A350100 Numbers k such that the prime gap between the consecutive primes p1 < k^2 < p2 sets a new record.

2, 3, 5, 11, 23, 30, 41, 50, 76, 100, 149, 159, 189, 345, 437, 509, 693, 1110, 1165, 5018, 14908, 18906, 19079, 28634, 38682, 80444, 105686, 185179, 265236, 269697, 409049, 558269, 1673629, 2965232, 3528015, 4292936, 34919969, 43957056, 148793437, 187220890, 424171123

Offset: 1

Author

Hugo Pfoertner, Dec 25 2021

Keywords

• nonn

Comments

a(51) (in b-file) > 1.5*10^11, corresponding to A378904(51) > 723. - Hugo Pfoertner, Jan 04 2025

Examples

			  n  a(n)  p1   a(n)^2   p2   gap=2*A378904(n)
  1   2     3      4      5    2
  2   3     7      9     11    4
  3   5    23     25     29    6
  4  11   113    121    127   14
  5  23   523    529    541   18
  6  30   887    900    907   20
  7  41  1669   1681   1693   24
  8  50  2477   2500   2503   26
		

Links

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

Crossrefs

A378904 are the corresponding gaps, divided by 2.

Cf. A001223, A005250, A058043.

Programs

Mathematica

Module[{nn=4242*10^5,pg},pg=Table[{n,NextPrime[n^2]-NextPrime[n^2,-1]},{n,2,nn}];DeleteDuplicates[pg,GreaterEqual[#1[[2]],#2[[2]]]&]][[All,1]] (* Harvey P. Dale, Jan 28 2023 *)

PARI

a350100(limit) = {my(pmax=0); for(k=2,limit, my(kk=k*k, pp=precprime(kk), pn=nextprime(kk), d=pn-pp); if(d>pmax, print1(k,", "); pmax=d))};
a350100(3000000)

Python

from itertools import count, islice
from sympy import prevprime, nextprime
def A350100_gen(): # generator of terms
    c = 0
    for k in count(2):
        a = nextprime(m:=k**2)-prevprime(m)
        if a>c:
            yield k
            c = a
A350100_list = list(islice(A350100_gen(),20)) # Chai Wah Wu, Dec 17 2024