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A125682 a(n) = 3*(6^n - 1)/5.

%I A125682 #60 Mar 29 2025 09:15:39
%S A125682 3,21,129,777,4665,27993,167961,1007769,6046617,36279705,217678233,
%T A125682 1306069401,7836416409,47018498457,282110990745,1692665944473,
%U A125682 10155995666841,60935974001049,365615844006297,2193695064037785,13162170384226713,78973022305360281,473838133832161689
%N A125682 a(n) = 3*(6^n - 1)/5.
%C A125682 The base-6 numbers 3_6, 33_6, 333_6, 3333_6, 33333_6, 333333_6, ... converted to base 10.
%C A125682 Also the total number of holes in a certain triangle fractal (start with 6 triangles, 3 holes) after n iterations. See illustration in Ngaokrajang link. - _Jens Ahlström_, Aug 29 2023
%H A125682 Bruno Berselli, <a href="/A125682/b125682.txt">Table of n, a(n) for n = 1..1000</a>
%H A125682 Kival Ngaokrajang, <a href="/A003464/a003464.pdf">Illustration of initial terms</a>.
%H A125682 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7,-6).
%F A125682 G.f.: 3*x/((1-x)*(1-6*x)). - _Bruno Berselli_, Apr 18 2012
%F A125682 a(n) = 7*a(n-1) - 6*a(n-2). - _Wesley Ivan Hurt_, Dec 25 2021
%F A125682 From _Elmo R. Oliveira_, Mar 29 2025: (Start)
%F A125682 E.g.f.: 3*exp(x)*(exp(5*x) - 1)/5.
%F A125682 a(n) = 3*A003464(n). (End)
%e A125682 Base 6        Base 10
%e A125682 3 ............. 3 = 3*6^0
%e A125682 33 ........... 21 = 3*6^1 + 3*6^0
%e A125682 333 ......... 129 = 3*6^2 + 3*6^1 + 3*6^0
%e A125682 3333 ........ 777 = 3*6^3 + 3*6^2 + 3*6^1 + 3*6^0, etc.
%p A125682 seq((6^n-1)*3/5, n=1..27);
%t A125682 a[n_]:=(6^n-1)*3/5; Table[a[n],{n,1,22}] (* _Robert P. P. McKone_, Aug 29 2023 *)
%o A125682 (Magma) [(6^n-1)*3/5: n in [1..22]]; // _Bruno Berselli_, Apr 18 2012
%Y A125682 Cf. A000400, A003464, A005610, A024062, A105281, A125687, A146884.
%K A125682 nonn,easy
%O A125682 1,1
%A A125682 _Zerinvary Lajos_, Jan 31 2007
%E A125682 Edited by _N. J. A. Sloane_, Feb 02 2007
%E A125682 Definition rewritten (with Lajos formula) from _Bruno Berselli_, Apr 18 2012