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%I #18 Sep 07 2013 14:28:55
%S 13,19,43,47,83,89,103,109,131,167,193,229,233,313,349,353,359,383,
%T 389,409,443,449,463,503,643,647,677,683,691,709,797,823,859,883,919,
%U 941,971,983,1013,1093,1097,1109,1171,1193,1217,1279,1283,1303,1373,1429,1433
%N Primes which are the integer harmonic mean of the previous prime and the following prime.
%C 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.
%C So, for two quantities p1 and p2, their harmonic mean may be written as H=(2*p1*p2)/(p1+p2).
%C 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.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Harmonic_mean">Harmonic Mean</a>
%e 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.
%e 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.
%p 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;
%t 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 *)
%Y Cf. A006562, A001008, A102928, A216098, A216124.
%K nonn
%O 1,1
%A _César Eliud Lozada_, Sep 01 2012
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