Formally, an n-Kaprekar triple k satisfies the pair of equations
k3 = p·102n + q·10n + r,
k = p + q + r,
where 0 ≤ r < 10n, 0 ≤ q < 10n, and p > 0 are integers. As the 3-Kaprekar triple 297 shows, p may have fewer than n digits, and so may q or r (note the leading zero in r = 073). The stipulation that p > 0 precludes many otherwise trivial examples such as
1003 = 0·108 + 100·104 + 0,
100 = 0 + 100 + 0,
i.e., 100 as a 4-Kaprekar triple. Having p > 0 also precludes 1 as a Kaprekar triple, in spite of its inclusion in sequence A006887 by Sloan [5].