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Base-7 digits are, in order, the first n terms of the periodic sequence with initial period 1,3,0,2.
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%I #16 Dec 24 2015 13:41:01

%S 1,10,70,492,3445,24118,168826,1181784,8272489,57907426,405351982,

%T 2837463876,19862247133,139035729934,973250109538,6812750766768,

%U 47689255367377,333824787571642,2336773513001494,16357414591010460,114501902137073221,801513314959512550

%N Base-7 digits are, in order, the first n terms of the periodic sequence with initial period 1,3,0,2.

%H Colin Barker, <a href="/A037705/b037705.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (7,0,0,1,-7).

%F a(n) = 7*a(n-1) + a(n-4) - 7*a(n-5).

%F From _Colin Barker_, Dec 24 2015: (Start)

%F a(n) = 1/200*(25*(-1)^n-(8+6*i)*(-i)^n-(8-6*i)*i^n+41*7^n-50) where i=sqrt(-1).

%F G.f.: x*(1+3*x+2*x^3) / ((1-x)*(1+x)*(1-7*x)*(1+x^2)).

%F (End)

%o (PARI) Vec(x*(1+3*x+2*x^3)/((1-x)*(1+x)*(1-7*x)*(1+x^2)) + O(x^30)) \\ _Colin Barker_, Dec 24 2015

%K nonn,base,easy

%O 1,2

%A _Clark Kimberling_