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%I #16 Mar 04 2024 01:14:05
%S 0,1,1101,1111201,1112231301,1112335361401,1112345871501501,
%T 1112346933021651601,1112347040235186811701,1112347051063753867981801,
%U 1112347052157439140666161901,1112347052267901353207282352001,1112347052279058036673935517552101
%N a(n+1) = a(n)*1000 + 101^n with a(0) = 0.
%C Numerators of fractions in partial sums of sum_{n=0...infinity} 101^n/1000^(n+1)converging to fraction 1/899 or sum_{n=0...infinity} 100^n/999^(n+1) whose alternating sum converges to fraction 1/1099 [= alternating sum_99^n/1000^(n+1)].
%H Colin Barker, <a href="/A178407/b178407.txt">Table of n, a(n) for n = 0..334</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1101,-101000).
%F From _Colin Barker_, Oct 02 2015: (Start)
%F a(n) = 1101*a(n-1) - 101000*a(n-2) for n>2.
%F G.f.: x / ((101*x-1)*(1000*x-1)).
%F (End)
%e As interlocking Pascal triangles starting a new triangle to the left of each row:
%e ................1
%e ............1.1.0.1
%e ........1.1.1.1.2.0.1
%e ....1.1.1.2.2.3.1.3.0.1
%e 1.1.1.2.3.3.5.3.6.1.4.0.1
%t RecurrenceTable[{a[0]==0,a[n]==1000*a[n-1]+101^(n-1)},a,{n,15}] (* _Harvey P. Dale_, Nov 18 2013 *)
%o (PARI) concat(0, Vec(x/((101*x-1)*(1000*x-1)) + O(x^30))) \\ _Colin Barker_, Oct 02 2015
%Y Cf. A000930, A007318.
%K nonn,easy
%O 0,3
%A _Mark Dols_, May 27 2010
%E Prepended zero (from definition), changed offset accordingly, and more terms from _Harvey P. Dale_, Nov 18 2013
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