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%I #22 Apr 06 2022 10:54:51
%S 1,10,83,665,5322,42579,340633,2725066,21800531,174404249,1395233994,
%T 11161871955,89294975641,714359805130,5714878441043,45719027528345,
%U 365752220226762,2926017761814099,23408142094512793,187265136756102346,1498121094048818771
%N Base 8 digits are, in order, the first n terms of the periodic sequence with initial period 1,2,3.
%H Colin Barker, <a href="/A037608/b037608.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (8,0,1,-8).
%F a(n) = 8*a(n-1)+a(n-3)-8*a(n-4). - _Colin Barker_, Nov 28 2014
%F G.f.: x*(3*x^2+2*x+1) / ((x-1)*(8*x-1)*(x^2+x+1)). - _Colin Barker_, Nov 28 2014
%e 1, 10, 83, 665, 5322, ... in base 8 are 1, 12, 123, 1231, 12312, ...
%t Table[FromDigits[PadRight[{},n,{1,2,3}],8],{n,30}] (* _Harvey P. Dale_, Apr 06 2022 *)
%o (PARI) Vec(x*(3*x^2+2*x+1)/((x-1)*(8*x-1)*(x^2+x+1)) + O(x^100)) \\ _Colin Barker_, Nov 28 2014
%Y Cf. A037610.
%K nonn,base,easy
%O 1,2
%A _Clark Kimberling_
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