A255381 Number of strings of k+n decimal digits that contain one string of exactly k consecutive "0" digits, where k >= n.

1, 18, 261, 3420, 42300, 504000, 5850000, 66600000, 747000000, 8280000000, 90900000000, 990000000000, 10710000000000, 115200000000000, 1233000000000000, 13140000000000000, 139500000000000000, 1476000000000000000, 15570000000000000000, 163800000000000000000

Offset: 0

Author

Jon E. Schoenfield, Feb 27 2015

Keywords

• nonn
• base
• easy

Comments

This sequence gives the first k+1 nonzero terms from A255371 (the k=1 case), A255372 (the k=2 case), etc., through A255380 (the k=10 case). Those sequences' definitions concern the number of strings of decimal digits that contain "at least one" string of exactly k consecutive "0" digits; the present sequence omits the words "at least" because, with k >= n, and thus 2k >= k+n, it is not possible to have more than one string of exactly k consecutive "0" digits in a string of k+n digits. (Two strings each having exactly k consecutive "0" digits would have to be separated by at least one nonzero digit, or else they would constitute a single string of exactly 2k consecutive "0" digits.)

Omitting the zero terms of each, A255371 through A255380 begin

1, 18, 252, 3177, 37764, 432315, 4821867, 52767711, ...

1, 18, 261, 3411, 42057, 499383, 5775480, 65506986, ...

1, 18, 261, 3420, 42291, 503757, 5845383, 66525399, ...

1, 18, 261, 3420, 42300, 503991, 5849757, 66595383, ...

1, 18, 261, 3420, 42300, 504000, 5849991, 66599757, ...

1, 18, 261, 3420, 42300, 504000, 5850000, 66599991, ...

1, 18, 261, 3420, 42300, 504000, 5850000, 66600000, ...

1, 18, 261, 3420, 42300, 504000, 5850000, 66600000, ...

1, 18, 261, 3420, 42300, 504000, 5850000, 66600000, ...

1, 18, 261, 3420, 42300, 504000, 5850000, 66600000, ...

and the terms of this present sequence give the limiting value for each column.

Examples

			Trivially, a(0)=1 because there is 1 string of k decimal digits that contains one string of exactly k consecutive "0" digits, where k >= 0: namely, the string of k consecutive "0" digits itself.
a(1)=18 because there are 18 strings of k+1 decimal digits that contain one string of exactly k consecutive "0" digits, where k >= 1. Letting "S" and "+" represent the string of exactly k consecutive "0" digits and any nonzero digit, respectively, the 18 strings comprise 9 of the form "S+" and 9 of the form "+S".
a(2)=261 because there are 261 strings of k+2 decimal digits that contain one string of exactly k consecutive "0" digits, where k >= 2. Letting "S", "+", and "." represent the string of exactly k consecutive "0" digits, any nonzero digit, and any digit (zero or nonzero), respectively, the 261 strings comprise 9*10=90 of the form "S+.", 9*9=81 of the form "+S+", and 10*9=90 of the form ".+S".
a(3)=3420 because there are 3420 strings of k+3 decimal digits that contain one string of exactly k consecutive "0" digits, where k >= 3. Using "S", "+", and "." as above, the 3420 strings comprise 9*10*10=900 of the form "S+..", 9*9*10=810 of the form "+S+.", 10*9*9=810 of the form ".+S+", and 10*10*9=900 of the form "..+S".
		

Links

• Colin Barker, Table of n, a(n) for n = 0..996
• Index entries for linear recurrences with constant coefficients, signature (20,-100).

Crossrefs

Cf. A255371, A255372, A255373, A255374, A255375, A255376, A255377, A255378, A255379, A255380.

Programs

PARI

Vec((x-1)^2/(10*x-1)^2 + O(x^100)) \\ Colin Barker, Feb 27 2015

Formula

a(0) = 1, a(n) = (81n + 99) * 10^(n-2) for n >= 1.

G.f.: (x-1)^2/(10*x-1)^2. - Alois P. Heinz, Feb 27 2015

a(n) = 20*a(n-1) - 100*a(n-2) for n>2. - Colin Barker, Feb 27 2015