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 A178407 a(n+1) = a(n)*1000 + 101^n with a(0)= 0. 1

%I

%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 a(n) = (1/899)*(1000^n-101^n), with n>=1. - _Paolo P. Lava_, Jun 10 2010

%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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Last modified December 14 22:42 EST 2019. Contains 329987 sequences. (Running on oeis4.)