CHARACTERIZING THE 2-ADIC ORDER OF THE LOGARITHM

T. Lengyel

Occidental College, 1600 Campus Road, Los Angeles, CA 90041 (Submitted March 1993)

1. Introduction

We define ν_{p}(x) as the highest power of prime p which divides the integer x. The function ν_{p}(x) is often called the p-adic order of x. In this paper we characterize the divisibility by 2 of the series ∑_{k=1}^{∞}(−1)^{k−1}x^{k}/k and ∑_{k=1}^{∞} \frac{x^{k}}{k}, i. e., we determine their 2-adic orders. The characterization generalizes previously known results on 2-adic orders and is based on elementary proofs.

\underline{2. Results}

For an integer x, the p-adic order ν_{p}(x) of x is the highest power of prime p which divides x. We can think of the relations p|x and p |̸x as ν_{p}(x) ≥ 1 and ν_{p}(x) = 0, respectively.

We set ν_{p}(0) = ∞ and ν_{p}(x/y) = ν_{p}(x) − ν_{p}(y) if both x and y are integers. Therefore, for all nonzero rational numbers the order is defined to be a finite integer. From now on all rational numbers will be meant in lowest terms.

For rational numbers a_{k}(k ≥ 0) and rational x, the p-adic order, ν_{p}(∑_{k=0}^{∞} a_{k}x^{k}) of the series ∑_{k=0}^{∞} a_{k}x^{k} can be introduced as lim_{n→∞} ν_{p}(∑_{k=0}^{n} a_{k}x^{k}) if the limit exists, in which case there exists an n_{0} such that ν_{p}(∑_{k=0}^{n} a_{k}x^{k}) = ν_{p}(∑_{k=0}^{∞} a_{k}x^{k}) for n ≥ n_{0}. To illustrate this, we consider the series \frac{x}{1−x} = x + x^{2} + x^{3} + . . .. The reader can easily verify that ν_{p}(\frac{x}{1−x}) = ν_{p}(x) if ν_{p}(x) ≥ 1 and the limit does not exist if ν_{p}(x) ≤ 0. Actually, ν_{p}(x + x^{2} + x^{3} + … + x^{n}) = nν_{p}(x) if ν_{p}(x) < 0. Notice that if ν_{2}(x) = 0 then ν_{2}(x+x^{2} +x^{3} +…+x^{2n+1}) = 0, while ν_{2}(x+x^{2} +x^{3} +…+x^{2^{n}}) ≥ n. Finding the p-adic order of functions helps analyzing the divisibility property of the underlying or related functions. We note that Clarke [1] has recently studied the p-adic order of the logarithm by using p-adic arguments in order to characterize the divisibility properties of the Stirling and partial Stirling numbers. The interested reader should consult a book on p-adic metrics (e.g, [2]) for a general treatise of p-adic power series.

In this paper we consider the series log (1 + x) = ∑_{k=1}^{∞}(−1)^{k−1}x^{k}/k and − log (1 − x) = ∑_{k=1}^{∞} \frac{x^{k}}{k} and determine their 2-adic orders by elementary arguments based on binomial expansion.

In most cases the p-adic order of log(1+x) can be derived by the well-known

Theorem A. (Yu [4]) We have ν_{p}( log(1 + x)) = ν_{p}(∑_{k=1}^{∞}(−1)^{k−1}x^{k}/k) = ν_{p}(x) if ν_{p}(x) > \frac{1}{p−1}, and ν_{p}( log(1 + x)) does not exist if ν_{p}(x) ≤ 0. In particular, for any integer x, ν_{p}( log(1+x)) = ν_{p}(x) if p ≥ 3 and p|x, or if p = 2 and 4|x, while for p |̸x the p-adic order ν_{p}( log(1 + x)) does not exist.

In fact, Theorem A completely describes the p-adic order for p ≥ 3. The purpose of this paper is to characterize the 2-adic orders of the two series in the case not covered by Theorem A, i. e., for every even integer x and p = 2. We note that the proof of Theorem A is based on the observation that under the conditions of Theorem A given for p and x, the p-adic order of the terms (−1)^{k−1} \frac{x^{k}}{k} , k ≥ 2, of the infinite series ∑_{k=1}^{∞}(−1)^{k−1} \frac{x^{k}}{k} exceeds that of the first term, x (cf. [2], p. 81).

If p = 2 and x = 2 then the following lemma (cf. [2], Exercise 7, p. 83) describes the 2-adic “behavior” of ∑_{k=1}^{n} \frac{2^{k}}{k}, i. e., the sum of the first n terms of the expansion − log(1 − x). Lemma B. The 2-adic order of the rational number ∑_{k=1}^{n} \frac{2^{k}}{k} approaches infinity as n increases.

An elementary proof can be given based on the observation that ν_{2}( ∑_{k=n+1}^{∞} \frac{2^{k}}{k} ) ≥ min_{k≥n+1}(k − ν_{2}(k)), which assures that ν_{2}( ∑_{k=n+1}^{∞} \frac{2^{k}}{k} ) becomes arbitrary large as n → ∞. One can prove that ν_{2}(∑_{k=1}^{n} \frac{2^{k}}{k} ) ≥ ν_{2}( ∑_{k=n+1}^{∞} \frac{2^{k}}{k} ) holds for infinitely many values n. In fact, a p-adic argument shows that equality holds for all n. We leave the details to the reader.

We set ν_{p}(∑_{k=0}^{∞} a_{k}x^{k}) = ∞ if, for every integer N ≥ 1 there exists an integer n_{0} such that p^{N} divides ∑_{k=0}^{n} a_{k}x^{k} for every n ≥ n_{0}. In this case ν_{p}(∑_{k=0}^{n} a_{k}x^{k}) = ν_{p}(∑_{k=n+1}^{∞} a_{k}x^{k}) holds. By the Lemma, we set ν_{2}(∑_{k=1}^{∞} \frac{2^{k}}{k} ) = ∞. We note that 0 and 2 play a special role in the 2-adic analysis of log(1 − x) for these are the values for which ν_{2}( log(1 − x)) = ∞ ([2]). Our results are summarized in the following two theorems.

Theorem 1. For any even positive integer x,

ν_{2}(∑_{k=1}^{∞}(−1)^{k−1} \frac{x^{k}}{k} ) = \{ \begin{matrix}2, & if x = 2, \\ 2, & if x ≡ 2 (mod 16), \\ 2, & if x ≡ 4 (mod 16), \\ 3, & if x ≡ 6 (mod 16), \\ 3, & if x ≡ 8 (mod 16), \\ 2, & if x ≡ 10 (mod 16), \\ 2, & if x ≡ 12 (mod 16), \\ ν_{2}(x + 2), & if x ≡ 14 (mod 16), \\ ν_{2}(x), & if x ≡ 0 (mod 16). \end{matrix}

Theorem 2. For any even positive integer x,

ν_{2}(∑_{k=1}^{∞} \frac{x^{k}}{k} ) = \{ \begin{matrix}∞, & if x = 2, \\ ν_{2}(x − 2), & if x ≡ 2 (mod 16), \\ 2, & if x ≡ 4 (mod 16), \\ 2, & if x ≡ 6 (mod 16), \\ 3, & if x ≡ 8 (mod 16), \\ 3, & if x ≡ 10 (mod 16), \\ 2, & if x ≡ 12 (mod 16), \\ 2, & if x ≡ 14 (mod 16), \\ ν_{2}(x), & if x ≡ 0 (mod 16). \end{matrix}

Remark 1. The above theorems could be restated in a more compact form:

ν_{2}(∑_{k=1}^{∞}(−1)^{k−1} \frac{x^{k}}{k} ) = \{ \begin{matrix}ν_{2}(x), & if x ≡ 0, 4, 8, 12 (mod 16), \\ ν_{2}(x + 2), & if x ≡ 2, 6, 10, 14 (mod 16), \end{matrix}

and

ν_{2}(∑_{k=1}^{∞} \frac{x^{k}}{k} ) = \{ \begin{matrix}ν_{2}(x), & if x ≡ 0, 4, 8, 12 (mod 16), \\ ν_{2}(x − 2), & if x ≡ 2, 6, 10, 14 (mod 16). \end{matrix}

Notice the sharp contrast between ν_{2}(∑_{k=1}^{∞}(−1)^{k−1}2^{k}/k) and ν_{2}(∑_{k=1}^{∞} \frac{2^{k}}{k} ). We can combine the cases x ≠ 2 of the two theorems by substituting −x in place of x and carrying out the modular calculations.

For a rational x = \frac{a}{b} with ν_{2}(x) = 1 and b > 1, there exists a sufficiently large integer m such that ν_{2}( log(1 + x)) < m. We set x′ = a ∗ b^{−1}, where b^{−1} is the unique solution to the equation b ∗ b^{−1} ≡ 1 (mod 2^{m}) with 0 < b^{−1} < 2^{m}. We can proceed to determine ν_{2}( log(1 + x′)) by Theorem 1 and observing that ν_{2}( log(1 + x)) = ν_{2}( log(1 + x′)). If x′ ≢ 14 (mod 16), then m = 4 is an appropriate choice. However, if it turns out that the remainder is 14, then one should check whether ν_{2}(x′ + 2) < m and try a larger m if it fails. A similar method works for determining ν_{2}( log(1 − x)), too.

For example, if x = 6/5 then ν_{2}( log(1 − 6/5)) = 2 follows easily with m = 4. We use m = 5 and have x′ = 6 ∗ 13 ≡ 14 (mod 16) in order to obtain ν_{2}( log(1 + 6/5)) = ν_{2}(6 ∗ 13 + 2) = 4. For x = 426/555, we start with m = 4. Since x′ = 426 ∗ 3 ≡ 14 (mod 16) and ν_{2}(426 ∗ 3 + 2) = 8, we note that we need a larger m. By using m = 10, we obtain x′ = 426 ∗ 131 ≡ 14 (mod 16) and ν_{2}( log(1 + 426/555)) = ν_{2}(426 ∗ 131 + 2) = 9.

Remark 2. Similarly to the proof of Theorem A, we observe that ν_{2}(2^{s}) < ν_{2}((2^{s})^{k}/k) if k ≥ 2 and s ≥ 2. Therefore, ν_{2}(∑_{k=1}^{∞} \frac{(2^{s})^{k}}{k} ) = ν_{2}(∑_{k=1}^{∞}(−1)^{k−1} \frac{(2^{s})^{k}}{k} ) = ν_{2}(2^{s}) = s if s ≥ 2.

\underline{3. Proofs}

Proof of Theorem 1. The case of x = 2 is easily verified by checking the first couple of terms of ∑_{k=1}^{∞}(−1)^{k−1}x^{k}/k. Indeed, ν_{2}(∑_{k=1}^{4}(−1)^{k−1}2^{k}/k) = 2 and ν_{2}(2^{k}/k) > 2 for k ≥ 5.

If x = 6 or 10 then by inspecting the sum of the first few terms we obtain, similarly to the case of x = 2, that the orders are 3 and 2, respectively.

We can extend these results for x ≡ 2, 6, and 10 (mod 16). From now on a denotes an arbitrary integer while b is an arbitrary odd integer. The basic idea is that if ν_{2}(∑_{k=1}^{∞}(−1)^{k−1} \frac{x^{k}}{k} ) = r < s then ν_{2}(∑_{k=1}^{∞}(−1)^{k−1} \frac{(x+a2^{s})^{k}}{k} ) = r too, since x^{k} ≡ (x + a2^{s})^{k} (mod 2^{s}). [Of course, the same applies if we omit the factors (−1)^{k−1}.] By the previous observations, we can set s = 4.

For x ≡ 0, 4, 8 or 12 (mod 16) the statement follows from Theorem A which claims that the order must be ν_{2}(x).

Instead of simply proving the remaining case x ≡ 14 (mod 16), we combine the cases x ≡ 2 and 14 (mod 16) to make this proof transparent to prove Theorem 2. Let s = 4. We calculate the 2-adic order of ∑_{k=1}^{∞}(−1)^{k−1}x^{k}/k using the binomial expansion of the terms x^{k} = (b2^{s} + 2c)^{k} where c is either 1 or −1. The expansion yields

(b2^{s} + 2c)^{k} = (2 (b2^{s−1} + c))^{k} = ∑_{l=0}^{k} 2^{k}\binom{k}{l}(b2^{s−1})^{l}c^{k−l}.

Note that the identity \binom{k}{l} = \frac{k}{l} \binom{k−1}{l−1} implies that \binom{k}{l}/k is an integer multiple of 1/l. Consider the sum ∑_{k=1}^{∞}(−1)^{k−1} \frac{(b2^{s}+2c)^{k}}{k} in three terms, one term for l = 0, another for l = 1, and the last one for all the remaining cases, l ≥ 2. We get

(1)  \begin{matrix}∑_{k=1}^{∞}(−1)^{k−1} \frac{(b2^{s} + 2c)^{k}}{k} = \\ = − ∑_{k=1}^{∞} \frac{(−2c)^{k}}{k} + ∑_{k=1}^{∞} b2^{k+s−1}(−c)^{k−1}+ \\ + ∑_{k=1}^{∞}(−1)^{k−1} ∑_{l=2}^{k} \frac{\binom{k−1}{l−1}b^{l}2^{l(s−1)+k}c^{k−l}}{l} . \end{matrix}

Obviously, the 2-adic order of the second term is s if b ≠ 0. Notice that the third term is always divisible by 2^{s+1} for s ≥ 3, since this condition implies that l(s − 1) + k − ν_{2}(l) ≥ l(s − 1) + k − log_{2} l ≥ s + 1. It turns out that the 2-adic order of the first term on the right side of identity (1) is 2 if c = 1 as we have seen it at the beginning of the proof. By Lemma B, the 2-adic order of the first term is ∞ if c = −1. It follows that ν_{2}(∑_{k=1}^{∞}(−1)^{k−1}(b2^{s} + 2c)^{k}/k) = s if c = −1 (and b ≠ 0), while it is 2 if c = 1.

Proof of Theorem 2. Basically, the proof of Theorem 1 can be repeated here except for x = 2, which case is the content of Lemma B. Careful inspection reveals that the 2-adic orders are switched for x ≡ 6 and 10 (mod 16).

Similarly to identity (1), we have

(2)  \begin{matrix}∑_{k=1}^{∞} \frac{(b2^{s} + 2c)^{k}}{k} = ∑_{k=1}^{∞} \frac{(2c)^{k}}{k} + \\ + ∑_{k=1}^{∞} b2^{k+s−1}c^{k−1} + ∑_{k=1}^{∞} ∑_{l=2}^{k} \frac{\binom{k−1}{l−1}b^{l}2^{l(s−1)+k}c^{k−l}}{l} \end{matrix}

where the last term is always divisible by 2^{s+1} for s ≥ 3.

By simply switching the cases c = 1 and c = −1 in the previous proof and using identity (2), we derive that ν_{2}(∑_{k=1}^{∞} \frac{(b2^{s}+2c)^{k}}{k} ) = s if c = 1 (and b ≠ 0), while it is 2 if c = −1.

We note that Clarke [1] has recently proved similar results by using p-adic arguments.

Lemma B points to the odd behavior of ν_{2}(∑_{k=1}^{n} \frac{x^{k}}{k} ) at x = 2. Analysis of this behavior gives rise to the question on the rate at which ν_{2}(∑_{k=1}^{n} \frac{2^{k}}{k} ) increases as n gets larger. We were unable to answer this question; however, numerical evidence suggests some pattern for the increase of the 2-adic order. The following conjecture has been proposed in [3], in the context of the divisibility by 2 of the Stirling numbers of the second kind, S(a 2^{n} − 1, 2^{m}) where n > m ≥ 4 and a is a positive integer.

Conjecture 3. For m ≥ 4, ν_{2}(∑_{k=1}^{2^{m}} \frac{2^{k}}{k} ) = 2^{m} + 2m − 2.

Acknowledgment

The author wishes to thank an anonymous referee for his/her remarks and suggestions which improved the clarity of the presentation of this paper.

\underline{References}

1. Clarke, F., Hensel’s lemma and the divisibility by primes of Stirling-like numbers, preprint, 1993

2. Koblitz, N., p-adic numbers, p-adic analysis, and zeta-functions, 2nd edition, New York: Springer 1984

3. Lengyel, T., On the divisibility by 2 of the Stirling numbers of the second kind. (to appear in The Fibonacci Quarterly)

4. Yu, K., Linear-forms in p-adic logarithms, Acta Arithmetica, 53(1989), 107-186.

AMS Classification Numbers: 11E95, 11B50

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