Basics

In decimal, unit fractions ${\displaystyle {\tfrac {1}{2}}}$ and ${\displaystyle {\tfrac {1}{5}}}$ have no repeating decimal, while ${\displaystyle {\tfrac {1}{3}}}$ repeats ${\displaystyle 0.3333\dots }$ indefinitely. The remainder of ${\displaystyle {\tfrac {1}{7}}}$, on the other hand, repeats over six digits as,

$${\displaystyle 0.{\mathbf {1}}42857{\mathbf {1}}42857{\mathbf {1}}\dots }$$

Consequently, multiples of one-seventh exhibit cyclic permutations of these six digits:^[1]

$${\displaystyle {\begin{aligned}1/7&=0.142857\dots \\2/7&=0.285714\dots \\3/7&=0.428571\dots \\4/7&=0.571428\dots \\5/7&=0.714285\dots \\6/7&=0.857142\dots \end{aligned}}}$$

If the digits are laid out as a square, each row and column sums to ${\displaystyle 1+4+2+8+5+7=27.}$ This yields the smallest base-10 non-normal, prime reciprocal magic square

${\displaystyle 1}$	${\displaystyle 4}$	${\displaystyle 2}$	${\displaystyle 8}$	${\displaystyle 5}$	${\displaystyle 7}$
${\displaystyle 2}$	${\displaystyle 8}$	${\displaystyle 5}$	${\displaystyle 7}$	${\displaystyle 1}$	${\displaystyle 4}$
${\displaystyle 4}$	${\displaystyle 2}$	${\displaystyle 8}$	${\displaystyle 5}$	${\displaystyle 7}$	${\displaystyle 1}$
${\displaystyle 5}$	${\displaystyle 7}$	${\displaystyle 1}$	${\displaystyle 4}$	${\displaystyle 2}$	${\displaystyle 8}$
${\displaystyle 7}$	${\displaystyle 1}$	${\displaystyle 4}$	${\displaystyle 2}$	${\displaystyle 8}$	${\displaystyle 5}$
${\displaystyle 8}$	${\displaystyle 5}$	${\displaystyle 7}$	${\displaystyle 1}$	${\displaystyle 4}$	${\displaystyle 2}$

In contrast with its rows and columns, the diagonals of this square do not sum to 27; however, their mean is 27, as one diagonal adds to 23 while the other adds to 31.

All prime reciprocals in any base with a ${\displaystyle p-1}$ period will generate magic squares where all rows and columns produce a magic constant, and only a select few will be full, such that their diagonals, rows and columns collectively yield equal sums.

Decimal expansions

In a full, or otherwise prime reciprocal magic square with ${\displaystyle p-1}$ period, the even number of ${\displaystyle k}$−th rows in the square are arranged by multiples of ${\displaystyle 1/p}$ — not necessarily successively — where a magic constant can be obtained.

For instance, an even repeating cycle from an odd, prime reciprocal of ${\displaystyle p}$ that is divided into ${\displaystyle n}$−digit strings creates pairs of complementary sequences of digits that yield strings of nines (9) when added together:

$${\displaystyle {\begin{aligned}1/7=&{\text{ }}0.142\;857\dots \\+&{\text{ }}0.857\;142\ldots =6/7\\&------------\\&{\text{ }}0.999\;999\ldots \\\\1/13=&{\text{ }}0.076\;923\;076\;923\dots \\+&{\text{ }}0.923\;076\;923\;076\ldots =12/13\\&------------\\&{\text{ }}0.999\;999\;999\;999\ldots \\\\1/19=&{\text{ }}0.052631578\;947368421\dots \\+&{\text{ }}0.947368421\;052631578\ldots =18/19\\&------------\\&{\text{ }}0.999999999\;999999999\dots \\\end{aligned}}}$$

This is a result of Midy's theorem.^[2][3] These complementary sequences are generated between multiples of prime reciprocals that add to 1.

More specifically, a factor ${\displaystyle n}$ in the numerator of the reciprocal of a prime number ${\displaystyle p}$ will shift the decimal places of its decimal expansion accordingly,

$${\displaystyle {\begin{aligned}1/23&=0.04347826\;08695652\;173913\ldots \\2/23&=0.08695652\;17391304\;347826\ldots \\4/23&=0.17391304\;34782608\;695652\ldots \\8/23&=0.34782608\;69565217\;391304\ldots \\16/23&=0.69565217\;39130434\;782608\ldots \\\end{aligned}}}$$

In this case, a factor of 2 moves the repeating decimal of ${\displaystyle {\tfrac {1}{23}}}$ by eight places.

A uniform solution of a prime reciprocal magic square, whether full or not, will hold rows with successive multiples of ${\displaystyle 1/p}$. Other magic squares can be constructed whose rows do not represent consecutive multiples of ${\displaystyle 1/p}$, which nonetheless generate a magic sum.

Magic constant

Magic squares based on reciprocals of primes ${\displaystyle p}$ in bases ${\displaystyle b}$ with periods ${\displaystyle p-1}$ have magic sums equal to,

$${\displaystyle M=(b-1)\times {\frac {p-1}{2}}.}$$

The table below lists some prime numbers that generate prime-reciprocal magic squares in given bases.

Variations

A ${\displaystyle {\tfrac {1}{17}}}$ prime reciprocal magic square with maximum period of 16 and magic constant of 72 can be constructed where its rows represent non-consecutive multiples of one-seventeenth:^[7][8]

$${\displaystyle {\begin{aligned}1/17&=0.{\color {blue}0}{\text{ }}5\;8\;8\;2\;3\;5\;2\;9\;4\;1\;1\;7\;6\;4\;{\color {blue}7}\dots \\5/17&=0.2\;{\color {blue}9}\;4\;1\;1\;7\;6\;4\;7\;0\;5\;8\;8\;2\;{\color {blue}3}\;5\dots \\8/17&=0.4\;7\;{\color {blue}0}\;5\;8\;8\;2\;3\;5\;2\;9\;4\;1\;{\color {blue}1}\;7\;6\dots \\6/17&=0.3\;5\;2\;{\color {blue}9}\;4\;1\;1\;7\;6\;4\;7\;0\;{\color {blue}5}\;8\;8\;2\dots \\13/17&=0.7\;6\;4\;7\;{\color {blue}0}\;5\;8\;8\;2\;3\;5\;{\color {blue}2}\;9\;4\;1\;1\dots \\14/17&=0.8\;2\;3\;5\;2\;{\color {blue}9}\;4\;1\;1\;7\;{\color {blue}6}\;4\;7\;0\;5\;8\dots \\2/17&=0.1\;1\;7\;6\;4\;7\;{\color {blue}0}\;5\;8\;{\color {blue}8}\;2\;3\;5\;2\;9\;4\dots \\10/17&=0.5\;8\;8\;2\;3\;5\;2\;{\color {blue}9}\;{\color {blue}4}\;1\;1\;7\;6\;4\;7\;0\dots \\16/17&=0.9\;4\;1\;1\;7\;6\;4\;{\color {blue}7}\;{\color {blue}0}\;5\;8\;8\;2\;3\;5\;2\dots \\12/17&=0.7\;0\;5\;8\;8\;2\;{\color {blue}3}\;5\;2\;{\color {blue}9}\;4\;1\;1\;7\;6\;4\dots \\9/17&=0.5\;2\;9\;4\;1\;{\color {blue}1}\;7\;6\;4\;7\;{\color {blue}0}\;5\;8\;8\;2\;3\dots \\11/17&=0.6\;4\;7\;0\;{\color {blue}5}\;8\;8\;2\;3\;5\;2\;{\color {blue}9}\;4\;1\;1\;7\dots \\4/17&=0.2\;3\;5\;{\color {blue}2}\;9\;4\;1\;1\;7\;6\;4\;7\;{\color {blue}0}\;5\;8\;8\dots \\3/17&=0.1\;7\;{\color {blue}6}\;4\;7\;0\;5\;8\;8\;2\;3\;5\;2\;{\color {blue}9}\;4\;1\dots \\15/17&=0.8\;{\color {blue}8}\;2\;3\;5\;2\;9\;4\;1\;1\;7\;6\;4\;7\;{\color {blue}0}\;5\dots \\7/17&=0.{\color {blue}4}\;1\;1\;7\;6\;4\;7\;0\;5\;8\;8\;2\;3\;5\;2\;{\color {blue}9}\dots \\\end{aligned}}}$$

As such, this full magic square is the first of its kind in decimal that does not admit a uniform solution where consecutive multiples of ${\displaystyle 1/p}$ fit in respective ${\displaystyle k}$−th rows.