A157185 - OEIS

%I #13 Aug 27 2023 14:13:32

%S 2,13,41,83,83,167,227,227,2273,2273,2273,5297,340007,837077,837077,

%T 837077,837077,837077,2004917,2004917,2004917,208305767,208305767

%N a(n) = least prime p such that there are n consecutive even numbers E < 2 sqrt(p) such that E^2-p is prime.

%C Inspired by a puzzle proposed by _J. M. Bergot_, cf. link.

%C The limit of 2*sqrt(p) imposed here is somewhat arbitrary, but this seemed the most natural choice. (Note that the E's must be > sqrt(p).) Changing it to 3*sqrt(p) would "reveal" the numbers 5,17 and the fact that p=227 is "good" up to n=13. Beyond a(14) both of these limits yield the same terms.

%C a(24) > 3*10^9. - _Donovan Johnson_, Nov 29 2010

%H J. M. Bergot, <a href="http://www.primepuzzles.net/puzzles/puzz_481.htm">Puzzle #481: E.E-p=q</a>, in C. Rivera's Prime puzzles, Feb. 2009.

%e The following table gives n (length of the run), a(n) (the prime), and the largest of the n consecutive even numbers E[1],...,E[n] such that E[i]^2-p is prime:

%e [1, 2, 2] since 2^2-2 = 2 is prime

%e [2, 13, 6] since 4^2-5 = 11 and 6^2-5 = 31 are prime

%e [3, 41, 12]

%e [4, 83, 16] is a truncation of the following chain:

%e [5, 83, 18] since {10,12,14,16,18} squared minus 83 are all prime

%e [6, 167, 24] since 20^2-83 which is prime is ignored because 20 > 2*sqrt(83)

%e [7, 227, 28]

%e [8, 227, 30]

%e [9, 2237, 64] since E=32,34,36,38,40 are > 2*sqrt(227),

%e [10, 2273, 66] although they would also yield primes E^2-p for p=227.

%e [11, 2273, 68]

%e [12, 5297, 96]

%e [13, 340007, 690]

%e [14, 837077, 942]

%e [15, 837077, 944]

%e [16, 837077, 946]

%e [17, 837077, 948]

%e [18, 837077, 950]

%e [19, 2004917, 2572]

%e [20, 2004917, 2574]

%e [21, 2004917, 2576]

%o (PARI) list_A157185( m=0, p=0, C=2, L=0 )={ until( L=0, forstep( j=(t=sqrt(p=nextprime(p+2)))\2*2+2,C*t\1,2, if(isprime(j^2-p), L++>m & print([m=L,p,j]), L&L=0)))}

%K more,nonn

%O 1,1

%A _M. F. Hasler_, Mar 02 2009

%E a(22)-a(23) from _Donovan Johnson_, Nov 29 2010