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a(1)=1, a(n) = a(n-1)*1000 + 11^(n-1) for n>=2.
1

%I #17 Mar 04 2024 01:14:02

%S 1,1011,1011121,1011122331,1011122345641,1011122345802051,

%T 1011122345803822561,1011122345803842048171,1011122345803842262529881,

%U 1011122345803842264887828691,1011122345803842264913766115601,1011122345803842264914051427271611

%N a(1)=1, a(n) = a(n-1)*1000 + 11^(n-1) for n>=2.

%C Numerators of fractions in partial sums of Sum_{n>=0} 11^n/1000^(n+1) converging to fraction 1/989 or Sum_{n>=0} 10^n/999^(n+1) whose alternating sum converges to fraction 1/1009 [= alternating sum_9^n/1000^(n+1)].

%H Colin Barker, <a href="/A178396/b178396.txt">Table of n, a(n) for n = 1..334</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1011,-11000).

%F From _Colin Barker_, Oct 02 2015: (Start)

%F a(n) = 1011*a(n-1) - 11000*a(n-2) for n>2.

%F G.f.: x / ((11*x-1)*(1000*x-1)).

%F (End)

%e As overlapping Pascal triangles starting a new triangle to the left of each row skipping one place:

%e ....................1

%e ...............1.0.1.1

%e ..........1.0.1.1.1.2.1

%e .....1.0.1.1.1.2.2.3.3.1

%e 1.0.1.1.1.2.2.3.4.5.6.4.1

%o (PARI) Vec(x/((11*x-1)*(1000*x-1)) + O(x^40)) \\ _Colin Barker_, Oct 02 2015

%Y Cf. A000931, A007318.

%K nonn,easy

%O 1,2

%A _Mark Dols_, May 27 2010

%E Definition adapted to offset by _Georg Fischer_, Jun 19 2021