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## When the square root of n is less than or equal to n's number of divisors

Just in case you ever needed to know, there are 55 positive integers whose square root √n is less than or equal to their number of divisors τ(n).

√n < τ(n) in 53 cases, and in two cases √n = τ(n): 1 and 9.

They are:
 n τ(n) √n 1 1 1 2 2 1.41 3 2 1.73 4 3 2 6 4 2.45 8 4 2.83 9 3 3 10 4 3.16 12 6 3.46 14 4 3.74 15 4 3.87 16 5 4 18 6 4.24 20 6 4.47 24 8 4.90 28 6 5.30 30 8 5.48 32 6 5.66 36 9 6 40 8 6.32 42 8 6.48 48 10 6.93 54 8 7.35 56 8 7.48 60 12 7.75 72 12 8.49 80 10 8.94 84 12 9.17 90 12 9.49 96 12 9.80 108 12 10.39 120 16 10.95 126 12 11.22 132 12 11.49 140 12 11.83 144 15 12 168 16 12.96 180 18 13.42 192 14 13.86 210 16 14.49 216 16 14.70 240 20 15.49 252 18 15.87 288 18 16.97 300 18 17.32 336 20 18.33 360 24 18.97 420 24 20.49 480 24 21.91 504 24 22.45 540 24 23.24 720 30 26.83 840 32 28.98 1260 36 35.50

Or to put it another way:

 n τ(n) √n 1 1 1 2 2 1.41 3 1.73 4 3 2 9 3 6 4 2.45 8 2.83 10 3.16 14 3.74 15 3.87 16 5 4 12 6 3.46 18 4.24 20 4.47 28 5.29 32 5.66 24 8 4.90 30 5.48 40 6.32 42 6.48 54 7.35 56 7.48 36 9 6 48 10 6.93 80 8.94 60 12 7.75 72 8.49 84 9.17 90 9.49 96 9.80 108 10.39 126 11.22 132 11.49 140 11.83 192 14 13.86 144 15 12 120 16 10.95 168 12.96 210 14.49 216 14.70 180 18 13.42 252 15.87 288 16.97 300 17.32 240 20 15.49 336 18.33 360 24 18.97 420 20.49 480 21.91 504 22.45 540 23.24 720 26.83 840 32 28.98 1260 36 35.50

The number whose square root is smallest relative to its number of divisors is 12, whose 6 divisors are almost twice its square root of 3.45.

There, aren't you glad I told you?

Edited to add: Looks like I miscounted and there are only 54.

## Comments

Feb. 19th, 2019 01:23 pm (UTC)
If a number factorises as n = p_1^α_1 × ... × p_r^α_r, then τ(n) = Π (α_i+1).

To have τ(n)^2 ≥ n, we therefore need Π (α_i+1)^2 ≥ Π p_i^α_i, which rearranges as Π (α_i+1)^{2/α_i} ≥ Π p_i.

The LHS is Π f(α_i), where f(α) = (α+1)^{2/α}. f is a decreasing function, whose maximum value is 4 (when α=1). So we can make the LHS as large as we can by setting all α_i to 1, in which case the value of the LHS is exactly 4^r (where r is the number of distinct prime factors of n).

But the RHS is the product of r distinct primes, so its value must be at least the product of the smallest r distinct primes. So if r = 5, for example, then the RHS is at least 2×3×5×7×11=2310, and the LHS is at most 4^5 = 1024, so the inequality can't work. And increasing r further can only make things worse, because you multiply a factor of 4 into the LHS and a factor of at least 13 into the RHS.

So n must have at most 4 distinct prime factors (because 2×3×5×7 < 4^4). Also, none of them can be too large: in particular, if any prime factor is bigger than 64 then that immediately makes the RHS too big in one go. So there's a finite list of possible sets of distinct prime factors of n.

That still leaves open the possibility that for some particular list of distinct prime factors, there might be infinitely many choices of (α_i) that will work. For that to be true, you'd need the product f(α_1) × ... × f(α_{r-1}) to be large enough to make the inequality work no matter how small f(α_r) might be. But that product is at most 4^{r-1}, and for no possible list of r prime factors does that come to more than the product of the primes, even if the primes are chosen as small as possible: 4^0 < 2, 4^1 < 2×3, 4^2 < 2×3×5, 4^3 < 2×3×5×7.

So for each of the finite collection of possible sets of distinct prime factors of n, there is also a finite set of possible tuples of (α_i) that can work. From there, it's just exhaustive enumeration.