%I #16 Jun 29 2023 16:03:56

%S 0,45,4950,499500,49995000,4999950000,499999500000,49999995000000,

%T 4999999950000000,499999999500000000,49999999995000000000,

%U 4999999999950000000000,499999999999500000000000,49999999999995000000000000,4999999999999950000000000000

%N a(n) = 10^n*(10^n-1) / 2.

%C Sum of all numbers with <= n digits.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (110, -1000).

%F Sum_{i=1..10^n-1} i = the (10^n-1)th triangular number (A000217). - _Marvin Ray Burns_

%F From _Chai Wah Wu_, Mar 17 2018: (Start)

%F a(n) = 110*a(n-1) - 1000*a(n-2) for n > 1.

%F G.f.: 45*x/((10*x - 1)*(100*x - 1)). (End)

%t a[n_] := 10^n(10^n - 1)/2; Table[ a[n], {n, 0, 12}] (* _Robert G. Wilson v_, Dec 24 2004 *)

%Y Partial sums of A101291.

%K nonn,easy

%O 0,2

%A _Marvin Ray Burns_