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, 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
Keywords
Examples
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.
Crossrefs
Cf. A090403.
Programs
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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}]
Comments