A117804 Natural position of n in the string 12345678910111213....

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120

Offset

1,2

Comments

The number of digits necessary to write down all the numbers 0, 1, 2, ..., n-1. Thus, the partial sums of A055642 are given by a(n+1). - Hieronymus Fischer, Jun 08 2012

From Daniel Forgues, Mar 21 2013: (Start)

From n = 10^0 + 1 to 10^1: a(n) - a(n-1) = 1 (9 * 10^0 terms);

From n = 10^1 + 1 to 10^2: a(n) - a(n-1) = 2 (9 * 10^1 terms);

From n = 10^2 + 1 to 10^3: a(n) - a(n-1) = 3 (9 * 10^2 terms);

...

From n = 10^k + 1 to 10^(k+1): a(n) - a(n-1) = k+1 (9 * 10^k terms). (End)

By the "number of digits" definition, a(n) = 1 + A058183(n-1) for n > 1. - David Fifield, Jun 02 2019

Links

Hieronymus Fischer, Table of n, a(n) for n = 1..10000

Formula

a(n) = d*n + 1 - (10^d - 1)/9 where d is the number of decimal digits in n, i.e., d = floor(log_10(n)) + 1.

From Hieronymus Fischer, Jun 08 2012: (Start)

a(n) = Sum_{j=0..n-1} A055642(j).

a(n) = 1 + A055642(n-1)*n - (10^A055642(n-1)-1)/9.

a(n) = 1 + A055642(n)*n - (10^A055642(n)-1)/9.

a(10^n) = (9*n-1)*(10^n-1)/9 + n + 1. (This is the total number of digits necessary to write down all the numbers with <= n places.)

G.f.: g(x) = x/(1-x) + (x/(1-x)^2)*Sum_{j>=0} x^10^j; corrected by Ilya Gutkovskiy, Jan 09 2017 (End)

Example

12 begins at the 14th place in 12345678910111213... (we are ignoring "early bird" occurrences here, cf. A116700), so a(12) = 14.
From Daniel Forgues, Mar 21 2013: (Start)
a(10^1) = 10. (1*10^1 - 0)
a(10^2) = 190. (2*10^2 - 10)
a(10^3) = 2890. (3*10^3 - 110)
a(10^4) = 38890. (4*10^4 - 1110)
a(10^5) = 488890. (5*10^5 - 11110)
a(10^6) = 5888890. (6*10^6 - 111110)
...
a(10^k) = k*10^k - R_k + 1, R_k := k-th repunit (cf. A002275)
(the number of digits necessary to write down the numbers 0..10^k-1). (End)

Crossrefs

Cf. A116700.

Cf. A055640, A055641, A055642, A102669-A102685.

Sequence in context: A067082 A102685 A032960 * A088235 A064223 A098952

Adjacent sequences: A117801 A117802 A117803 * A117805 A117806 A117807

Keyword

nonn,base

Author

Warut Roonguthai, Jul 23 2007

Status

approved