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A096696
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Consider the n-th prime, p_n, as the beginning of 2k+1 consecutive primes; then a(n) = p_(n+k) a balanced prime of order k, k maximized, or 0 if no such prime exists.
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0
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0, 5, 29, 37, 0, 0, 0, 0, 0, 149, 53, 0, 53, 71, 137, 227, 0, 0, 89, 79, 0, 0, 0, 0, 179, 0, 0, 173, 173, 0, 0, 419, 0, 157, 0, 157, 173, 0, 173, 0, 263, 0, 0, 0, 0, 211, 229, 0, 353, 397, 0, 0, 353, 359, 409, 577, 0, 353, 383, 353, 0, 0, 0, 0, 0, 0, 349, 349, 0, 0, 0, 397, 373
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OFFSET
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1,2
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COMMENTS
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a(n) either equals 0 or belongs to A090403.
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LINKS
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EXAMPLE
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a(2) = 5 because beginning with the second prime, 3, there is a run of three prime, (3,5,7) the average and median of which is 5.
a(5) = 0 because there does not exist a run of 2k + 1 primes such that the arithmetic mean and the median are the same.
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MATHEMATICA
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f[n_] := Block[{k = 1, p = 0}, While[k < 10^4, If[(Plus @@ Table[Prime[i], {i, n, n + 2k}]) == (2k + 1)Prime[n + k], p = Prime[n + k]]; k++ ]; p]; Table[ f[n], {n, 74}]
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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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