A037182 a(n) = 10^n*(10^n-1) / 2.

0, 45, 4950, 499500, 49995000, 4999950000, 499999500000, 49999995000000, 4999999950000000, 499999999500000000, 49999999995000000000, 4999999999950000000000, 499999999999500000000000, 49999999999995000000000000, 4999999999999950000000000000

Offset: 0

Author

Marvin Ray Burns

Keywords

• nonn
• easy

Comments

Sum of all numbers with <= n digits.

Links

• Index entries for linear recurrences with constant coefficients, signature (110, -1000).

Crossrefs

Partial sums of A101291.

Programs

Mathematica

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

Formula

Sum_{i=1..10^n-1} i = the (10^n-1)th triangular number (A000217). - Marvin Ray Burns

From Chai Wah Wu, Mar 17 2018: (Start)

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

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