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A247189
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a(n) is the least integer k such that prime(k) < (prime(n)+ prime(n+1) + ... + prime(n+k))/k <= prime(k+1).
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1
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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
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OFFSET
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1,2
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COMMENTS
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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.)
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LINKS
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International Mathematical Olympiad 2014 Problem No 1, Cape Town - South Africa.
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EXAMPLE
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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.
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MAPLE
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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:
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MATHEMATICA
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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], k++]; AppendTo[lst, k-n], {n, 1, 50}]; lst
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PROG
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(PARI) s=[]; for(n=1, 50, k=n+1; while(((sum(j=n, k, prime(j))/(k-n))>prime(k+1))||((sum(j=n, k, prime(j))/(k-n))<prime(k)), k++); s=concat(s, k-n)); s
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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