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A234039 (2^(3^(n-1)) + 1)/3^n, n >= 1. 2
1, 1, 19, 1657009, 9950006745799417075771, 19389268200585836264288587113776883575610248525384021488302948711030121 (list; graph; refs; listen; history; text; internal format)
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).

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

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

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Last modified May 12 17:10 EDT 2021. Contains 343825 sequences. (Running on oeis4.)