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A216101
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Primes which are the integer harmonic mean of the previous prime and the following prime.
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1
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13, 19, 43, 47, 83, 89, 103, 109, 131, 167, 193, 229, 233, 313, 349, 353, 359, 383, 389, 409, 443, 449, 463, 503, 643, 647, 677, 683, 691, 709, 797, 823, 859, 883, 919, 941, 971, 983, 1013, 1093, 1097, 1109, 1171, 1193, 1217, 1279, 1283, 1303, 1373, 1429, 1433
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OFFSET
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1,1
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COMMENTS
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The harmonic mean of N numbers p1,p2,..,pN is the number H whose reciprocal is the arithmetic mean of the reciprocals of p1,p2,..,pN; that is to say, 1/H = ((1/p1)+(1/p2)+..+(1/pN))/N.
So, for two quantities p1 and p2, their harmonic mean may be written as H=(2*p1*p2)/(p1+p2).
The harmonic mean (sometimes called the subcontrary mean) is one of several kinds of average. Typically, it is appropriate for situations when the average of rates is desired.
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LINKS
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EXAMPLE
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The primes before and after the prime p=13 are p1=11 and p2=17. So, the harmonic mean of p1 and p2 is 2*11*17/(11+17)=13.35714285... whose integer part is p=13. Then p=13 belongs to the sequence.
The primes before and after the prime p=17 are p1=13 and p2=19. The harmonic mean of p1 and p2 is 2*13*19/(13+19)=15.4375, having 15 as its integer part. Therefore, as 15<>p, p=17 is not in the sequence.
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MAPLE
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A := {}: for n from 2 to 1000 do p1 := ithprime(n-1): p := ithprime(n): p2 := ithprime(n+1): if p = floor(2*p1*p2/(p1+p2)) then A := `union`(A, {p}) end if end do; A := A;
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MATHEMATICA
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t = {}; Do[p = Prime[n]; If[Floor[HarmonicMean[{Prime[n - 1], Prime[n + 1]}]] == p, AppendTo[t, p]], {n, 2, 200}]; t (* T. D. Noe, Sep 04 2012 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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