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%I #36 Sep 08 2022 08:46:04
%S 0,1,22,463,9724,204205,4288306,90054427,1891142968,39714002329,
%T 833994048910,17513875027111,367791375569332,7723618886955973,
%U 162195996626075434,3406115929147584115,71528434512099266416
%N a(n) = (21^n - 1)/20.
%C Partial sums of powers of 21 (A009965); q-integers for q=21: diagonal k=1 in triangle A022185.
%C Partial sums are in A014905. Also, the sequence is related to A014938 by A014938(n) = n*a(n)- sum_{i=0..n-1}a(i) for n>0. - _Bruno Berselli_, Nov 06 2012
%C For n >= 1, 4*a(n) is the total number of holes in a certain box fractal (start with 21 boxes, 4 holes) after n iterations. See illustration in links. -_Kival Ngaokrajang_, Jan 27 2015
%H Vincenzo Librandi, <a href="/A218724/b218724.txt">Table of n, a(n) for n = 0..700</a>
%H Kival Ngaokrajang, <a href="/A218724/a218724.pdf">Illustration of initial terms</a>
%H <a href="/index/Par#partial">Index entries related to partial sums</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (22,-21).
%F a(n) = floor(21^n/20).
%F G.f.: x/((1-x)*(1-21*x)). - _Bruno Berselli_, Nov 06 2012
%F a(n) = 22*a(n-1) - 21*a(n-2). - _Vincenzo Librandi_, Nov 07 2012
%F a(n) = 21*a(n-1) + 1. - _Kival Ngaokrajang_, Jan 27 2015
%F a(n) = a(n-1) + 21^(n-1), n >= 1, a(0) = 0. - _Wolfdieter Lang_, Feb 02 2015
%t LinearRecurrence[{22, -21}, {0, 1}, 40] (* _Vincenzo Librandi_, Nov 07 2012 *)
%o (PARI) A218724(n)=21^n\20
%o (Maxima) A218724(n):=(21^n-1)/20$ makelist(A218724(n),n,0,30); /* _Martin Ettl_, Nov 05 2012 */
%o (Magma) [n le 2 select n-1 else 22*Self(n-1) - 21*Self(n-2): n in [1..20]]; // _Vincenzo Librandi_, Nov 07 2012
%Y Cf. similar sequences of the form (k^n-1)/(k-1): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218725-A218734, A132469, A218736-A218753, A133853, A094028, A218723.
%K nonn,easy
%O 0,3
%A _M. F. Hasler_, Nov 04 2012
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