cp's OEIS Frontend

Essentially the same as A070632. - R. J. Mathar, Jun 13 2024

The sequence of the number of digits of a(n) is 1, 1, 2, 7, 22, 71, 217, 655, 1971, 5921, 17771, 53321, 159974, 479933, 1439810, ...

The proof that a(n) = (2^(3^(n-1)) + 1)/3^n, n >= 1, is indeed a natural number uses 2 = 3 - 1 and the binomial theorem.

Euler's theorem shows, in particular, that (4^(3^(n-1)) - 1)/3^n is a natural number (see A152007).

From Jianing Song, Dec 27 2022: (Start)

Note that a(n)/a(n-1) = 1 + ((2^(3^(n-2)) - 2)/3) * a(n-1) * 3^(n-1) for n >= 2. As a result:

(a) 19 is the only prime in this sequence;

(b) a(m) == a(n) (mod 3^n) for all m >= n. This means that this sequence converges to ...210120102112201 in the ring of 3-adic integers. In particular, all terms are congruent to 1 modulo 9. But a(m) !== a(n) (mod 3^n) for all m < n unless m = 1 and n = 2, because 1 < m < n implies that a(m) !== a(m+1) == a(n) (mod 3^(m+1)). (End)