A055081 - OEIS

%I #21 Dec 23 2023 20:26:04

%S 1,2,3,3,3,7,3,4,5,6,3,10,3,6,10,5,3,11,3,10,9,6,3,13,5,6,7,10,3,20,3,

%T 6,9,6,10,16,3,6,9,13,3,20,3,9,17,6,3,16,5,10,9,9,3,15,9,13,9,6,3,30,

%U 3,6,16,7,9,20,3,9,9,19,3,22,3,6,16,9,10,19,3,16,9,6,3,30,9,6,9,13,3,33

%N Number of positive integers whose harmonic mean with n is a positive integer.

%C Also the number of factors of 2n^2 which are less than 2n, since the harmonic mean of n and 2n^2/k-n is 2n-k and these are all positive integers iff k<2n is a factor of 2n^2. So a(n)=3 iff n=4 or n is an odd prime.

%C For any n>2, there are three distinct trivial Diophantine solutions of H(n,x)=y, H being the harmonic mean: [x=n,y=n],[x=n(n-1),y=2(n-1)],[x=n(2n-1),y=2n-1]. Existence of any other solution proves that n is not a prime. - _Stanislav Sykora_, Feb 03 2016

%C a(n)=4 only for n=8. a(n)=5 iff n is 16 or the square of an odd prime. - _Robert Israel_, Feb 07 2016

%H Ivan Neretin, <a href="/A055081/b055081.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) >= min(n,3). - _Stanislav Sykora_, Feb 03 2016

%F a(2^n) = n+1, a(p^n) = 2n+1 if p>=3 is prime. - _Benoit Cloitre_, Nov 26 2023

%e a(6)=7 since the pairwise harmonic means of 6 with 2, 3, 6, 12, 18, 30 and 66 are 3, 4, 6, 8, 9, 10 and 11 respectively.

%p seq(nops(select(`<`,numtheory:-divisors(2*n^2),2*n)),n=1..100); # _Robert Israel_, Feb 07 2016

%t Count[Divisors[2 #^2], x_ /; x < 2 #] & /@ Range[90] (* _Ivan Neretin_, May 04 2015 *)

%o (PARI) a(n) = {my(c=0); for(y=1, 2*n-1, if((y*n)%(2*n-y)==0, c++)); return(c);} \\ _Stanislav Sykora_, Feb 03 2016

%Y The smallest and largest positive integers whose harmonic means with n are positive integers are A053626 and A000384 with harmonic means of A053627 and A004273.

%K nonn

%O 1,2

%A _Henry Bottomley_, Jun 13 2000