A prime reciprocal magic square is a magic square using the decimal digits of the reciprocal of a prime number.
In decimal, unit fractions 1/2 and 1/5 have no repeating decimal, while 1/3 repeats 0.3333 … {\displaystyle 0.3333\dots } indefinitely. The remainder of 1/7, on the other hand, repeats over six digits as, 0. 1 42857 1 42857 1 … {\displaystyle 0.{\mathbf {1}}42857{\mathbf {1}}42857{\mathbf {1}}\dots }
Consequently, multiples of one-seventh exhibit cyclic permutations of these six digits:[1]
1 / 7 = 0.142857 … 2 / 7 = 0.285714 … 3 / 7 = 0.428571 … 4 / 7 = 0.571428 … 5 / 7 = 0.714285 … 6 / 7 = 0.857142 … {\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 1 + 4 + 2 + 8 + 5 + 7 = 27. This yields the smallest base-10 non-normal, prime reciprocal magic square
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 {\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.
In a full, or otherwise prime reciprocal magic square with p − 1 {\displaystyle p-1} period, the even number of k−th rows in the square are arranged by multiples of 1 / p {\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 p that is divided into n−digit strings creates pairs of complementary sequences of digits that yield strings of nines (9) when added together:
1 / 7 = 0.142 857 … + 0.857 142 … = 6 / 7 − − − − − − − − − − − − 0.999 999 … 1 / 13 = 0.076 923 076 923 … + 0.923 076 923 076 … = 12 / 13 − − − − − − − − − − − − 0.999 999 999 999 … 1 / 19 = 0.052631578 947368421 … + 0.947368421 052631578 … = 18 / 19 − − − − − − − − − − − − 0.999999999 999999999 … {\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 n in the numerator of the reciprocal of a prime number p will shift the decimal places of its decimal expansion accordingly,
1 / 23 = 0.04347826 08695652 173913 … 2 / 23 = 0.08695652 17391304 347826 … 4 / 23 = 0.17391304 34782608 695652 … 8 / 23 = 0.34782608 69565217 391304 … 16 / 23 = 0.69565217 39130434 782608 … {\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 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 {\displaystyle 1/p} . Other magic squares can be constructed whose rows do not represent consecutive multiples of 1 / p {\displaystyle 1/p} , which nonetheless generate a magic sum.
Magic squares based on reciprocals of primes p in bases b with periods p − 1 {\displaystyle p-1} have magic sums equal to,[citation needed]
M = ( b − 1 ) × p − 1 2 . {\displaystyle M=(b-1)\times {\frac {p-1}{2}}.}
The 1 19 {\displaystyle {\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 {\displaystyle k} −th rows:[4][5]
1 / 19 = 0. 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 … 2 / 19 = 0.1 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 … 3 / 19 = 0.1 5 7 8 9 4 7 3 6 8 4 2 1 0 5 2 6 3 … 4 / 19 = 0.2 1 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 … 5 / 19 = 0.2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 0 5 … 6 / 19 = 0.3 1 5 7 8 9 4 7 3 6 8 4 2 1 0 5 2 6 … 7 / 19 = 0.3 6 8 4 2 1 0 5 2 6 3 1 5 7 8 9 4 7 … 8 / 19 = 0.4 2 1 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 … 9 / 19 = 0.4 7 3 6 8 4 2 1 0 5 2 6 3 1 5 7 8 9 … 10 / 19 = 0.5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 0 … 11 / 19 = 0.5 7 8 9 4 7 3 6 8 4 2 1 0 5 2 6 3 1 … 12 / 19 = 0.6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 0 5 2 … 13 / 19 = 0.6 8 4 2 1 0 5 2 6 3 1 5 7 8 9 4 7 3 … 14 / 19 = 0.7 3 6 8 4 2 1 0 5 2 6 3 1 5 7 8 9 4 … 15 / 19 = 0.7 8 9 4 7 3 6 8 4 2 1 0 5 2 6 3 1 5 … 16 / 19 = 0.8 4 2 1 0 5 2 6 3 1 5 7 8 9 4 7 3 6 … 17 / 19 = 0.8 9 4 7 3 6 8 4 2 1 0 5 2 6 3 1 5 7 … 18 / 19 = 0. 9 4 7 3 6 8 4 2 1 0 5 2 6 3 1 5 7 8 … {\displaystyle {\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[6]
The smallest prime number to yield such magic square in binary is 59 (1110112), while in ternary it is 223 (220213); these are listed at A096339, and A096660.
A 1 17 {\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]
1 / 17 = 0. 0 5 8 8 2 3 5 2 9 4 1 1 7 6 4 7 … 5 / 17 = 0.2 9 4 1 1 7 6 4 7 0 5 8 8 2 3 5 … 8 / 17 = 0.4 7 0 5 8 8 2 3 5 2 9 4 1 1 7 6 … 6 / 17 = 0.3 5 2 9 4 1 1 7 6 4 7 0 5 8 8 2 … 13 / 17 = 0.7 6 4 7 0 5 8 8 2 3 5 2 9 4 1 1 … 14 / 17 = 0.8 2 3 5 2 9 4 1 1 7 6 4 7 0 5 8 … 2 / 17 = 0.1 1 7 6 4 7 0 5 8 8 2 3 5 2 9 4 … 10 / 17 = 0.5 8 8 2 3 5 2 9 4 1 1 7 6 4 7 0 … 16 / 17 = 0.9 4 1 1 7 6 4 7 0 5 8 8 2 3 5 2 … 12 / 17 = 0.7 0 5 8 8 2 3 5 2 9 4 1 1 7 6 4 … 9 / 17 = 0.5 2 9 4 1 1 7 6 4 7 0 5 8 8 2 3 … 11 / 17 = 0.6 4 7 0 5 8 8 2 3 5 2 9 4 1 1 7 … 4 / 17 = 0.2 3 5 2 9 4 1 1 7 6 4 7 0 5 8 8 … 3 / 17 = 0.1 7 6 4 7 0 5 8 8 2 3 5 2 9 4 1 … 15 / 17 = 0.8 8 2 3 5 2 9 4 1 1 7 6 4 7 0 5 … 7 / 17 = 0. 4 1 1 7 6 4 7 0 5 8 8 2 3 5 2 9 … {\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 1 / p {\displaystyle 1/p} fit in respective k {\displaystyle k} −th rows.
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