# Prime reciprocal magic square

A prime reciprocal magic square is a magic square using the decimal digits of the reciprocal of a prime number.

## Formulation
### Basics

In decimal, unit fractions ${\tfrac {1}{2}}$ and ${\tfrac {1}{5}}$ have no repeating decimal, while ${\tfrac {1}{3}}$ repeats $0.3333\dots$ indefinitely. The remainder of ${\tfrac {1}{7}}$, on the other hand, repeats over six digits as, $0.{\mathbf {1}}42857{\mathbf {1}}42857{\mathbf {1}}\dots$
Consequently, multiples of one-seventh exhibit cyclic permutations of these six digits:
${\begin{aligned}1/7&=0.142857\dots \\2/7&=0.285714\dots \\3/7&=0.428571\dots \\4/7&=0.571428\dots \\5/7&=0.714285\dots \\6/7&=0.857142\dots \end{aligned}}$
If the digits are laid out as a square, each row and column sums to $1+4+2+8+5+7=27.$ This yields the smallest base-10 non-normal, prime reciprocal magic square
| $1$ | $4$ | $2$ | $8$ | $5$ | $7$ |
| $2$ | $8$ | $5$ | $7$ | $1$ | $4$ |
| $4$ | $2$ | $8$ | $5$ | $7$ | $1$ |
| $5$ | $7$ | $1$ | $4$ | $2$ | $8$ |
| $7$ | $1$ | $4$ | $2$ | $8$ | $5$ |
| $8$ | $5$ | $7$ | $1$ | $4$ | $2$ |
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 $p-1$ 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 $p-1$ period, the even number of $k$−th rows in the square are arranged by multiples of $1/p$ — not necessarily successively — where a magic constant can be obtained.
For instance, an even repeating cycle from an odd, prime reciprocal of $p$ that is divided into $n$−digit strings creates pairs of complementary sequences of digits that yield strings of nines (9) when added together:
${\begin{aligned}1/7=&{\text{ }}0.142\;857\dots \\+&{\text{ }}0.857\;142\ldots =6/7\\&------------\\&{\text{ }}0.999\;999\ldots \\\\1/13=&{\text{ }}0.076\;923\;076\;923\dots \\+&{\text{ }}0.923\;076\;923\;076\ldots =12/13\\&------------\\&{\text{ }}0.999\;999\;999\;999\ldots \\\\1/19=&{\text{ }}0.052631578\;947368421\dots \\+&{\text{ }}0.947368421\;052631578\ldots =18/19\\&------------\\&{\text{ }}0.999999999\;999999999\dots \\\end{aligned}}$
This is a result of Midy's theorem. These complementary sequences are generated between multiples of prime reciprocals that add to 1.
More specifically, a factor $n$ in the numerator of the reciprocal of a prime number $p$ will shift the decimal places of its decimal expansion accordingly,
${\begin{aligned}1/23&=0.04347826\;08695652\;173913\ldots \\2/23&=0.08695652\;17391304\;347826\ldots \\4/23&=0.17391304\;34782608\;695652\ldots \\8/23&=0.34782608\;69565217\;391304\ldots \\16/23&=0.69565217\;39130434\;782608\ldots \\\end{aligned}}$
In this case, a factor of 2 moves the repeating decimal of ${\tfrac {1}{23}}$ by eight places.
A uniform solution of a prime reciprocal magic square, whether full or not, will hold rows with successive multiples of $1/p$. Other magic squares can be constructed whose rows do not represent consecutive multiples of $1/p$, which nonetheless generate a magic sum.

## Magic constant

Magic squares based on reciprocals of primes $p$ in bases $b$ with periods $p-1$ have magic sums equal to,
$M=(b-1)\times {\frac {p-1}{2}}.$
The table below lists some prime numbers that generate prime-reciprocal magic squares in given bases.
| Prime | Base | Magic sum |
| 19 | 10 | 81 |
| 53 | 12 | 286 |
| 59 | 2 | 29 |
| 67 | 2 | 33 |
| 83 | 2 | 41 |
| 89 | 19 | 792 |
| 211 | 2 | 105 |
| 223 | 3 | 222 |
| 307 | 5 | 612 |
| 383 | 10 | 1,719 |
| 397 | 5 | 792 |
| 487 | 6 | 1,215 |
| 593 | 3 | 592 |
| 631 | 87 | 27,090 |
| 787 | 13 | 4,716 |
| 811 | 3 | 810 |
| 1,033 | 11 | 5,160 |
| 1,307 | 5 | 2,612 |
| 1,499 | 11 | 7,490 |
| 1,877 | 19 | 16,884 |
| 2,011 | 26 | 25,125 |
| 2,027 | 2 | 1,013 |

## Full magic squares

The ${\mathbf {\tfrac {1}{19}}}$ magic square with maximum period 18 contains a row-and-column total of 81, that is also obtained by both diagonals. This makes it the first full, non-normal base-10 prime reciprocal magic square whose multiples fit inside respective $k$−th rows:
${\begin{aligned}1/19&=0.{\color {red}0}{\text{ }}5{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}5{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}4{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}4{\text{ }}2{\text{ }}{\color {red}1}\dots \\2/19&=0.1{\text{ }}{\color {red}0}{\text{ }}5{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}5{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}4{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}{\color {red}4}{\text{ }}2\dots \\3/19&=0.1{\text{ }}5{\text{ }}{\color {red}7}{\text{ }}8{\text{ }}9{\text{ }}4{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}4{\text{ }}2{\text{ }}1{\text{ }}0{\text{ }}5{\text{ }}{\color {red}2}{\text{ }}6{\text{ }}3\dots \\4/19&=0.2{\text{ }}1{\text{ }}0{\text{ }}{\color {red}5}{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}5{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}4{\text{ }}7{\text{ }}{\color {red}3}{\text{ }}6{\text{ }}8{\text{ }}4\dots \\5/19&=0.2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}{\color {red}5}{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}4{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}{\color {red}4}{\text{ }}2{\text{ }}1{\text{ }}0{\text{ }}5\dots \\6/19&=0.3{\text{ }}1{\text{ }}5{\text{ }}7{\text{ }}8{\text{ }}{\color {red}9}{\text{ }}4{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}4{\text{ }}{\color {red}2}{\text{ }}1{\text{ }}0{\text{ }}5{\text{ }}2{\text{ }}6\dots \\7/19&=0.3{\text{ }}6{\text{ }}8{\text{ }}4{\text{ }}2{\text{ }}1{\text{ }}{\color {red}0}{\text{ }}5{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}{\color {red}1}{\text{ }}5{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}4{\text{ }}7\dots \\8/19&=0.4{\text{ }}2{\text{ }}1{\text{ }}0{\text{ }}5{\text{ }}2{\text{ }}6{\text{ }}{\color {red}3}{\text{ }}1{\text{ }}5{\text{ }}{\color {red}7}{\text{ }}8{\text{ }}9{\text{ }}4{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8\dots \\9/19&=0.4{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}4{\text{ }}2{\text{ }}1{\text{ }}{\color {red}0}{\text{ }}{\color {red}5}{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}5{\text{ }}7{\text{ }}8{\text{ }}9\dots \\10/19&=0.5{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}5{\text{ }}7{\text{ }}8{\text{ }}{\color {red}9}{\text{ }}{\color {red}4}{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}4{\text{ }}2{\text{ }}1{\text{ }}0\dots \\11/19&=0.5{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}4{\text{ }}7{\text{ }}3{\text{ }}{\color {red}6}{\text{ }}8{\text{ }}4{\text{ }}{\color {red}2}{\text{ }}1{\text{ }}0{\text{ }}5{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1\dots \\12/19&=0.6{\text{ }}3{\text{ }}1{\text{ }}5{\text{ }}7{\text{ }}8{\text{ }}{\color {red}9}{\text{ }}4{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}{\color {red}8}{\text{ }}4{\text{ }}2{\text{ }}1{\text{ }}0{\text{ }}5{\text{ }}2\dots \\13/19&=0.6{\text{ }}8{\text{ }}4{\text{ }}2{\text{ }}1{\text{ }}{\color {red}0}{\text{ }}5{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}5{\text{ }}{\color {red}7}{\text{ }}8{\text{ }}9{\text{ }}4{\text{ }}7{\text{ }}3\dots \\14/19&=0.7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}{\color {red}4}{\text{ }}2{\text{ }}1{\text{ }}0{\text{ }}5{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}{\color {red}5}{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}4\dots \\15/19&=0.7{\text{ }}8{\text{ }}9{\text{ }}{\color {red}4}{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}4{\text{ }}2{\text{ }}1{\text{ }}0{\text{ }}5{\text{ }}2{\text{ }}{\color {red}6}{\text{ }}3{\text{ }}1{\text{ }}5\dots \\16/19&=0.8{\text{ }}4{\text{ }}{\color {red}2}{\text{ }}1{\text{ }}0{\text{ }}5{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}5{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}4{\text{ }}{\color {red}7}{\text{ }}3{\text{ }}6\dots \\17/19&=0.8{\text{ }}{\color {red}9}{\text{ }}4{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}4{\text{ }}2{\text{ }}1{\text{ }}0{\text{ }}5{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}{\color {red}5}{\text{ }}7\dots \\18/19&=0.{\color {red}9}{\text{ }}4{\text{ }}7{\text{ }}3{\text{ }}6{\text{ }}8{\text{ }}4{\text{ }}2{\text{ }}1{\text{ }}0{\text{ }}5{\text{ }}2{\text{ }}6{\text{ }}3{\text{ }}1{\text{ }}5{\text{ }}7{\text{ }}{\color {red}8}\dots \\\end{aligned}}$
The first few prime numbers in decimal whose reciprocals can be used to produce a non-normal, full prime reciprocal magic square of this type are
    {19, 383, 32327, 34061, 45341, 61967, 65699, 117541, 158771, 405817, ...} (sequence A072359 in the OEIS).
The smallest prime number to yield such magic square in binary is 59 (111011$_2$), while in ternary it is 223 (22021$_3$); these are listed at A096339, and A096660.

### Variations

A ${\tfrac {1}{17}}$ 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:
${\begin{aligned}1/17&=0.{\color {blue}0}{\text{ }}5\;8\;8\;2\;3\;5\;2\;9\;4\;1\;1\;7\;6\;4\;{\color {blue}7}\dots \\5/17&=0.2\;{\color {blue}9}\;4\;1\;1\;7\;6\;4\;7\;0\;5\;8\;8\;2\;{\color {blue}3}\;5\dots \\8/17&=0.4\;7\;{\color {blue}0}\;5\;8\;8\;2\;3\;5\;2\;9\;4\;1\;{\color {blue}1}\;7\;6\dots \\6/17&=0.3\;5\;2\;{\color {blue}9}\;4\;1\;1\;7\;6\;4\;7\;0\;{\color {blue}5}\;8\;8\;2\dots \\13/17&=0.7\;6\;4\;7\;{\color {blue}0}\;5\;8\;8\;2\;3\;5\;{\color {blue}2}\;9\;4\;1\;1\dots \\14/17&=0.8\;2\;3\;5\;2\;{\color {blue}9}\;4\;1\;1\;7\;{\color {blue}6}\;4\;7\;0\;5\;8\dots \\2/17&=0.1\;1\;7\;6\;4\;7\;{\color {blue}0}\;5\;8\;{\color {blue}8}\;2\;3\;5\;2\;9\;4\dots \\10/17&=0.5\;8\;8\;2\;3\;5\;2\;{\color {blue}9}\;{\color {blue}4}\;1\;1\;7\;6\;4\;7\;0\dots \\16/17&=0.9\;4\;1\;1\;7\;6\;4\;{\color {blue}7}\;{\color {blue}0}\;5\;8\;8\;2\;3\;5\;2\dots \\12/17&=0.7\;0\;5\;8\;8\;2\;{\color {blue}3}\;5\;2\;{\color {blue}9}\;4\;1\;1\;7\;6\;4\dots \\9/17&=0.5\;2\;9\;4\;1\;{\color {blue}1}\;7\;6\;4\;7\;{\color {blue}0}\;5\;8\;8\;2\;3\dots \\11/17&=0.6\;4\;7\;0\;{\color {blue}5}\;8\;8\;2\;3\;5\;2\;{\color {blue}9}\;4\;1\;1\;7\dots \\4/17&=0.2\;3\;5\;{\color {blue}2}\;9\;4\;1\;1\;7\;6\;4\;7\;{\color {blue}0}\;5\;8\;8\dots \\3/17&=0.1\;7\;{\color {blue}6}\;4\;7\;0\;5\;8\;8\;2\;3\;5\;2\;{\color {blue}9}\;4\;1\dots \\15/17&=0.8\;{\color {blue}8}\;2\;3\;5\;2\;9\;4\;1\;1\;7\;6\;4\;7\;{\color {blue}0}\;5\dots \\7/17&=0.{\color {blue}4}\;1\;1\;7\;6\;4\;7\;0\;5\;8\;8\;2\;3\;5\;2\;{\color {blue}9}\dots \\\end{aligned}}$
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 $1/p$ fit in respective $k$−th rows.