OFFSET
1,2
COMMENTS
Numbers k such that antiharmonic mean of divisors of k is an integer. Antiharmonic mean of divisors of number m = Product (p_i^e_i) is A001157(m)/A000203(m) = Product ((p_i^(e_i+1)+1)/(p_i+1)). So a(n) = k, for some n, if A001157(k)/A000203(k) is an integer. - Jaroslav Krizek, Mar 09 2009
Squares are antiharmonic, since (p^(2*e+1)+1)/(p+1) = p^(2*e) - p^(2*e-1) + p^(2*e-2) - ... + 1 is an integer. The nonsquare antiharmonic numbers are A227771. They include the primitive antiharmonic numbers A228023, except for its first term. - Jonathan Sondow, Aug 02 2013
Sequence is infinite, see A227771. - Charles R Greathouse IV, Sep 02 2013
The term "antiharmonic" is also known as "contraharmonic". - Pahikkala Jussi, Dec 11 2013
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (term 1..1000 from Paolo P. Lava)
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, The Biharmonic mean, Mathematical Reports, Vol. 18(68), No. 4 (2016), pp. 483-495, preprint, arXiv:1601.03081 [math.NT], 2016.
EXAMPLE
a(3) = 9 = 3^2; antiharmonic mean of divisors of 9 is (3^(2+1) + 1)/(3 + 1) = 7; 7 is an integer. - Jaroslav Krizek, Mar 09 2009
MATHEMATICA
Select[Range[2000], Divisible[DivisorSigma[2, #], DivisorSigma[1, #]]&] (* Jean-François Alcover, Nov 14 2017 *)
PROG
(Magma) [n: n in [1..1300] | IsZero(DivisorSigma(2, n) mod DivisorSigma(1, n))]; // Bruno Berselli, Apr 10 2013
(PARI) is(n)=sigma(n, 2)%sigma(n)==0 \\ Charles R Greathouse IV, Jul 02 2013
(Haskell)
a020487 n = a020487_list !! (n-1)
a020487_list = filter (\x -> a001157 x `mod` a000203 x == 0) [1..]
-- Reinhard Zumkeller, Jan 21 2014
(Python)
from sympy import divisor_sigma
def ok(n): return divisor_sigma(n, 2)%divisor_sigma(n, 1) == 0
print([k for k in range(1, 1300) if ok(k)]) # Michael S. Branicky, Feb 25 2024
(Python) # faster for producing initial segment of sequence
from math import prod
from sympy import factorint
def ok(n):
f = factorint(n)
sigma1 = prod((p**( e+1)-1)//(p-1) for p, e in f.items())
sigma2 = prod((p**(2*e+2)-1)//(p**2-1) for p, e in f.items())
return sigma2%sigma1 == 0
print([k for k in range(1, 1300) if ok(k)]) # Michael S. Branicky, Feb 25 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved