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A157884 For each natural number m exist TWO primes Q=Q(m) and P=P(m) with: (2m-1)^2 + 1 <= Q <= (2m)^2 - (2m-1) <= 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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified June 16 05:18 EDT 2019. Contains 324145 sequences. (Running on oeis4.)