3, 10, 42, 60, 63, 840, 1260, 12642, 18480, 18900, 18963, 154350, 228480, 252840, 379260, 3458700, 5562480, 5688900, 68772480, 1041068700, 15032156160, 53621568000, 4524679004160, 9812746944000
a(25) > 10^13. - Giovanni Resta, Feb 17 2020
nonn,more
approved
Hemi-imperfect numbers: numbers such that the denominator of k/A206369(k) is equal to 2.
3, 10, 42, 60, 63, 840, 1260, 12642, 18480, 18900, 18963, 154350, 228480, 252840, 379260, 3458700, 5562480, 5688900, 68772480, 1041068700, 15032156160, 53621568000, 4524679004160, 9812746944000
1,1
This is to rho (A206369) what hemiperfect numbers are to sigma (A000203).
After 3, 10 and 42, whose quotients are resp. 3/2, 5/2 and 7/2, 373316437260251755241798182764378479569038727298776522806597255168000000 is an instance of a term with quotient 9/2. - Michel Marcus, Dec 17 2017
a(25) > 10^13. - Giovanni Resta, Feb 17 2020
Douglas E. Iannucci, <a href="http://www.integers-ejcnt.org/g41/g41.Abstract.html">On a variation of perfect numbers</a>, INTEGERS: Electronic Journal of Combinatorial Number Theory, 6 (2006), #A41.
László Tóth, <a href="https://arxiv.org/abs/1111.4842">A survey of the alternating sum-of-divisors function</a>, arXiv:1111.4842 [math.NT], 2011-2014.
Wikipedia, <a href="https://en.wikipedia.org/wiki/Hemiperfect_number">Hemiperfect number</a>
3 is a term since rho(3) = 2, so 3/rho(3) is 3/2.
10 is a term since rho(10) = 4, so 10/rho(10) is 5/2.
42 is a term since rho(42) = 12, so 42/rho(42) is 7/2.
rho:= proc(n) local f;
mul((f[1]^(f[2]+1)+(-1)^f[2])/(f[1]+1), f = ifactors(n)[2]);
end proc:
select(t -> denom(t/rho(t)) = 2, [$1..10^6]); # Robert Israel, Nov 20 2017
(* b = A209369 *) b[n_] := n*DivisorSum[n, LiouvilleLambda[#]/# &];
Select[Range[10^6], If[Denominator[#/b[#]] == 2, Print[#]; True, False]&] (* Jean-François Alcover, Dec 04 2017 *)
(PARI) rho(n) = {my(f = factor(n), res = q = 1); for(i=1, #f~, q = 1; for(j = 1, f[i, 2], q = -q + f[i, 1]^j); res *= q); res; }
isok(n) = denominator(n/rho(n))==2;
nonn,
Michel Marcus, Nov 19 2017
approved