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%I #14 Mar 04 2024 01:13:57
%S 1,1011,1021111,1031322111,1041635332111,1052051685432111,
%T 1062572202286432111,1073197924309296432111,1083929903552389396432111,
%U 1094769202587913290396432111,1105716894613792423300396432111,1116774063559930347533400396432111
%N Partial sums of sum_{n=0...infinity} 1010^n.
%C For n = infinite: a(n)/1010^n = 1/1009 (Sequence A000930 reversed in decimal expansion).
%H G. C. Greubel, <a href="/A178349/b178349.txt">Table of n, a(n) for n = 1..330</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1011,-1010).
%F a(n) = a(n-1)* 1010 + 1
%F G.f.: x / ( (1010*x-1)*(x-1) ). - _R. J. Mathar_, Nov 03 2016
%e As overlapping Pascal triangles:
%e ......1
%e ....1.0.1.1
%e ...1.0.2.1.1.1.1
%e ..1.0.3.1.3.2.2.1.1.1
%e .1.0.4.1.6.3.5.3.3.2.1.1.1
%e 1.0.5.2.0.5.1.6.8.5.4.3.2.1.1.1
%t Accumulate[1010^Range[0,20]] (* _Harvey P. Dale_, Oct 02 2014 *)
%o (PARI) for(n=1,25, print1((1010^n-1)/1009, ", ")) \\ _G. C. Greubel_, May 02 2017
%Y Cf. A000930, A007318.
%K nonn,easy
%O 1,2
%A _Mark Dols_, May 25 2010
%E Corrected and extended by _Harvey P. Dale_, Oct 02 2014
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