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 A247189 a(n) is the least integer k such that prime(k) < (prime(n)+ prime(n+1) + ... + prime(n+k))/k <= prime(k+1). 1
 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 7, 7, 6, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 7, 8, 9, 8, 8, 7, 7, 8, 8, 9, 8, 8, 9, 9, 9, 10, 10, 9, 9, 8, 9, 10, 9, 9, 9, 8, 10, 10, 10, 10, 10, 11, 10, 11, 11, 10, 11, 10, 11, 12, 12, 11, 12, 13, 12 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS General property: Let b(0) < b(1) < b(2) < ... be an infinite sequence of strictly positive integers. So there exists a unique integer k >= 1 such that: b(k) < (b(0) + b(1) + ... + b(k))/k <= b(k+1). (See the reference.) LINKS Michel Lagneau, Table of n, a(n) for n = 1..10000 International Mathematical Olympiad 2014 Problem No 1, Cape Town - South Africa. EXAMPLE a(3)=2 because the infinite sequence {b(0),b(1),b(2),...} ={prime(3),prime(4),...} = {5, 7, 11, 13, ...} => b(2) < b(0) + b(1) + b(2))/2 <= b(3) => 11 < (5+7+11)/2 < 13 => 11 < 11.5 < 13. Hence a(3)=2. MAPLE for n from 1 to 80 do: ii:=0:   for k from n+1 to 10^8 while(ii=0)do:     s:=sum('ithprime(i)', 'i'=n..k):s1:=evalf(s/(k-n)):      if s1<= ithprime(k+1) and s1>ithprime(k)       then       printf(`%d, `, k-n):ii:=1:       else      fi:   od: od: MATHEMATICA lst={}; Do[k=n+1; While[Sum[Prime[j]/(k-n), {j, n, k}]>Prime[k+1]||Sum[Prime[j]/(k-n), {j, n, k}]prime(k+1))||((sum(j=n, k, prime(j))/(k-n))

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Last modified May 25 13:00 EDT 2022. Contains 354071 sequences. (Running on oeis4.)