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Partial sums of 80^n.
1

%I #34 Apr 08 2024 17:48:05

%S 1,81,6481,518481,41478481,3318278481,265462278481,21236982278481,

%T 1698958582278481,135916686582278481,10873334926582278481,

%U 869866794126582278481,69589343530126582278481,5567147482410126582278481,445371798592810126582278481

%N Partial sums of 80^n.

%C Related to backward decimal expansion of fraction 1/79 and Pell numbers. [In which way? - _Joerg Arndt_, May 17 2011]

%H Vincenzo Librandi, <a href="/A178513/b178513.txt">Table of n, a(n) for n = 0..150</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (81,-80).

%F a(n) = 80*a(n-1) + 1.

%F a(n) = (80^(n+1)-1)/79.

%F G.f.: 1/((1-80*x)*(1-x)).

%t Accumulate[80^Range[0,20]] (* _Harvey P. Dale_, Aug 11 2014 *)

%Y Cf. A000129, A021083, A038207, A178510, A178511, A178512.

%K nonn,easy

%O 0,2

%A _Mark Dols_, May 29 2010