cp's OEIS Frontend

A152007 a(n) = (2^phi(3^n)-1)/3^n.

%I A152007 #29 Sep 08 2022 08:45:39
%S A152007 1,1,7,9709,222399981598543,
%T A152007 24057640120673299065081231814259802792690247621
%N A152007 a(n) = (2^phi(3^n)-1)/3^n.
%C A152007 The next term is too large to display.
%C A152007 With the exception of 7 there are no primes in this sequence.
%C A152007 All numbers in this sequence are squarefree.
%C A152007 a(n) is divisible by a(k) for every k < n.
%C A152007 The sequence of number of digits of a(n), n >= 1, is 1, 1, 1, 4, 15, 47, 144, 436, 1313, 3946, 11846, 35546, 106648, 319954, 959872, 2879628, 8638896, 25916701, 77750117, 233250368, 699751120,... - _Wolfdieter Lang_, Feb 21 2014
%C A152007 Each a(n) is by definition the same as the repetend of 1/3^n, viewed as a binary integer. E.g., 1/9 = .000111000111...; consequently a(2) = 000111 (base 2) = 7 (base 10) - _Joe Slater_, Nov 29 2016
%H A152007 Vincenzo Librandi, <a href="/A152007/b152007.txt">Table of n, a(n) for n = 0..7</a>
%H A152007 W. Lang, <a href="http://arxiv.org/abs/1404.2710">On Collatz' Words, Sequences and Trees</a>, arXiv preprint arXiv:1404.2710 [math.NT], 2014  and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Lang/lang6.html">J. Int. Seq. 17 (2014) # 14.11.7</a>.
%F A152007 a(n) = (4^(3^(n-1)) - 1)/3^n for n>=1, a(0) = 1, with EulerPhi(1) = 1 = A000010(1). - _Wolfdieter Lang_, Feb 21 2014
%t A152007 Table[(2^EulerPhi[3^n] - 1)/3^n, {n, 0, 10}]
%o A152007 (Magma) [(2^EulerPhi(3^n)-1)/3^n: n in [0..6]]; // _Vincenzo Librandi_, Feb 23 2014
%o A152007 (PARI) a(n)=(2^eulerphi(3^n)-1)/3^n \\ _Charles R Greathouse IV_, Nov 29 2016
%Y A152007 Cf. A008776, A152008, A234039.
%K A152007 nonn
%O A152007 0,3
%A A152007 _Artur Jasinski_, Nov 19 2008
%E A152007 Edited by _N. J. A. Sloane_, Nov 28 2008
%E A152007 Offset corrected from _Wolfdieter Lang_, Feb 21 2014
%E A152007 Definition clarified by _Joerg Arndt_, Feb 23 2014