A234039 (2^(3^(n-1)) + 1)/3^n, n >= 1.

1, 1, 19, 1657009, 9950006745799417075771, 19389268200585836264288587113776883575610248525384021488302948711030121

Offset

1,3

Comments

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)

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..8

Formula

a(n) = (2^(3^(n-1)) + 1)/3^n, n >= 1.

a(n) = (2^(phi(3^n)/2) + 1)/3^n, n >= 1, with Euler's phi(k)= A000010(k).

Mathematica

 Table[(2^(3^(n - 1)) + 1)/3^n, {n, 1, 10}] (* Vincenzo Librandi, Feb 23 2014 *)

Prog

(Magma) [(2^(3^(n-1)) + 1)/3^n: n in [1..8]]; // Vincenzo Librandi, Feb 23 2014

Crossrefs

Cf. A152007.

Sequence in context: A269446 A172824 A070632 * A099114 A013810 A172875

Adjacent sequences: A234036 A234037 A234038 * A234040 A234041 A234042

Keyword

nonn

Author

Wolfdieter Lang, Feb 21 2014

Status

approved