINKS

Table of n, a(n) for n=1..50.


EXAMPLE

Examples: m=1: (*) is: 2 <= Q <= 3 <= P <= 4 this gives Q(1)= 2 and P(1)=3 => a(1)=2, a(2)=3 m=2: (*) is: 10 <= Q <= 13 <= P <= 16 this gives smallest prime in the interval Q(2)= 11 and P(2)=13 => a(3)=11, a(4)=13 So for m=1,2,...15: a(1)=2, a(2)=3, a(3)=11, a(4)=13, a(5)=29, a(6)=31, a(7)=53, a(8)=59, a(9)=83, a(10)=97, a(11)=127, a(12)=137, a(13)=173, a(14)=191, a(15)=227, a(16)=277, a(17)=293, a(18)=347, a(19)=367, a(20)=431, a(21)=443, a(22)=509, a(23)=541, a(24)=607, a(25)=631, a(26)=709, a(27)=733, a(28)=827, a(29)=853, a(30)=937


CROSSREFS

A145354
Sequence in context: A045317 A215378 A078763 * A234530 A235632 A085306
Adjacent sequences: A157881 A157882 A157883 * A157885 A157886 A157887


KEYWORD

nonn


AUTHOR

Ulrich Krug (leuchtfeuer37(AT)gmx.de), Mar 08 2009


EXTENSIONS

Replaced 277 by 241, 347 by 307, 431 by 383 etc  R. J. Mathar, Nov 01 2010


STATUS

approved



