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%I #23 Mar 15 2024 02:23:31
%S 0,1,111,11221,1123431,112357741,11235935151,1123595286661,
%T 112359548153271,11235955029685981,1123595505326545791,
%U 112359550558592003701,11235955056144512040711,1123595505617589632447821,112359550561793485956926031,11235955056179728345526186341
%N a(n) = 100*a(n-1) + 11^(n-1) for n>0, a(0)=0.
%C Generalization of A000225. - _Mark Dols_, Jan 28 2010
%H G. C. Greubel, <a href="/A165155/b165155.txt">Table of n, a(n) for n = 0..495</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1100).
%F G.f.: x/((1-11*x)*(1-100*x)). - _R. J. Mathar_, Nov 02 2016
%F E.g.f.: (1/89)*(exp(100*x) - exp(11*x)). - _G. C. Greubel_, Feb 09 2023
%e From _Mark Dols_, Jan 28 2010: (Start)
%e As triangle:
%e ........... 1
%e ......... 1 1 1
%e ....... 1 1 2 2 1
%e ..... 1 1 2 3 4 3 1
%e ... 1 1 2 3 5 7 7 4 1
%e . 1 1 2 3 5 9 3 5 1 5 1
%e 1 1 2 3 5 9 5 2 8 6 6 6 1
%e (Mirrored version of A162741) (End)
%t RecurrenceTable[{a[0]==0,a[n]==100a[n-1]+11^(n-1)},a,{n,40}] (* _Harvey P. Dale_, Feb 20 2016 *)
%o (Magma) [(1/89)*(100^n-11^n): n in [0..40]] // _Vincenzo Librandi_, Dec 05 2010
%o (SageMath) [(10^(2*n) - 11^n)/89 for n in range(41)] # _G. C. Greubel_, Feb 09 2023
%Y Cf. A000225, A021093, A094704, A162741, A164913.
%K nonn,easy
%O 0,3
%A _Mark Dols_, Sep 05 2009
%E a(0) prepended by _Bruno Berselli_, Oct 02 2015
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