Prime numbers making constant : 1.2527

Question

Reading "Excursion in calculus" (Robert M. Young, 1992), exercice 13 on page 71 ask the reader to show there is a constant $c\approx 1.25$ such that

1. $a_0=2^c$
2. $a_{n+1}=2^{a_n}$
3. $\forall n\; \lfloor a_n\rfloor$ is prime

This can be proved using Bertrand's postulate.

I found that $c\approx 1.2527$ is a good candidate as

• $\lfloor a_0\rfloor=2$
• $\lfloor a_1\rfloor=5$
• $\lfloor a_2\rfloor=37$
• $\lfloor a_3\rfloor=153703792531$ that is prime too.

There must be several different constants that have this property, but we can suppose that $c$ is minimal.

Does that constant has a name ? Does someone computed or is trying to compute more decimal ?

Answer 1

This number is called Bertrand's number in OEIS; the decimal expansion is OEIS A$079614$ and is

$$1.251647597790463017594432053623346969\ldots\;.$$

The sequence of associated primes is OEIS A$051501$. There are references in both OEIS entries.