Is $1+\left (\frac{m}{m-1}\right )^{\frac{\log(p-1)}{\log(2)}}\cdot(p-1)^{\frac{\log(m)}{\log(2)}+1}<\frac{m}{m-1}p^{\frac{\log(m)}{\log(2)}+1}$?

Question

For $p>2,m>2$, is $$1+\left (\frac{m}{m-1}\right )^{\frac{\log(p-1)}{\log(2)}}\cdot(p-1)^{\frac{\log(m)}{\log(2)}+1}<\frac{m}{m-1}p^{\frac{\log(m)}{\log(2)}+1}$$ ?

Motivation:

I am trying to prove that:

$$f_n(m) < \left( \frac{m}{m-1}\right)^{\Omega(n)} \cdot n^{\frac{\log(m)}{\log(2)}+1}$$

where $f_n(x)$ is defined as: Define inductively for prime numbers: $f_1(x) := 1$, $f_2(x):=x$, $f_p(x) := 1+\prod_{q\mid p-1} f_q(x)^{v_q(p-1)}$.

Let $f_n(x):= \prod_{p\mid n} f_p(x)^{v_p(n)}$ for each natural number $n$.

I want to prove this so that I can give a criterion when the Dirichlet series :

$$\zeta(s,m) := \sum_{n=1}^{\infty} \frac{f_n(m)}{n^s} =^{\text{ Euler product}} \prod_{p \text{prime}} \frac{1}{1-\frac{f_p(m)}{p^s}}$$

converges for $s>2$ and $ m \ge 2$.

I have reduced the inequality to the case where $n=p$ is a prime number $p>2$ and $m>2$.

Notice that for $m \ge 2$ we have $\zeta(s,m) \ge \zeta(s-1)$ , hence for $s=2$ the following product and sums diverge:

$$\sum_{n=1}^{\infty} \frac{f_n(m)}{n^2}$$ and $$\prod_{p \text{ prime}} \frac{1}{1-\frac{f_p(m)}{p^2}}$$

My naive wish is that through this divergence study one can say something for example for $m=3$ and the prime numbers attained in the sequence $f_p(m)$ where $p$ runs through all primes. Here is a small list of those $p$ where $f_p(3)$ is a prime, to get an idea:

p,f_p(3)
2 3
7 13
11 31
13 37
41 271
43 157
47 283
67 373
73 433
109 577
139 1129
151 1201
163 769
191 1471
193 2917
199 1489
223 1741
227 3163
233 3187
251 3001
281 3511
311 3631
347 4243
373 4357
401 8101
439 5197
443 9103
461 8461
499 9769
571 5881
619 11821
641 21871
643 12037
661 11161
673 12637
701 11701
709 12781
719 22639
739 13009
757 7489
811 7681
821 24391
829 13537
881 25111
911 14431
941 25471
971 29191
1013 26227
1019 16879
1051 15601
1063 17041
1069 30169
1087 17293
1093 17317
1151 28201
1171 17761
1201 32401
1237 35461
1259 35671
1301 33301
1427 34123
1429 38377
1451 35401
1459 12289
1483 21757
1579 40009
1601 72901
1621 23041

Thanks for your help!

Answer 1

Even the weaker inequality $$\left (\frac{m}{m-1}\right )^{\frac{\log(p-1)}{\log(2)}}\cdot(p-1)^{\frac{\log(m)}{\log(2)}+1}<\frac{m}{m-1}p^{\frac{\log(m)}{\log(2)}+1}$$ taking logarithms writes $$ {\frac{\log(p-1)}{\log(2)}} \log\big(\frac m{m-1}\big)+ \Big( \frac{\log(m)}{\log(2)}+1 \Big)\log(p-1)<\\<\log\big(\frac{m}{m-1}\big) + \Big(\frac{\log(m)}{\log(2)}+1\Big)\log p$$ that is, separating the variables, $$ \frac{\log\big(\frac m{m-1}\big)}{\frac{\log(m)}{\log(2)}+1 }<\frac{\log\big( \frac p{p-1}\big)}{{\frac{\log(p-1)}{\log(2)}}-1},$$ which for every $m>2$ can’t be true for all $p>2$ since the LHS is positive and the RHS is $o(1)$ for $p\to\infty$ .

Answer 2

This is not true. Indeed, the difference (say $f(p,m)$) between the right- and left-hand sides of the conjectured inequality at $(p,m)=(6,2)$ is $-54<0$. More generally, $f(p,2)=-p (3 - 5 p + p^2)\to-\infty$ as $p\to\infty$.