%I #49 Jun 29 2023 11:28:28
%S 1,2,12,72,432,2592,15552,93312,559872,3359232,20155392,120932352,
%T 725594112,4353564672,26121388032,156728328192,940369969152,
%U 5642219814912,33853318889472,203119913336832,1218719480020992,7312316880125952,43873901280755712,263243407684534272
%N a(n) = phi(6^n).
%H Vincenzo Librandi, <a href="/A167747/b167747.txt">Table of n, a(n) for n = 0..1000</a>
%H D. Bevan, D. Levin, P. Nugent, J. Pantone, L. Pudwell, M. Riehl and M. L. Tlachac, <a href="http://arxiv.org/abs/1510.08036">Pattern avoidance in forests of binary shrubs</a>, arXiv preprint arXiv:1510:08036 [math.CO], 2015-2016.
%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (6).
%F a(n+1) = 2*6^n. - _Charles R Greathouse IV_, Nov 12 2009
%F G.f.: (1-4x)/(1-6x). - _Philippe Deléham_, Oct 10 2011
%F a(n) = ((8*n-4)*a(n-1)-12*(n-2)*a(n-2))/n, a(0)=1, a(1)=2. - _Sergei N. Gladkovskii_, Jul 19 2012
%F Sum_{n>=0} 1/a(n) = 8/5. - _Amiram Eldar_, Jan 02 2021
%F a(n) = A000010(A000400(n)). - _Michel Marcus_, Jan 02 2021
%F From _Amiram Eldar_, May 08 2023: (Start)
%F Sum_{n>=0} (-1)^n/a(n) = 4/7.
%F Product_{n>=1} (1 - 1/a(n)) = A132022. (End)
%t Table[EulerPhi[6^n],{n,0,40}]
%o (PARI) a(n) = eulerphi(6^n); \\ _Michel Marcus_, Jan 02 2021
%Y Cf. A000010, A000400, A132022.
%K nonn,easy
%O 0,2
%A _Vladimir Joseph Stephan Orlovsky_, Nov 10 2009
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