|
%I #25 Mar 10 2017 03:00:22
%S 3,21,148,1038,7269,50883,356182,2493276,17452935,122170545,855193816,
%T 5986356714,41904497001,293331479007,2053320353050,14373242471352,
%U 100612697299467,704288881096269,4930022167673884
%N Base-7 digits are, in order, the first n terms of the periodic sequence with initial period 3,0,1,2.
%H Vincenzo Librandi, <a href="/A037761/b037761.txt">Table of n, a(n) for n = 1..200</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (7,0,0,1,-7).
%F From _Colin Barker_, Apr 30 2014: (Start)
%F a(n) = (-100-25*(-1)^n-(24-i*32)*(-i)^n-(24+32*i)*i^n+173*7^n)/400, where i=sqrt(-1).
%F a(n) = 7*a(n-1)+a(n-4)-7*a(n-5).
%F G.f.: x*(2*x^3+x^2+3) / ((x-1)*(x+1)*(7*x-1)*(x^2+1)). (End)
%t CoefficientList[Series[(2 x^3 + x^2 + 3)/((x - 1) (x + 1) (7 x - 1) (x^2 + 1)), {x, 0, 30}], x] (* _Vincenzo Librandi_, May 01 2014 *)
%o (PARI) Vec(x*(2*x^3+x^2+3)/((x-1)*(x+1)*(7*x-1)*(x^2+1)) + O(x^100)) \\ _Colin Barker_, Apr 30 2014
%K nonn,base,easy
%O 1,1
%A _Clark Kimberling_
|
