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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 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 December 7 12:12 EST 2022. Contains 358656 sequences. (Running on oeis4.)