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A088968
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a(n) is the number of consecutive primes x-3,x+3 such that x=j*(p(n)#/3)/p(k), where 1 <= j < p(n+1) and 3 <= k <= n and p(k) doesn't divide j.
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3
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0, 0, 0, 2, 3, 5, 6, 7, 13, 4, 10, 9, 2, 12, 8, 14, 6, 8, 16, 8, 9, 8, 19, 10, 15, 18, 17, 8, 10, 14, 9, 13, 10, 15, 14, 11, 15, 10, 13, 20, 15, 13, 14, 16, 16, 15, 19, 17, 14, 18, 13, 13, 15, 15, 7, 14, 16, 21, 12, 11, 13, 20, 7, 19, 18, 13, 8, 19
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OFFSET
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1,4
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COMMENTS
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p(n) is the n-th prime; # denotes primorial (A002110).
a(n) seems to grow like 2*log(p(n)).
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LINKS
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EXAMPLE
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a(5) = 3 because for (j,k) = (1,3),(10,4),(8,5), j*(11#/3)/p(k)+-3 are consecutive primes.
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PROG
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(PARI) a(n) = {my(p=vector(n, i, prime(i)), x, y=2*prod(i=3, n, p[i])); sum(j=1, prime(n+1)-1, sum(k=3, n, j%p[k]>0 && ispseudoprime(x=j*y/p[k]-3) && nextprime(x+1)==x+6)); } \\ Jinyuan Wang, Mar 20 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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