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%I #39 Mar 04 2024 01:13:54
%S 1,1101,1211101,1332211101,1465432211101,1611975432211101,
%T 1773172975432211101,1950490272975432211101,2145539300272975432211101,
%U 2360093230300272975432211101,2596102553330300272975432211101,2855712808663330300272975432211101
%N a(n) = Sum_{k=0..n} 1100^k.
%H Colin Barker, <a href="/A178348/b178348.txt">Table of n, a(n) for n = 0..328</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1101,-1100).
%F a(n) = 1100*a(n-1) + 1.
%F Lim_{n -> infinity} a(n)/1100^n = 1100/1099.
%F From _Colin Barker_, Oct 02 2015: (Start)
%F a(n) = 1101*a(n-1) - 1100*a(n-2) for n>=2.
%F G.f.: 1 / ((x-1)*(1100*x-1)). (End)
%e As overlapping Pascal triangles:
%e .....1
%e ....1.1.0.1
%e ...1.2.1.1.1.0.1
%e ..1.3.3.2.2.1.1.1.0.1
%e .1.4.6.5.4.3.2.2.1.1.1.0.1
%t Table[Sum[1100^k, {k, 0, n}], {n, 0, 11}] (* _Michael De Vlieger_, Oct 02 2015 *)
%o (PARI) Vec(1/((x-1)*(1100*x-1)) + O(x^25)) \\ _Colin Barker_, Oct 02 2015
%o (PARI) vector(100, n, n--; sum(k=0, n, 1100^k)) \\ _Altug Alkan_, Oct 06 2015
%o (Magma) [(1/1099)*(1100^n-1): n in [0..20]]; // _Vincenzo Librandi_, Oct 07 2015
%Y Cf. A000931, A007318.
%K nonn,easy
%O 0,2
%A _Mark Dols_, May 25 2010
%E Offset corrected by _Joerg Arndt_, Oct 03 2015
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