This can't be a coincidence as well

Number Theory Level 5

Consider the decimal expansion of $1 / 998998999$ :

$0. \quad 000 \quad 000 \quad 001 \quad 001 \quad 002 \\ \ \ \quad 004 \quad 007 \quad 013 \quad 024 \quad \ldots$

Notice that those numbers in units of three digits - let's call them $\lbrace a_{n} \rbrace$ - satisfy $a_{n} = a_{n-1} + a_{n-2} + a_{n-3}$ .

If $a_{1} = a_{2} = 1$ and $a_{3} = 2$ , find the least $n$ for which the recurrence is no longer satisfied.

Inspired by this problem .

The answer is 13.

1 solution

Vishnu C
Jul 6, 2015

The series $\{ a_n \}$ is nothing but the tribonacci series, and its generating function is given by: $f(z) = \frac z {1-z-z^2-z^3} =\displaystyle \sum ^{\infty} _{n=1} {a_n \cdot z^n}$

1/998998999 = $f(\frac 1 {1000})$ = 0. 000 000 001 001 002 004 ....

The 13th tribonacci number = 927 and after that, you have 4 digit tribonacci numbers. But because there are only places for 3 digits, the carry of the next tribonacci (1705) gets added to the units of 927, making it 928. So, the sequence stops there at n=13.

This can't be a coincidence as well

The answer is 13.

1 solution

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