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.
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
KEYWORD
nonn
AUTHOR
Peter C. Heinig (algorithms(AT)gmx.de), Apr 18 2007, Apr 22 2007, Apr 23 2007
STATUS
approved