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A157185
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a(n) = least prime p such that there are n consecutive even numbers E < 2 sqrt(p) such that E^2-p is prime.
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0
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2, 13, 41, 83, 83, 167, 227, 227, 2273, 2273, 2273, 5297, 340007, 837077, 837077, 837077, 837077, 837077, 2004917, 2004917, 2004917, 208305767, 208305767
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OFFSET
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1,1
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COMMENTS
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Inspired by a puzzle proposed by J. M. Bergot, cf. link.
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.
a(24) > 3*10^9. - Donovan Johnson, Nov 29 2010
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LINKS
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Table of n, a(n) for n=1..23.
J. M. Bergot, Puzzle #481: E.E-p=q, in C. Rivera's Prime puzzles, Feb. 2009.
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EXAMPLE
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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:
[1, 2, 2] since 2^2-2 = 2 is prime
[2, 13, 6] since 4^2-5 = 11 and 6^2-5 = 31 are prime
[3, 41, 12]
[4, 83, 16] is a truncation of the following chain:
[5, 83, 18] since {10,12,14,16,18} squared minus 83 are all prime
[6, 167, 24] since 20^2-83 which is prime is ignored because 20 > 2sqrt(83)
[7, 227, 28]
[8, 227, 30]
[9, 2237, 64] since E=32,34,36,38,40 are > 2sqrt(227),
[10, 2273, 66] although they would also yield primes E^2-p for p=227.
[11, 2273, 68]
[12, 5297, 96]
[13, 340007, 690]
[14, 837077, 942]
[15, 837077, 944]
[16, 837077, 946]
[17, 837077, 948]
[18, 837077, 950]
[19, 2004917, 2572]
[20, 2004917, 2574]
[21, 2004917, 2576]
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PROG
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(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)))}
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CROSSREFS
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Sequence in context: A042795 A179925 A263980 * A219054 A154354 A138089
Adjacent sequences: A157182 A157183 A157184 * A157186 A157187 A157188
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KEYWORD
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more,nonn
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AUTHOR
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M. F. Hasler, Mar 02 2009
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EXTENSIONS
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a(22)-a(23) from Donovan Johnson, Nov 29 2010
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STATUS
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approved
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