Solution

It suffices first to prove the elementary primorial bound for positive integers

n

, by strong induction. The case

n = 1

is immediate.

n = 2 k

, every prime

p

with

k < p \leq 2 k

divides

(k 2 k)

: the numerator

(2 k)!

contains

p

, whereas the two copies of

k!

do not. Since these primes are distinct,

P (2 k) \leq P (k) (k 2 k) .

The induction hypothesis and the binomial theorem give

P (2 k) \leq 4^{k} (k 2 k) \leq 4^{k} 2^{2 k} = 4^{2 k} .

n = 2 k + 1

, part (b) shows that every prime

k + 2 \leq p \leq 2 k + 1

divides

(k + 1 2 k + 1)

. Hence

P (2 k + 1) \leq P (k + 1) (k + 1 2 k + 1) .

The two central coefficients of

(1 + 1)^{2 k + 1}

are equal, so

2 (k + 1 2 k + 1) \leq 2^{2 k + 1},

and therefore

P (2 k + 1) \leq 4^{k + 1} 4^{k} = 4^{2 k + 1} .

This completes the induction. For real

X \geq 1

P (X) = P (⌊ X ⌋) \leq 4^{⌊ X ⌋} \leq 4^{X} .

Solved by gpt-5.6-sol high.

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