User:Arthur Rubin/Parasitic number

Working page for possible inclusion in Parasitic number

Direct approach

Using $\circ$ for concatentation of digit strings, the definition of Parasitic number reads:

$n\times (y\circ d)=(d\circ y)$

$x=(y\circ d)\,$

$y={\frac {x-d}{10}}=nx-d10^{m-1}$

Multiplying by 10, and separating:

$\left(10^{m}-1\right)d=\left(10n-1\right)x$

Or:

$x=d{\frac {10^{m}-1}{10n-1}}$

As $GCD(10n-1,10)=1$, there is an m such that this is an integer. (Or, equivalently, ${\frac {1}{10n-1}}$ is a repeating decimal with no initial term, and with period dividing m.)

Generalizations

For a k-digit right shift, the equation becomes:

$n\times (y\circ d)=(d\circ y)$

$x=(y\circ d)\,$

(where d is no longer a "digit", but a k-digit number)

Mathematically, that reads:

$y={\frac {x-d}{10^{k}}}=nx-d10^{m-k}$

Multiplying by 10^k, and separating:

$\left(10^{m}-1\right)d=\left(10^{k}n-1\right)x$

Or:

$x=d{\frac {10^{m}-1}{10^{k}n-1}}$

To avoid leading 0's, $d\geq 10^{k-1}n$.

Generalizations

For a k-digit left shift:

$n\times (d\circ y)=(y\circ d)$

$x=(d\circ y)\,$

(where d is no longer a "digit", but a k-digit number)

$y={\frac {nx-d}{10^{k}}}=x-d10^{m-k}$

Multiplying by 10^k, and separating:

$\left(10^{m}-1\right)d=\left(10^{k}-n\right)x$

Or:

$x=d{\frac {10^{m}-1}{10^{k}-n}}$

Note also that, regardless of the value of m, $GCD\left(10^{\infty },10^{k}-n)\right)|d.$

Here, for the count to be correct, we have the additional condition $y<10^{m-k}$, which corresponds to

$d{\frac {n10^{m-k}-1}{10^{k}-n}}\leq 10^{m-k}-1,$ or

$d\leq \left(10^{m-k}-1\right){\frac {10^{k}-n}{n10^{m-k}-1}}$, from which follows

$d<{\frac {10^{k}-n}{n}},$ or

$d<{\frac {10^{k}}{n}}-1.$

To avoid leadling 0s, $d\geq 10^{k-1}.$

For k=1, this only has non-trivial solutions for:

n = 3, x = 142857 or 285714 (and repeats, of course)