Is there a Bell number $B_n$ of the form $2^k$ for some $k>1$? If there is, are there infinitely many?
$\begingroup$
$\endgroup$
asked Aug 16, 2023 at 21:23
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
No. It's easy to see that $B_n=4$ is impossible, and $B_n$ is never divisible by 8. This follows from the fact that $B_n$ is periodic modulo 8 with period 24. See, e.g., W. F. Lunnon, P. A. B. Pleasants, and N. M. Stephens, Arithmetic properties of Bell numbers to a composite modulus I, Acta Arithmetica 35 (1979), 1–16, Theorem 6.4.
The Bell numbers $B_n \bmod 8$ for $n$ from 0 to 23 are 1, 1, 2, 5, 7, 4, 3, 5, 4, 3, 7, 2, 5, 5, 2, 1, 3, 4, 7, 1, 4, 7, 3, 2. So $B_n$ is never 0 or 6 modulo 8.
