login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A067743
Number of divisors of n not in the half-open interval [sqrt(n/2), sqrt(n*2)).
5
0, 1, 2, 2, 2, 2, 2, 3, 2, 4, 2, 4, 2, 4, 2, 4, 2, 5, 2, 4, 4, 4, 2, 6, 2, 4, 4, 4, 2, 6, 2, 5, 4, 4, 2, 8, 2, 4, 4, 6, 2, 6, 2, 6, 4, 4, 2, 8, 2, 5, 4, 6, 2, 6, 4, 6, 4, 4, 2, 10, 2, 4, 4, 6, 4, 6, 2, 6, 4, 6, 2, 9, 2, 4, 6, 6, 2, 8, 2, 8, 4, 4, 2, 10, 4, 4, 4, 6, 2, 10, 2, 6, 4, 4, 4, 10, 2, 5, 4, 8, 2, 8
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
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
a(n) = A000005(n) - A067742(n).
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