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For an integer $n \geq 3$, we define the sequence $\alpha_1,\alpha_2,\ldots,\alpha_k$ as the sequence of exponents in the prime factorization of $n! = p_1^{\alpha_1}p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, where $p_1 < p_2 < \cdots < p_k$ are primes. Determine all integers $n \geq 3$ for which $\alpha_1,\alpha_2,\ldots,\alpha_k$ is a geometric progression.

For $n \leq 10$ I found the following whose exponents form a geometric progression: \begin{align*}3! &= 2 \cdot 3\\4! &= 2^3 \cdot 5\\6! &= 2^4 \cdot 3^2 \cdot 5\\10! &= 2^8 \cdot 3^4 \cdot 5^2 \cdot 7.\end{align*} We have $11! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7 \cdot 11$ and for $n \geq 11$ I found that $n!$ seems to have at least two prime factors which are single powers in the prime decomposition. This means the exponents can't form a geometric progression. How can we prove this in general?

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    $\begingroup$ when does the exponent on each prime change ? $\endgroup$ – user451844 Aug 26 '17 at 18:04
  • $\begingroup$ the answer to my previous question was when the n in n! is a multiple of the prime. $\endgroup$ – user451844 Aug 26 '17 at 18:57
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From the book "ELEMENTARY THEORY OF NUMBERS" ;
written by W. SIERPINSKI ,
Edited by A. SCH1NZEL ,
Chapter III. Prime Numbers
section 10. Proof of Bertrand's Postulate (Theorem of Tchebycheff)
page 144

Theorem7. If $n$ is a natural number $> 5$, then between $n$ and $2n$ there are at least two different prime numbers.
i.e. there are primes $p_1 \neq p_2$ such that $n < p_1 , p_2 \color{Blue}{<} 2n$.


corollary: If $n$ is a natural number $> 11$, then between $\dfrac{n}{2}$ and $n$ there are at least two different prime numbers.
i.e. there are primes $p_1 \neq p_2$ such that $\dfrac{n}{2} < p_1, p_2 \color{Red}{\leq} n$.


Remark: Let $p$ be a prime number such that $\dfrac{n}{2} < p \color{Red}{\leq} n$;
then $p$ will appears by single power in the prime decomposition of $n!$.

Proof: Only notice that for every integer $2 \leq k$ we have:
$$n < 2p < kp;$$

so every positive multiple of $p$; except $p$; is stricly greater than $n$.





By the above corollary we can conclude that for any integer $11 \leq n$;
there are at least two different prime numbers $p_1, p_2$ such that $\dfrac{n}{2} < p_1 , p_2 \color{Red}{\leq} n$.
Notice that by the above remark both $p_1$ and $p_2$ will appear
in the prime decomposition of $n!$ by single power.

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