Introduction

• Many questions in number theory revolve around divisors of a number.
• Two classical functions help us understand how “divisor‑rich” a number is:
  • $τ(n)$ — the number of positive divisors of $n$
    • $τ$ is the Greek letter tau, and rhymes with "cow"
  • $σ(n)$ — the sum of all positive divisors of $n$
    • $σ$ is the Greek letter sigma
• These functions reveal surprising patterns:
  • Why do some numbers have many divisors?
  • Why are some numbers “perfect”?
  • How does prime factorisation control everything?

This article revisits these ideas with fresh examples and intuition.

Review: Divisors and Prime Factorisation

• A divisor of $n$ is a positive integer that divides $n$ with no remainder.
• Example: divisors of $12$ are $1,2,3,4,6,12$.
• Every positive integer has a unique prime factorisation: $$n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$$
• Prime factorisation is the key to understanding $τ(n)$ and $σ(n)$.

The Functions τ(n) and σ(n): Definitions and First Examples

• Number of divisors:
  • $τ(n)$ counts how many positive divisors $n$ has.
  • Example: $τ(12)=6$.
• Sum of divisors:
  • $σ(n)$ adds all positive divisors.
  • Example: $σ(12)=1+2+3+4+6+12=28$.
• Quick examples:
  • $τ(7)=2$ (prime numbers have exactly two divisors)
  • $σ(7)=1+7=8$

How Prime Factorisation Controls Divisors

• If $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$:
  • Each divisor is formed by choosing an exponent for each prime:
    • For $p_1$: choose $0,1,\dots,a_1$
    • For $p_2$: choose $0,1,\dots,a_2$
    • …
• This leads to:
  • $(a_1+1)$ choices for the first prime
  • $(a_2+1)$ choices for the second prime
  • …
• Multiply the choices to count all divisors.

Formulas for τ(n) and σ(n)

Number of divisors

If $$n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k},$$ then $$τ(n) = (a_1+1)(a_2+1)\cdots(a_k+1).$$

Sum of divisors

For a single prime power: $$σ(p^a) = 1 + p + p^2 + \cdots + p^a.$$ This is a geometric series: $$σ(p^a) = \frac{p^{a+1}-1}{p-1}.$$ For a general $n$: $$σ(n) = σ(p_1^{a_1}) \, σ(p_2^{a_2}) \cdots σ(p_k^{a_k}).$$

Working Through Examples Step by Step

Example 1: $n=18$

• Factorisation: $18 = 2^1 \cdot 3^2$
• Number of divisors:
  • $τ(18) = (1+1)(2+1) = 2 \cdot 3 = 6$
• Sum of divisors:
  • $σ(2^1)=1+2=3$
  • $σ(3^2)=1+3+9=13$
  • $σ(18)=3 \cdot 13 = 39$

Example 2: $n=72$

• Factorisation: $72 = 2^3 \cdot 3^2$
• $τ(72) = (3+1)(2+1) = 4 \cdot 3 = 12$
• $σ(2^3)=1+2+4+8=15$
• $σ(3^2)=1+3+9=13$
• $σ(72)=15 \cdot 13 = 195$

Multiplicativity of σ and τ

Both $σ$ and $τ$ are multiplicative functions, meaning:

• If $\gcd(a,b)=1$, then $$τ(ab)=τ(a)\,τ(b),$$ $$σ(ab)=σ(a)\,σ(b).$$

This is not true if $a$ and $b$ share a prime factor.

Why multiplicativity holds:

• Divisors of $ab$ correspond to pairs of divisors $(d_1,d_2)$ with $d_1\mid a$ and $d_2\mid b$.
• When $a$ and $b$ are coprime, these pairs uniquely describe all divisors.

Why Multiplicativity Matters

• It allows fast computation of $σ(n)$ and $τ(n)$ from prime factorisation.
• It explains why numbers with many small prime factors tend to have many divisors.
• It helps classify special types of numbers (perfect, abundant, highly composite).

Behaviour of σ(n): Abundant, Perfect, and Deficient Numbers

Using the sum of proper divisors: $$s(n) = σ(n) - n.$$ We classify numbers:

• Perfect: $s(n)=n$
  • Example: $6$ and $28$
• Abundant: $s(n)>n$
  • Example: $12$ (proper divisors sum to $16$)
• Deficient: $s(n)<n$
  • Example: $8$ (proper divisors sum to $7$)

Patterns:

• Powers of primes are always deficient.
• Numbers with many small prime factors tend to be abundant.

Behaviour of τ(n): Highly Composite and Related Ideas

• $τ(n)$ measures how “divisor‑rich” a number is.
• A highly composite number is a number with more divisors than any smaller number.
  • Examples: $1,2,4,6,12,24,36,48,60,120,\dots$
• These numbers tend to:
  • Use small primes
  • Use exponents in decreasing order (e.g., $2^3 3^2 5^1$)

Comparing Growth Rates of σ(n) and τ(n)

• $τ(n)$ grows slowly; even very large numbers rarely have more than a few thousand divisors.
• $σ(n)$ grows faster because it adds divisors, not just counts them.
• Rough intuition:
  • $τ(n)$ is usually around the size of a small power of $\log n$.
  • $σ(n)$ is often a bit larger than $n$, but can be much larger for numbers with many small prime factors.

Common Pitfalls and Misconceptions

• Forgetting that $σ$ and $τ$ are multiplicative only for coprime arguments.
• Thinking $σ(n)$ is always larger than $n$ (false for deficient numbers).
• Assuming prime numbers have $τ(p)=1$ (they have $2$ divisors).
• Forgetting to include the exponent $0$ when counting divisors.

Summary and Key Takeaways

• $τ(n)$ counts divisors; $σ(n)$ sums them.
• Prime factorisation completely determines both functions.
• Both functions are multiplicative for coprime inputs.
• $σ(n)$ helps classify numbers as perfect, abundant, or deficient.
• $τ(n)$ highlights numbers with unusually many divisors.

Calculator

Exercises

1. Compute $τ(45)$ and $σ(45)$.

   Solution

   Compute $τ(45)$ and $σ(45)$

   • Step 1: Factorise
     • $45 = 3^2 \cdot 5^1$
   • Step 2: Number of divisors
     • Exponents: $2$ and $1$
     • $$τ(45) = (2+1)(1+1) = 3 \cdot 2 = 6$$
   • Step 3: Sum of divisors
     • $$σ(3^2) = 1 + 3 + 9 = 13$$
     • $$σ(5^1) = 1 + 5 = 6$$
     • $$σ(45) = 13 \cdot 6 = 78$$

   Answer: $τ(45)=6$, $σ(45)=78$.

2. Compute $τ(84)$ and $σ(84)$.

   Solution

   Compute $τ(84)$ and $σ(84)$

   • Step 1: Factorise
     • $84 = 2^2 \cdot 3^1 \cdot 7^1$
   • Step 2: Number of divisors
     • Exponents: $2,1,1$
     • $$τ(84) = (2+1)(1+1)(1+1) = 3 \cdot 2 \cdot 2 = 12$$
   • Step 3: Sum of divisors
     • $$σ(2^2) = 1 + 2 + 4 = 7$$
     • $$σ(3^1) = 1 + 3 = 4$$
     • $$σ(7^1) = 1 + 7 = 8$$
     • $$σ(84) = 7 \cdot 4 \cdot 8 = 224$$

   Answer: $τ(84)=12$, $σ(84)=224$.

3. Compute $σ(2^5)$ and $σ(3^3)$ using the geometric‑series formula.

   Solution

   Compute $σ(2^5)$ and $σ(3^3)$

   Use the formula $$σ(p^k) = 1 + p + p^2 + \dots + p^k = \frac{p^{k+1}-1}{p-1}.$$

   • For $2^5$
     • $$σ(2^5) = \frac{2^{6}-1}{2-1} = \frac{64-1}{1} = 63$$
   • For $3^3$
     • $$σ(3^3) = \frac{3^{4}-1}{3-1} = \frac{81-1}{2} = \frac{80}{2} = 40$$

   Answer: $σ(2^5)=63$, $σ(3^3)=40$.

4. Determine whether $n=20$ is perfect, abundant, or deficient.

   Solution

   Classify $n=20$ as perfect, abundant, or deficient

   • Step 1: Divisors of $20$
     • $1,2,4,5,10,20$
   • Step 2: Proper divisors (exclude $20$)
     • $1,2,4,5,10$
     • Sum: $1+2+4+5+10 = 22$
   • Step 3: Compare with $20$
     • $s(20) = 22 > 20$

   Answer: $20$ is abundant.

5. S Determine whether $n=28$ is perfect, abundant, or deficient.

   Solution

   Classify $n=28$ as perfect, abundant, or deficient

   • Step 1: Divisors of $28$
     • $1,2,4,7,14,28$
   • Step 2: Proper divisors
     • $1,2,4,7,14$
     • Sum: $1+2+4+7+14 = 28$
   • Step 3: Compare with $28$
     • $s(28) = 28$

   Answer: $28$ is perfect.

6. Explain why $τ(p^k)=k+1$ for any prime $p$.

   Solution

   Explain why $τ(p^k)=k+1$ for any prime $p$

   • The divisors of $p^k$ are exactly: $$1, p, p^2, \dots, p^k.$$
   • Each divisor is $p^e$ for some exponent $e$.
   • Possible exponents: $e = 0,1,2,\dots,k$.
   • That is a total of $k+1$ choices.

   Conclusion: $τ(p^k)=k+1$.

7. Explain why $σ(p^k)=\frac{p^{k+1}-1}{p-1}$.

   Solution

   Explain why $σ(p^k)=\dfrac{p^{k+1}-1}{p-1}$

   • The sum of divisors of $p^k$ is: $$σ(p^k) = 1 + p + p^2 + \dots + p^k.$$
   • This is a geometric series with:
     • first term $1$,
     • ratio $p$,
     • number of terms $k+1$.
   • The sum of a geometric series is: $$1 + r + r^2 + \dots + r^k = \frac{r^{k+1}-1}{r-1}.$$
   • Here $r=p$, so: $$σ(p^k) = \frac{p^{k+1}-1}{p-1}.$$
8. Why do numbers with many small prime factors tend to have many divisors?

   Solution

   Why do numbers with many small prime factors tend to have many divisors?

   • For $n = p_1^{a_1} \cdots p_k^{a_k}$, we have: $$τ(n) = (a_1+1)(a_2+1)\cdots(a_k+1).$$
   • To make $τ(n)$ large, we want:
     • many primes (large $k$),
     • reasonably large exponents $a_i$.
   • Using small primes (like $2,3,5$) lets us:
     • multiply several prime powers together
     • without $n$ becoming too large.
   • This allows many combinations of exponents, hence many divisors.

   Intuition: packing $n$ with small primes and their powers creates many ways to form divisors.

9. Give an example of two numbers $a$ and $b$ where $τ(ab)\neq τ(a)τ(b)$ and explain why.

   Solution

   Example where $τ(ab)\neq τ(a)τ(b)$

   We need $a$ and $b$ not coprime.

   • Take $a=4$ and $b=6$.
     • $4 = 2^2$, so $τ(4) = 2+1 = 3$.
     • $6 = 2 \cdot 3$, so $τ(6) = (1+1)(1+1) = 4$.
     • So $$τ(a)τ(b) = 3 \cdot 4 = 12$$
   • Now look at $ab=24$:
     • $24 = 2^3 \cdot 3^1$.
     • $τ(24) = (3+1)(1+1) = 4 \cdot 2 = 8.$

   So: $$τ(4 \cdot 6) = τ(24) = 8 \neq 12 = τ(4)τ(6).$$ Reason: $4$ and $6$ share the prime factor $2$, so they are not coprime, and multiplicativity does not apply.

10. Find a number between $1$ and $100$ that has the largest value of $τ(n)$.

    Solution

    Find a number between $1$ and $100$ with the largest $τ(n)$

    A full proof would require checking all $n$, but we can reason and then (in practice) verify by table.

    • Highly divisor‑rich numbers under $100$ include:
      • $60 = 2^2 \cdot 3 \cdot 5$
      • $72 = 2^3 \cdot 3^2$
      • $84 = 2^2 \cdot 3 \cdot 7$
      • $90 = 2 \cdot 3^2 \cdot 5$

    Compute $τ(n)$ for these:

    • $60 = 2^2 \cdot 3^1 \cdot 5^1$
      • $$τ(60) = (2+1)(1+1)(1+1) = 3 \cdot 2 \cdot 2 = 12$$
    • $72 = 2^3 \cdot 3^2$
      • $$τ(72) = (3+1)(2+1) = 4 \cdot 3 = 12$$
    • $84 = 2^2 \cdot 3^1 \cdot 7^1$
      • $$τ(84) = (2+1)(1+1)(1+1) = 3 \cdot 2 \cdot 2 = 12$$
    • $90 = 2^1 \cdot 3^2 \cdot 5^1$
      • $$τ(90) = (1+1)(2+1)(1+1) = 2 \cdot 3 \cdot 2 = 12$$

    No number $\le 100$ has more than $12$ divisors; several tie for the maximum.

    Answer: One valid choice is $60$ (others include $72,84,90$), each with $τ(n)=12$.