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A096696
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.
0
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
OFFSET
1,2
COMMENTS
a(n) either equals 0 or belongs to A090403.
EXAMPLE
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.
MATHEMATICA
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}]
CROSSREFS
Cf. A090403.
Sequence in context: A321701 A243012 A053244 * A115279 A279393 A182288
KEYWORD
nonn
AUTHOR
Robert G. Wilson v, Jul 02 2004
STATUS
approved