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A087941
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a(n) is the number of consecutive primes x-2,x+2 such that x=j*(p(n)#/2)/p(k), where 1 <= j < p(n+1) and 2 <= k <= n and p(k) doesn't divide j.
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2
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0, 0, 1, 3, 7, 4, 4, 6, 7, 9, 7, 8, 8, 6, 9, 9, 7, 7, 6, 10, 9, 10, 5, 9, 10, 5, 8, 10, 13, 8, 15, 7, 6, 13, 8, 7, 8, 14, 13, 13, 11, 11, 7, 11, 10, 8, 11, 5, 11, 12, 14, 6, 16, 14, 15, 15, 12, 9, 7, 7, 13, 11, 10, 12, 12, 10, 13, 11, 7, 14, 14, 13, 14, 10, 13
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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(4) = 3 because for (j,k) = (1,3),(1,4),(3,4), j*(7#/2)/p(k)+-2 are consecutive primes.
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PROG
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(PARI) a(n) = {my(p=vector(n, i, prime(i)), x, y=prod(i=2, n, p[i])); sum(j=1, prime(n+1)-1, sum(k=2, n, j%p[k]>0 && ispseudoprime(x=j*y/p[k]-2) && nextprime(x+1)==x+4)); } \\ 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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