OFFSET
1,2
COMMENTS
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.
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
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
LINKS
Ivan Neretin, Table of n, a(n) for n = 1..10000
FORMULA
a(n) >= min(n,3). - Stanislav Sykora, Feb 03 2016
a(2^n) = n+1, a(p^n) = 2n+1 if p>=3 is prime. - Benoit Cloitre, Nov 26 2023
EXAMPLE
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.
MAPLE
seq(nops(select(`<`, numtheory:-divisors(2*n^2), 2*n)), n=1..100); # Robert Israel, Feb 07 2016
MATHEMATICA
Count[Divisors[2 #^2], x_ /; x < 2 #] & /@ Range[90] (* Ivan Neretin, May 04 2015 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Henry Bottomley, Jun 13 2000
STATUS
approved