A227771 - OEIS

%I #18 Dec 10 2016 19:31:00

%S 20,50,117,180,200,242,325,450,468,500,578,605,650,800,968,980,1025,

%T 1058,1280,1300,1445,1476,1620,1682,1700,1800,1872,2178,2312,2340,

%U 2420,2450,2600,2645,2925,3200,3362,3380,3757,3872,4050,4100,4205,4232,4352,4418,4500,4693,5200

%N Antiharmonic numbers that are not squares.

%C Given prime factorization m = product (p_i^e_i), the antiharmonic (or contraharmonic) mean of the divisors of m is sigma_2(m)/sigma_1(m) = product (p_i^(e_i+1)+1)/(p_i+1). If this is an integer, then m is called antiharmonic.

%C All squares are trivially antiharmonic, since (p^(2*e+1)+1)/(p+1) = p^(2*e) - p^(2*e-1) + p^(2*e-2) - ... + 1 is an integer. Sequence gives the nontrivial antiharmonic numbers.

%C The antiharmonic means of their divisors are A227986.

%C Sequence is infinite, since if n is in the sequence and gcd(n, k) = 1 then nk^2 is also in the sequence. - _Charles R Greathouse IV_, Aug 02 2013

%C Removing such terms nk^2 leaves the primitive antiharmonic numbers A228023. - _Jonathan Sondow_, Aug 04 2013

%D R. Guy, Unsolved Problems in Number Theory, B2 (see harmonic number).

%H Charles R Greathouse IV, <a href="/A227771/b227771.txt">Table of n, a(n) for n = 1..10000</a>

%F A001157(a(n))/A000203(a(n)) = A227986(n).

%e sigma_2(20)/sigma_1(20) = (1^2 + 2^2 + 4^2 + 5^2 + 10^2 + 20^2)/(1 + 2 + 4 + 5 + 10 + 20) = 546/42 = 13 is an integer, 20 is not a square, and no smaller number has these properties, so a(1) = 20.

%o (PARI) is(n)=if(issquare(n),return(0)); my(f=factor(n)); denominator(prod(i=1,#f~,(f[i,1]^(f[i,2]+1)+1)/(f[i,1]+1)))==1 \\ _Charles R Greathouse IV_, Aug 02 2013

%Y Subsequence of A020487. Cf. A000203, A001157, A176797, A227771, A227986, A228023, A228024, A228036.

%K nonn

%O 1,1

%A _Jonathan Sondow_, Aug 02 2013