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%I #35 Dec 17 2023 08:07:15
%S 0,11,143,1727,20735,248831,2985983,35831807,429981695,5159780351,
%T 61917364223,743008370687,8916100448255,106993205379071,
%U 1283918464548863,15407021574586367,184884258895036415
%N a(n) = 12^n - 1.
%C In base 12 these are 0, B, BB, BBB, ... . - _David Rabahy_, Dec 12 2016
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (13,-12).
%F From _Mohammad K. Azarian_, Jan 14 2009: (Start)
%F G.f.: 1/(1-12*x) - 1/(1-x).
%F E.g.f.: exp(12*x) - exp(x). (End)
%F a(n) = 12*a(n-1) + 11 for n>0, a(0)=0. - _Vincenzo Librandi_, Nov 18 2010
%F a(n) = Sum_{i=1..n} 11^i*binomial(n,n-i) for n>0, a(0)=0. - _Bruno Berselli_, Nov 11 2015
%F From _Elmo R. Oliveira_, Dec 15 2023: (Start)
%F a(n) = 13*a(n-1) - 12*a(n-2) for n>1.
%F a(n) = A001021(n)-1 = A178248(n)-2.
%F a(n) = 11*(A016125(n) - 1)/12. (End)
%t 12^Range[0,20]-1 (* or *) LinearRecurrence[{13,-12},{0,11},20] (* _Harvey P. Dale_, Feb 01 2019 *)
%Y Cf. A001021, A016125, A178248.
%Y Cf. Similar sequences of the type k^n-1: A000004 (k=1), A000225 (k=2), A024023 (k=3), A024036 (k=4), A024049 (k=5), A024062 (k=6), A024075 (k=7), A024088 (k=8), A024101 (k=9), A002283 (k=10), A024127 (k=11), this sequence (k=12).
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_