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A066837
Least number k such that for the average of n consecutive primes, beginning with the k-th prime, is divisible by the average of their indices.
0
1, 4, 2700, 3, 69, 29, 68, 185, 67
OFFSET
1,2
COMMENTS
a(10)>2*10^7.
EXAMPLE
a(1) = 1 because P_1/1 = 2, a(2) = 4 because (P_4 + P_5)/(4 + 5) = 2, a(3) = 2700 because (P_2700 + P_2701 + P_2702)/(2700 + 2701 + 2702) = 9, a(4) = 3 because (P_3 + P_4 + P_5 + P_6)/(3 + 4 + 5 + 6) = 2, etc.
MATHEMATICA
Do[ a = Table[ Prime[i], {i, 1, n} ]; k = n; While[ !IntegerQ[ Apply[ Plus, a]/(n(k - n/2 + 1/2)) ], k++; a = Append[a, Prime[k]]; a = Drop[a, 1]]; Print[k - n + 1], {n, 1, 25} ]
CROSSREFS
Sequence in context: A079187 A131587 A066850 * A275683 A297008 A349074
KEYWORD
nonn
AUTHOR
Robert G. Wilson v, Jan 21 2002
STATUS
approved