A128869 a(n) = the largest number one can subtract from 10^n such that the square of the result is strictly greater than 10^(2*n-1).

6, 68, 683, 6837, 68377, 683772, 6837722, 68377223, 683772233, 6837722339, 68377223398, 683772233983, 6837722339831, 68377223398316, 683772233983162, 6837722339831620, 68377223398316206, 683772233983162066

Offset

1,1

Comments

Let a(n) := max{ a in N: (10^n-a)^2 > 10^(2*n-1) } then a(n) = floor(10^n*(1-sqrt(0.1))) is easily obtained by solving a quadratic equation. When proving that a(n) satisfies the definition, one has to use the following inequality, due to the fact that 10^n*(1-sqrt(0.1)) is never an integer: floor(10^n*(1-sqrt(0.1))) < 10^n*(1-sqrt(0.1)).

The exponent (2*n - 1) in the definition of the sequence was chosen because with a number of n decimal digits, the greater of the two possible lengths obtained when squaring it is 2*n. Since 10^(2*n-1) is the smallest decimal number with 2*n digits, (10^n - a(n)) gives the minimum number to obtain the maximum length by squaring it.

Links

Table of n, a(n) for n=1..18.

Formula

a(n) = floor(10^n*(1-sqrt(0.1))) That means that a(n) gives the decimal expansion of 1-sqrt(0.1).

Example

a(1) = 6 because (10-6)^2 = 4^2 > 10 = 10^(2*1-1) whereas (10-7)^2 = 3^2 <= 10.
a(2) = 68 because (100-68)^2 = 32^2 > 1000 = 10^(2*2-1) whereas
(100-69)^2 = 31^2 <= 1000.

Maple

for n from 1 to 31 do floor(10^n * (1-sqrt(0.1))); od;

Crossrefs

Sequence in context: A183470 A281051 A006737 * A186669 A258134 A256238

Adjacent sequences: A128866 A128867 A128868 * A128870 A128871 A128872

Keyword

nonn

Author

Peter C. Heinig (algorithms(AT)gmx.de), Apr 18 2007, Apr 22 2007, Apr 23 2007

Status

approved