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A293682
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a(n) = least odd number k > 1 such that p = prime(n) is the middle of k consecutive primes which have arithmetic mean p, or 1 if no such k exists.
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0
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1, 1, 3, 1, 1, 1, 7, 1, 1, 15, 1, 17, 1, 1, 1, 3, 1, 1, 1, 13, 1, 5, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 37, 1, 51, 17, 3, 1, 1, 3, 33, 1, 1, 7, 1, 1, 3, 1, 67, 7, 1, 1, 1, 1, 3, 3, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 23, 37, 1, 3, 1, 35, 1, 13, 1, 13, 99, 11, 1
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OFFSET
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1,3
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LINKS
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EXAMPLE
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There are no primes before prime(1) so a(1) = 1.
a(3) = 3 as prime(3) = 5, which is the arithmetic mean of the three consecutive primes {3, 5, 7}.
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PROG
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(PARI) a(n) = {my(s = pprev = pnxt = p = prime(n), q=1); for(i=1, n-1, pprev = precprime(pprev - 1); pnxt = nextprime(pnxt + 1); s += (pprev + pnxt); q += 2; if(p * q == s, return(q))); return(1)}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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