142857, the six repeating digits of 1/7, $0.{\overline {142857}}$ , is the best-known cyclic number in base 10. If it is multiplied by 2, 3, 4, 5, or 6, the answer will be a cyclic permutation of itself, and will correspond to the repeating digits of 2/7, 3/7, 4/7, 5/7, or 6/7 respectively.

142,857 is a Kaprekar number and a Harshad number (in base 10).

If you multiply by an integer greater than 7, there is a simple process to get to a cyclic permutation of 142857. By adding the rightmost six digits (ones through hundred thousands) to the remaining digits and repeating this process until you have only the six digits left, it will result in a cyclic permutation of 142857:

Multiplying by a multiple of 7 will result in 999999 through this process:

If you square the last three digits and subtract the square of the first three digits, you also get back a cyclic permutation of the number.

It is the repeating part in the decimal expansion of the rational number 1/7 = 0.142857. Thus, multiples of 1/7 are simply repeated copies of the corresponding multiples of 142857:

There is an interesting pattern of doubling, shifting and addition that gives 1/7.

${\begin{aligned}1/7\ &=0.142857142857142857\ldots \\[6pt]&=0.14+0.0028+0.000056+0.00000112+0.0000000224+0.000000000448+0.00000000000896+\cdots \\[6pt]&={\frac {14}{100}}+{\frac {28}{100^{2}}}+{\frac {56}{100^{3}}}+{\frac {112}{100^{4}}}+{\frac {224}{100^{5}}}+\cdots +{\frac {7\times 2^{N}}{100^{N}}}+\cdots \\[6pt]&=\left({\frac {7}{50}}+{\frac {7}{50^{2}}}+{\frac {7}{50^{3}}}+{\frac {7}{50^{4}}}+{\frac {7}{50^{5}}}+\cdots +{\frac {7}{50^{N}}}+\cdots \right)\\[6pt]&=\sum _{k=1}^{\infty }{\frac {7}{50^{k}}}\end{aligned}}$

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