

A216101


Primes which are the integer harmonic mean of the previous prime and the following prime.


1



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

1,1


COMMENTS

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.


LINKS

Table of n, a(n) for n=1..51.
Wikipedia, Harmonic Mean


EXAMPLE

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.


MAPLE

A := {}: for n from 2 to 1000 do p1 := ithprime(n1): 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;


MATHEMATICA

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 *)


CROSSREFS

Cf. A006562, A001008, A102928, A216098, A216124.
Sequence in context: A107188 A029478 A252021 * A096455 A124199 A119869
Adjacent sequences: A216098 A216099 A216100 * A216102 A216103 A216104


KEYWORD

nonn


AUTHOR

César Eliud Lozada, Sep 01 2012


STATUS

approved



