SIXES AND SEVENS

Carl Pomerance

University of Georgia

Cooper and Kennedy [1] have posed the following interesting question. For n = 0, 1, …, let a_{n} ∈ {6, 7} be such that the base 10 number a_{n−1}a_{n−2} … a_{0} is divisible by 2^{n}. It is clear that this condition defines the sequence {a_{n}}. The first five terms are 6, 7, 7, 7, 6. Cooper and Kennedy ask if {a_{n}} must contain infinitely many 6’s and infinitely many 7’s. We show a more general result which immediately answers their question in the affirmative.

\underline{Theorem}. Let b, c be integers at least 2 and let \{a_{n}\}_{n=0}^{∞} be a sequence with each a_{n} ∈ {0, 1, …, b − 1}. Suppose that for each positive integer m there is some integer N_{m}, such that c^{m} divides a_{0} + a_{1}b + ⋯ + a_{n}b^{n} for each n ≥ N_{m}. If {a_{n}} is eventually periodic, then {a_{n}} is identically zero.

\underline{Proof}. It clearly suffices to prove the result in the case that c = p, a prime. First note that the hypothesis that N_{m} always exists implies that either {a_{n}} is identically zero or p|b. Indeed, let

A_{n} := ∑_{i=0}^{n−1} a_{i}b^{i}

for n = 1, 2, . . .. Let m be so large that p^{m} ≥ b. If n > N_{m}, then

p^{m}|A_{n}, p^{m}|A_{n+1} = A_{n} + a_{n}b^{n},

so that p^{m}|a_{n}b^{n}. But 0 ≤ a_{n} ≤ b − 1 and p^{m} ≥ b. Thus, either {a_{n}} is eventually zero or p|b. But if {a_{n}} is eventually zero, then there is some N with A_{n} = A_{N} for all n ≥ N. Thus, p^{m}|A_{N} for all m, so that A_{N} = 0 and {a_{n}} is identically zero.

Thus, we may assume p|b. Suppose {a_{n}} is eventually periodic, so that there are integers N, k with a_{n} = a_{n+k} for all n ≥ N. Let

B := ∑_{i=0}^{k−1} a_{N+i}b^{i}.

Thus, for j = 1, 2, …, we have

A_{N+jk} = A_{N} + Bb^{N}(1 + b^{k} + ⋯ + b^{(j−1)k}) = A_{N} + Bb^{N} \frac{b^{jk} − 1}{b^{k} − 1},

so that

(1)  A_{N}(b^{k} − 1) − Bb^{N} = A_{N+jk}(b^{k} − 1) − Bb^{N+jk}.

By our hypothesis and by p|b, the right side of (1) is divisible by arbitrarily high powers of p as j → ∞. But the left side of (1) is constant, so it must be 0. Thus,

(2)  A_{N}(b^{k} − 1) = Bb^{N}.

We conclude that b^{N}|A_{N}. But 0 ≤ A_{N} < b^{N}, so that A_{N} = 0. Hence (2) implies that B = 0, from which it follows that {a_{n}} is identically zero. This completes the proof of the theorem.

\underline{Reference}

1. C. Cooper and R. E. Kennedy, Private communication, 14 January, 1994.