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Let $n>36$ be some colossally abundant number. Then $n$ must have a prime factorisation of the form $2^{a_1}3^{a_2}\cdots {p_i}^{a_i}$, where $a_{1} \geq a _{2} \geq \cdots \geq a_i \geq 1$. What is the best known upper bound for $a_1$ in terms of $n$ ?

Naively, one finds that $a_1 < \log_{2}n$ since $2^{a_1} < n$, but surely there should be a far much better bound than this ?

Let $n>36$ be some colossally abundant number. Then $n$ must have a prime factorisation of the form $2^{a_1}3^{a_2}\cdots {p_i}^{a_i}$, where $a_{1} \geq a _{2} \geq \cdots \geq a_i \geq 1$. What is the best known upper bound for $a_1$ ?

Naively, one finds that $a_1 < \log_{2}n$ since $2^{a_1} < n$, but surely there should be a far much better bound than this ?

Let $n>36$ be some colossally abundant number. Then $n$ must have a prime factorisation of the form $2^{a_1}3^{a_2}\cdots {p_i}^{a_i}$, where $a_{1} \geq a _{2} \geq \cdots \geq a_i \geq 1$. What is the best known upper bound for $a_1$ in terms of $n$ ?

Naively, one finds that $a_1 < \log_{2}n$ since $2^{a_1} < n$, but surely there should be a far much better bound than this ?

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user160539
user160539

On the maximal power of $2$ that divides a colossally abundant number

Let $n>36$ be some colossally abundant number. Then $n$ must have a prime factorisation of the form $2^{a_1}3^{a_2}\cdots {p_i}^{a_i}$, where $a_{1} \geq a _{2} \geq \cdots \geq a_i \geq 1$. What is the best known upper bound for $a_1$ ?

Naively, one finds that $a_1 < \log_{2}n$ since $2^{a_1} < n$, but surely there should be a far much better bound than this ?