A067743 Number of divisors of n not in the half-open interval [sqrt(n/2), sqrt(n*2)).

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

Author

Marc LeBrun, Jan 29 2002

Keywords

• easy
• nonn

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

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

Examples

			a(6)=2 because 2 divisors of 6 (i.e., 1 and 6) fall outside sqrt(3) to sqrt(12).
		

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.

Crossrefs

Cf. A067742, A000005, A001105, A302433.

Programs

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 *)

PARI

A067743(n)=sumdiv( n,d, d*d= 2*n ) \\ M. F. Hasler, May 12 2008

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

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