cp's OEIS Frontend

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

%I A234039 #25 Jun 13 2024 07:44:00
%S A234039 1,1,19,1657009,9950006745799417075771,
%T A234039 19389268200585836264288587113776883575610248525384021488302948711030121
%N A234039 a(n) = (2^(3^(n-1)) + 1)/3^n, n >= 1.
%C A234039 Essentially the same as A070632. - _R. J. Mathar_, Jun 13 2024
%C A234039 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, ...
%C A234039 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.
%C A234039 Euler's theorem shows, in particular, that (4^(3^(n-1)) - 1)/3^n is a natural number (see A152007).
%C A234039 From _Jianing Song_, Dec 27 2022: (Start)
%C A234039 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:
%C A234039 (a) 19 is the only prime in this sequence;
%C A234039 (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)
%H A234039 Vincenzo Librandi, <a href="/A234039/b234039.txt">Table of n, a(n) for n = 1..8</a>
%F A234039 a(n) = (2^(3^(n-1)) + 1)/3^n, n >= 1.
%F A234039 a(n) = (2^(phi(3^n)/2) + 1)/3^n, n >= 1, with Euler's phi(k)= A000010(k).
%t A234039  Table[(2^(3^(n - 1)) + 1)/3^n, {n, 1, 10}] (* _Vincenzo Librandi_, Feb 23 2014 *)
%o A234039 (Magma) [(2^(3^(n-1)) + 1)/3^n: n in [1..8]]; // _Vincenzo Librandi_, Feb 23 2014
%Y A234039 Cf. A152007.
%K A234039 nonn
%O A234039 1,3
%A A234039 _Wolfdieter Lang_, Feb 21 2014