OFFSET
1,3
COMMENTS
From Max Alekseyev, May 13 2008: (Start)
Direct proof of Joerg Arndt's g.f. (see formula section).
We need to count divisors d|n such that d^2<=n/2 or d^2>2n. In the latter case, let's switch to co-divisor, replacing d with n/d.
Then we need to find the total count of: 1) divisors d|n such that 2d^2<=n; 2) divisors d|n such that 2d^2<n.
Let d|n and 2d^2<=n. Then n-2d^2 must be a multiple of d, i.e., n-2d^2=td for some integer t>=0.
Moreover it is easy to see that 1) is equivalent to n = 2d^2 + td for some integer t>=0. Therefore the answer for 1) is the coefficient of z^n in Sum_{d>=1} Sum_{t>=0} x^(2d^2 + td) = Sum_{d>=1} x^(2d^2)/(1 - x^d).
Similarly, the answer for 2) is Sum_{d>=1} x^(2d^2)/(1 - x^d) * x^d.
Therefore the g.f. for A067743 is Sum_{d>=1} x^(2d^2)/(1 - x^d) + Sum_{d>=1} x^(2d^2)/(1 - x^d) * x^d = Sum_{d>=1} x^(2d^2)/(1 - x^d) * (1 + x^d), as proposed. (End)
a(n) is odd if and only if n is in A001105. - Robert Israel, Oct 05 2020
Number of nonmiddle divisors of n. - Omar E. Pol, Jun 11 2022
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
Robin Chapman, Kimmo Ericksson, Richard P. Stanley and Reiner Martin, On the Number of Divisors of n in a Special Interval: Problem 10847, The American Mathematical Monthly, Vol. 109, No. 1 (Jan., 2002), p. 80.
FORMULA
G.f.: Sum_{k>=1} z^(2*k^2)*(1+z^k)/(1-z^k). - Joerg Arndt, May 12 2008
EXAMPLE
a(6)=2 because 2 divisors of 6 (i.e., 1 and 6) fall outside sqrt(3) to sqrt(12).
MAPLE
f:=proc(n) nops(select(t -> t^2 < n/2 or t^2 >= 2*n, numtheory:-divisors(n))) end proc:
map(f, [$1..200]); # Robert Israel, Oct 05 2020
MATHEMATICA
hoi[n_]:=Length[DeleteCases[Divisors[n], _?(Sqrt[n/2]<=#<Sqrt[2*n]&)]]; Array[ hoi, 110] (* Harvey P. Dale, Aug 22 2020 *)
PROG
(PARI) A067743(n)=sumdiv( n, d, d*d<n/2 || d*d >= 2*n ) \\ M. F. Hasler, May 12 2008
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Marc LeBrun, Jan 29 2002
STATUS
approved