The problem asks whether ( S(k) = T(k) ), where:

• ( \beta(n,k) = \max_{d \leq k}(d \mid n) ) (the largest divisor of ( n ) that is at most ( k )),
• ( S(k) = \sum_{n=1}^{k!} \beta(n,k) ),
• ( T(k) = # { i \cdot j \mid 1 \leq i \leq k,\ 1 \leq j \leq k! } ) (the number of distinct products ( i \cdot j ) with ( i \leq k ) and ( j \leq k! )).

Key Analysis:

1. **Understandi...

Yes. In fact there is a clean bijective proof that S(k) = T(k), and it works more generally whenever you replace k! by any positive integer M that is divisible by every integer 1, 2, …, k (in particular, M = k! or M = lcm(1,2,…,k) = exp(ψ(k))).

Set-up

• β(n,k) = max{d ≤ k : d | n}.
• S(k) = ∑_{n=1}^{k!} β(n,k).
• T(k) = #{i·j : 1 ≤ i ≤ k, 1 ≤ j ≤ k!} (distinct values).

A bijection Consider the s...

Yes, the two quantities are always equal.

Notation
• k is a fixed positive integer, k ≥ 1.
• n always satisfies 1 ≤ n ≤ k!.
• For an integer m set
β(m,k) = max{ d ≤ k : d | m }.
• Put
S(k)=∑_{n=1}^{k!}β(n,k).
• Let
U(k)= { i·j : 1 ≤ i ≤ k, 1 ≤ j ≤ k! } ⊂ ℕ,
T(k)=|U(k)|.

We shall construct a natural bijection between a set with S(k) elements and U(k); this ...

<think> I need to understand this conjecture about the Sieve of Eratosthenes. The question defines a few things: β(n,k) is the largest divisor of n that is less than or equal to k. Then S(k) is the sum of β(n,k) for n from 1 to k!. T(k) is defined as the number of pairs (i,j) such that i is from 1 to k and j is from 1 to k!, and i divides j? The notation is a bit messy. It says:

T(k) = # { i·j | ...

Answering the question: Does ?

Yes, the identity is true. This is a known result in number theory, sometimes called the "Sieve of Eratosthenes identity", which has been discovered multiple times. Let's break down the proof.

1. Definitions

First, let's restate the definitions clearly. Let .

• $\beta(n, k) = \max{d \in \mathbb{Z}^+ \mid d \le k \text{ and...

I'll analyze this conjecture about the identity between and .

First, let me clarify what each function represents:

Understanding :

• is the largest divisor of that doesn't exceed
• sums these maximum divisors for all from 1 to

Understanding :

• $T(k) = #{i \cdot j : 1 \leq...

Proving the Conjecture on Sieve of Eratosthenes Collisions

This question asks us to prove or disprove the conjecture that for all positive integers , where:

Let me first clarify what these expressions mean:

• is the largest ...

Yes – for every positive integer k the two quantities coincide.

We first re-write the definition of T(k) a little more systematically and then show that the set that is being counted in T(k) can be split up into β(n,k) distinct elements for every n ≤ k!.
Because S(k) is the sum of these β(n,k)’s, the two numbers are equal.

---

1....

First, let us restate the two definitions in “set‐theoretic’’ form and then show by the same inclusion–exclusion expansion that both give exactly the same number.

1) The “union‐of‐multiples’’ form T(k).

For each i=1,2,…,k define the set
  A_i={ i·1, i·2, …, i·(k!) }⊂ℕ.
Then
  T(k)=∣A_1∪A_2∪⋯∪A_k∣.
By ordinary inclusion–exclusion,

  T(k)
  =∑{r=1}^k (−1)^{,r−1}
    ∑{1≤i_1<...