Generating Fraction for Lucas Numbers
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Theorem
The fraction:
- $\dfrac {199} {9899}$
has a decimal expansion which contains within it the start of the Lucas sequence:
- $0 \cdotp 02010 \, 30407 \, 11 \ldots$
Corollary
The fraction:
- $\dfrac {1999} {998 \, 999}$
has a decimal expansion which contains within it the start of the Lucas sequence:
- $0 \cdotp 00200 \, 10030 \, 04007 \, 011 \ldots$
and in general, the fraction:
- $\dfrac {2 \times 10^n - 1} {10^{2 n} - 10^n - 1}$
contains the Lucas sequence spread out with $n$ digits between each term.
Proof
By long division:
0.0201030407111829 ---------------------- 9899 ) 199.0000000000000000 197 98 --- -- 1 0200 9899 - ---- 30100 29697 ----- 40300 39596 ----- 70400 69293 ----- 11070 9899 ----- 11710 9899 ----- 18110 9899 ----- 82110 79192 ----- 29180 19798 ----- 93820 89091 ----- 4729
By Generating Function for Lucas Numbers:
\(\ds \sum_{k \mathop \ge 0} L_k 10^{-2 k - 2}\) | \(=\) | \(\ds \frac {2 - 10^{-2} } {1 - 10^{-2} - 10^{-4} } \times 10^{-2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \times 10^2 - 1} {10^4 - 10^2 - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {199} {9899}\) |
The first few terms are contained in the decimal expansion, as long as $L_{k + 1} < 100$, where there is no carry.
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $0 \cdotp 02010 \, 30407 \, 11 \ldots$