PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 97(111) (2015), 139–148 DOI: 10.2298/PIM140618001H

IRRATIONALITY MEASURES FOR CONTINUED FRACTIONS WITH ARITHMETIC FUNCTIONS

Jaroslav Hančl and Kalle Leppälä

Abstract. Let f(n) or the base-2 logarithm of f(n) be either d(n) (the divisor function), σ(n) (the divisor-sum function), φ(n) (the Euler totient function), ω(n) (the number of distinct prime factors of n) or Ω(n) (the total number of prime factors of n). We present good lower bounds for | \frac{M}{N} − α| in terms of N, where α = [0; f(1), f(2), …].

1. Introduction and notations

For α ∈ ℝ and N ∈ ℤ_{+} denote J_{α} (N) = N∥Nα∥, where ∥ · ∥ means the distance to the nearest integer. The function J_{α} (N) is connected to the rougher concept of the irrationality exponent μ(α), the infimum of exponents μ such that J(N) ⩽ N^{2−μ} holds for infinitely many N. For almost all α we have μ(α) = 2, although μ(α) = 1 for rational numbers and μ(α) ∈ (2, ∞] for a zero-measure subset of irrational numbers. For more information, see [1] for example. In all of our examples we have the usual μ(α) = 2 but we go further by studying the more refined function J_{α} (N). For irrational α we are interested in finding lower bounds J_{α} (N) ⩾ f(N) for N ⩾ N_{0}. To emphasize that our results are in some sense sharp we also give bounds J_{α} (N) ⩽ g(N) holding for infinitely many N. Throughout the work, this kind of pair of bounds is denoted by

J_{α} (N) ∈ (f(N), g(N)⟩.

Another short-hand notation used in the statement of Theorem 2.1 is to write

L_{k}(N) = \underbrace{loglog⋯ log}_{k times}N.

2010 Mathematics Subject Classification: 11J82; 11J70.

Key words and phrases: continued fraction, arithmetic functions, measure of irrationality.

This work was supported by the European Regional Development Fund in the IT4Innovations Centre of Excellence project (CZ.1.05/1.1.00/02.0070), the grant P201/12/2351 and the Finnish Academy of Science and Letters, the fund of Vilho, Yrjö and Kalle Väisälä.

The authors would like to express their gratitude to the referee for valuable suggestions and Jean-Louis Nicolas for kindly providing rare materials.

Communicated by Aleksandar Ivić.

Because of the law of best approximations, the simple continued fraction expansion of α is ideal for bounding J_{α} (N). Recall that if α = [a_{0}; a_{1}, …] is the simple continued fraction expansion of the irrational number α and p_{n}/q_{n} = [a_{0}; a_{1}, …, a_{n}] is the n-th convergent for each n ∈ ℕ, we have the recursion formulae

p_{0} = a_{0}, q_{0} = 1, p_{1} = a_{1}a_{0} + 1, q_{1} = a_{1},

p_{n+2} = a_{n+2} p_{n+1} + p_{n}, q_{n+2} = a_{n+2} q_{n+1} + q_{n} (1.1)

for n ∈ ℕ, and the estimates

(1.2)  \frac{1}{q_{n}^{2}(a_{n+1} + 2)} < |a − \frac{p_{n}}{q_{n}} | < \frac{1}{q_{n}^{2}a_{n+1}}

for n ∈ ℕ. We shall use the notation [0; \overline{f(j)} ]_{j=1}^{∞} = [0; f(1), f(2), …], where f: ℤ_{+} → ℤ_{+} is a function. For more details on continued fractions please see the book of Hardy and Wright [6] for example.

The case where the asymptotic geometric mean of the sequence {a_{j} }_{j=0}^{∞} tends to infinity is easiest to deal with, although it is untypical in the metric sense. In that case the lowest behavior of the function J_{α} (N) is basically governed by the maximal order and the asymptotic geometric mean of the sequence {a_{j} }_{j=0}^{∞} . In our examples we have chosen a_{n} = f(n) or a_{n} = 2^{f(n)} with some arithmetic function f(n), because maximal orders of arithmetic functions have been extensively studied by Landau, Ramanujan, Nicolas, etc. (see [2, 6, 7, 8, 9] for example). On the other hand, the behavior of the asymptotic geometric mean of arithmetic functions is generally not known. This is not a big problem however; as long as there is enough error in the maximal order of f(n), it suffices to bound the asymptotic geometric mean by asymptotic arithmetic mean, which again is usually known (more typically called the average order of f(n), see [6, 10] for example). And of course in the cases a_{n} = 2^{f(n)} the base-2 logarithm of the geometric mean of a_{n} simply corresponds to the arithmetic mean of f(n).

Finally we note that we have already introduced bounds like (2.2) as an example in our currently unpublished work [4]. However, the result of (2.2) is slightly sharper and presented here for completeness.

2. Results

The following theorem contains our bounds in all of the ten examples. Note that the lower bounds are always asymptotically bigger than any negative power of N, implying that in each case the irrationality exponent is 2.

Theorem 2.1. Let d(n) be the number of positive divisors of n. Then (2.1)

J_{[0;\overline{d(j)} ]_{j=1}^{∞}} (N) ∈ ( 2^{− \frac{L_{2} (N)(L_{3}^{2} (N)+L_{3} (N)+4.7624)}{L_{3}^{3} (N)}}  , 2^{− \frac{L_{2} (N)}{L_{3} (N)−1+O(L_{4} (N)/L_{3} (N))}}  ⟩ ,

(2.2) J_{[0;\overline{2^{d(j)}} ]_{j=1}^{∞}} (N)

∈ ( 2^{−2 \frac{L_{2} (N)(L_{3}^{2} (N)+L_{3} (N)+4.7624)}{L_{3}^{3} (N)}}  , 2^{−2 \frac{L_{2} (N)}{L_{3} (N)−1+O(L_{4} (N)/L_{3} (N))}}  ⟩ ,

where N is big enough.

Let σ(n) be the sum of positive divisors of n. Then

(2.3)  J_{[0;\overline{σ(j)} ]_{j=1}^{∞}} (N) ∈ ( \frac{L_{3}(N)(L_{2}(N) − L_{3}(N))}{L_{1}(N)(e^{γ}L_{3}^{2}(N) + 0.6483)} , \frac{L_{2}(N)(1 + o(1))}{e^{γ}L_{1}(N)L_{3}(N)} ⟩ ,

(2.4) J_{[0;\overline{2^{σ(j)}} ]_{j=1}^{∞}} (N)

∈ ( 2^{− \frac{2\sqrt{3L_{1} (N)} (e^{γ} (L_{3} (N)−L_{1} (2))L_{3} (N)+0.6483)}{π\sqrt{L_{1} (2)} L_{3} (N)}} , 2^{− \frac{2e^{γ}}{π} \sqrt{\frac{3L_{1} (N)}{L_{1} (2)}} L_{3} (N)(1+o(1))} ⟩ ,

where γ is the Euler–Mascheroni constant and N is big enough.

Let φ(n) be the number of positive integers less than and prime to n, denote

A = ∑_{p prime} \frac{1}{p} log( 1 + \frac{1}{p}) ,

let z(x) stand for the inverse of the function y(x) = x logx, and define z_{0}(x) = x and z_{k+1} (x) = x/ log(z_{k}(x)) for each k (see [3]). Then

(2.5) J_{[0;\overline{φ(j)} ]_{j=1}^{∞}} (N)

∈ ( \frac{2L_{1}(N)L_{1}(z(e^{A−1} L_{1}(N))) + L_{2}^{2}(N) − 2L_{1}(N)L_{3}(N) + O(L_{2}(N))}{2L_{1}^{2}(N) + 4L_{1}(N)} ,

\frac{2L_{1}(N)L_{1}(z(e^{A−1} L_{1}(N))) + L_{2}^{2}(N) + O(L_{2}(N))}{2L_{1}^{2}(N)} ⟩ ,

or if one prefers the use of elementary functions only then one can replace L_{1}(z(e^{A−1} L_{1}(N))) by L_{1}(z_{k}(e^{A−1} L_{1}(N))) + O(L_{3}(N)/L_{2}^{k}(N)) for any k, and

(2.6) J_{[0;\overline{2^{φ(j)}} ]_{j=1}^{∞}} (N)

∈ ( 2^{−π\sqrt{\frac{L_{1} (N)}{3 log 2}} +O(L_{2}^{2/3} (N)L_{3}^{3/4} (N))} , 2^{−π\sqrt{\frac{L_{1} (N)}{3 log 2}} +O(L_{2}^{2/3} (N)L_{3}^{3/4} (N))} ⟩ .

Let ω(n) be the number of different prime factors of n, counted without multiplicities. Then

(2.7) J_{[0;\overline{ω(j)} ]_{j=1}^{∞}} (N)

∈ ( \frac{L_{3}^{3}(N)}{L_{2}(N)(L_{3}^{2}(N) + L_{3}(N) + 2.89727)} , \frac{L_{3}(N) − 1 + O( \frac{L_{4} (N)}{L_{3} (N)} )}{L_{2}(N)} ⟩ ,

(2.8) J_{[0;\overline{2^{ω(j)}} ]_{j=1}^{∞}} (N)

∈ ( 2^{− \frac{L_{2} (N)(L_{3}^{2} (N)+L_{3} (N)+2.89727)}{L_{3}^{3} (N)}}  , 2^{− \frac{L_{2} (N)}{L_{3} (N)−1+O(L_{4} (N)/L_{3} (N))}}  ⟩ ,

where N is big enough.

Let Ω(n) be the number of different prime factors of n, counted with multiplicities, and denote

B = 1 − loglog2 + ∫_{2}^{∞} \frac{∑ _{p⩽t} \frac{1}{p} logp − logt dt}{t(logt)^{2}} + ∑_{p prime} \frac{1}{p(p − 1)} .

Then

(2.9) J_{[0;\overline{Ω(j)} ]_{j=1}^{∞}} (N)

∈ ( \frac{L_{1}(2)}{L_{2}(N) − L_{5}(N) + 2L_{1}(2) + o(1)} , \frac{L_{1}(2)}{L_{2}(N) − L_{5}(N) + o(1/L_{4}(N))} ⟩ ,

(2.10) J_{[0;\overline{2^{Ω(j)}} ]_{j=1}^{∞}} (N) ∈ ( \frac{L_{1}(2)(L_{3}(N) + B + o(1))}{L_{1}(N)} , \frac{L_{1}(2)(L_{3}(N) + B + o(1))}{L_{1}(N)} ⟩ .

Proof of Equation (2.1). The recursive formula (1.1) gives us

q_{n} = a_{n}q_{n−1} ( 1 + \frac{q_{n−2}}{a_{n}q_{n−1}} ) = ⋯= ∏_{j=1}^{n} a_{j} ∏_{j=1}^{n−1} ( 1 + \frac{q_{j−1}}{a_{j+1} q_{j}} ) = e^{O(n)} ∏_{j=1}^{n} a_{j} ,

since 1 + q_{j−1} /(a_{j+1} q_{j}) ∈ (1, 2) for all j. By using the asymptotic arithmetic mean

(2.11)  \frac{1}{n} ∑_{j=1}^{n} d(j) = logn + O(1)

(Theorem 320 on page 347 of [6]) and the arithmetic–geometric inequality we get an upper bound logq_{n} ⩽ n loglogn + O(n). On the other hand, the recursive formula (1.1) implies a trivial lower bound logq_{n} ⩾ bn with some positive constant b. Now we can solve n from those inequalities to get

n ⩾ \frac{logq_{n}}{loglogn + O(1)} ⩾ \frac{logq_{n}}{loglog(\frac{1}{b} logq_{n}) + O(1)} = \frac{logq_{n}}{log(loglogq_{n} − logb) + O(1)}

= \frac{logq_{n}}{logloglogq_{n} + O(1/ loglogq_{n}) + O(1)} = \frac{logq_{n}}{logloglogq_{n} + O(1)}

and n ⩽ logq_{n}/b.

To show that the lower bound of the claim always holds we use a result

(2.12)  d(n) ⩽ 2^{^{\frac{log n}{log log n}} (1+ \frac{1}{log log n} + \frac{4.7623...}{(log log n)^{2}} )}

of Robin (Proposition 8 in [8] with the constant rounded up to 4.7624; the fact that the constant is in fact strictly smaller is implicitly in [9]) and the left inequality in (1.2), and we simplify this. Note that rounding the constant 4.7623 … up to 4.7634 causes an error big enough to make many other terms, including the ones with b, negligible.

To show that the upper bound of the claim holds infinitely often we use a slightly weaker estimate. It is easy to show with the estimate p_{n} ∼ n logn of the n-th prime number (Theorem 8 on page 12 of [6]) and Abel’s partial summation formula

(2.13)  ∑_{j=1}^{n} a_{j} b_{j} = b_{n} ∑_{j=1}^{n} a_{j} − ∑_{j=1}^{n−1} (b_{j+1} − b_{j}) ∑_{i=1}^{j} a_{i},

that when n is a product of the first primes we have

(2.14)  d(n) = 2^{\frac{log n}{log log n−1+O(log log log n/ log log n)}} .

The result follows after using the right inequality of (1.2) and simplifying. □

Proof of Equation (2.2). From the asymptotic arithmetic mean (2.11) we deduce

\frac{logq_{n}}{log2} = n logn + O(n) = n logn + O(n loglogn).

Because of the error in bounds (2.12) and (2.14), making our error bigger like this will not matter, but instead simplifies things. Solving for n gives

n = \frac{logq_{n}}{log2 loglogq_{n} + O(logloglogq_{n})} ,

and after simplifying, the claims follow from estimations (2.12), (2.14) and (1.2). □

Proof of Equation (2.3). This time we bound the asymptotic geometric mean by the asymptotic arithmetic mean

(2.15)  \frac{1}{n} ∑_{j=1}^{n} σ(j) = \frac{π^{2}}{12}n + O(logn)

(Theorem 324 on page 351 of [6]) from above, and by the trivial estimate σ(j) ⩾ j+1 together with Stirling’s formula from below to show that logq_{n} = n logn + O(n). Solving for n yields

n = \frac{logq_{n}}{loglogq_{n} − logloglogq_{n} + O(1)} .

To verify our claim on the lower bound we use the estimate

(2.16)  σ(n) ⩽ n( e^{γ} loglogn + \frac{0.6482 …}{loglogn})

of Nicolas (Proposition 11 in [8]), where γ is the Euler–Mascheroni constant, and inequality (1.2). For the upper bound we use Grönwall’s theorem

(2.17)  lim sup_{n→∞} \frac{σ(n)}{n loglogn} = e^{γ}

(equation (25) in [2]), implying that σ(n) = e^{γ}n loglogn(1 + o(1)) for infinitely many values of n, and inequality (1.2). □

Proof of Equation (2.4). Now the asymptotic arithmetic mean (2.15) directly gives us the estimate

logq_{n} = \frac{π^{2} log2}{12}n^{2} + O(n logn).

Solving for n yields

n = \frac{2}{π} \sqrt{\frac{3 logq_{n}}{log2}} + O(loglogq_{n}).

After simplifying, the claims follow from estimations (2.16), (2.17) and (1.2) again. □

Proof of Equation (2.5). We happen to know exactly how the function φ(n) behaves at its biggest. Still we are allowed to have some slack in the estimates of the convergents because of the difficulty of solving n in terms of q_{n}.

We write

∏_{j=1}^{n} φ(j) = ∏_{j=1}^{n} j ∏_{\begin{matrix}p prime \\ p|j \end{matrix}} ( 1 − \frac{1}{p}) = n! ∏_{\begin{matrix}p prime \\ p⩽n \end{matrix}} ( 1 − \frac{1}{p})^{⌊n/p⌋} .

Let us deal with the product over the primes first. By using the prime number theorem (Theorem 6 on page 10 of [6]) and partial integration, we see that

∑_{\begin{matrix}p prime \\ p⩽n \end{matrix}} \frac{1}{p} log( 1 − \frac{1}{p}) = A + O( \frac{1}{n logn} ) ,

where

A = ∑_{p prime} \frac{1}{p} log( 1 − \frac{1}{p})

is a known (negative) constant. So estimating ⌊n/p⌋ ⩽ n/p gives

(2.18)  log( ∏_{\begin{matrix}p prime \\ p⩽n \end{matrix}} ( 1 − \frac{1}{p})^{⌊n/p⌋} ) ⩾ An + O( \frac{1}{logn}) .

On the other hand, by Mertens’s theorem

∏_{\begin{matrix}p prime \\ p⩽n \end{matrix}} ( 1 − \frac{1}{p}) = \frac{1 + o(1)}{e^{γ} logn}

(Theorem on 429 on page 466 of [6]), where γ is the Euler–Mascheroni constant, estimating ⌊n/p⌋ ⩾ n/p − 1 gives

(2.19)  log( ∏_{\begin{matrix}p prime \\ p⩽n \end{matrix}} ( 1 − \frac{1}{p})^{⌊n/p⌋} ) ⩽ An + loglogn + γ + o(1).

Next note that by recursive formula (1.1) and Landau’s theorem

lim inf_{n→∞} \frac{φ(n) loglogn}{n} = \frac{1}{e^{γ}}

(pages 217–219 of [7]), where γ is the Euler–Mascheroni constant again, we see

∏_{j=1}^{n−1} ( 1 + \frac{q_{j−1}}{φ(j + 1)q_{j}} ) = O(1)

as a beginning of a converging product. Finally we use Stirling’s formula and bounds (2.18) and (2.19) to get

logq_{n} ⩽ n logn + (A − 1)n + \frac{1}{2} log(2πn) + loglogn + O(1)
logq_{n} ⩾ n logn + (A − 1)n + \frac{1}{2} log(2πn) + O(1).

In any case we have at least

n = \frac{logq_{n}}{logn + A − 1 + O(1/n · logn)} ,

which is equivalent to

e^{A−1} n log(e^{A−1} n) = e^{A−1} logq_{n} + O(logn).

Now we apply the function z(x) (the inverse of x logx) and the logarithm function to both sides of the equation to get

logn + A − 1 = logz (e^{A−1} logq_{n} + O(logn)) .

The mean-value theorem implies

logn + A − 1 = logz (e^{A−1} logq_{n} + O(logn))
= log( z(e^{A−1} logq_{n}) + O( \frac{logn}{logz(logq_{n})}) )
= logz(e^{A−1} logq_{n}) + O( \frac{logn}{logq_{n}}) ,

Using this we can solve n:

n ⩾ \frac{logq_{n}}{logz(e^{A−1} logq_{n}) + \frac{(log log q_{n} )^{2}}{2 log q_{n}} + O( \frac{log log q_{n}}{log q_{n}} )}

n ⩽ \frac{logq_{n}}{logz(e^{A−1} logq_{n}) + \frac{(log log q_{n} )^{2}}{2 log q_{n}} − logloglogq_{n} + O( \frac{log log q_{n}}{log q_{n}} )} .

Now the claims follow from using (1.2), since φ(n) ⩽ n − 1 with equality whenever n is a prime.

We can also derive bounds that use only elementary functions but have bigger error. By using the fact that logz(e^{A−1} logq_{n}) as well as logz_{k}(e^{A−1} logq_{n}) for any

146 HANČL AND LEPPÄLÄ

k is O(loglogq_{n}), we see that

logz(e^{A−1} logq_{n}) = log( \frac{e^{A−1} logq_{n}}{logz(e^{A−1} logq_{n})})
= A − 1 + loglogq_{n} + O(logloglogq_{n})
= logz_{0}(e^{A−1} logq_{n}) + O(logloglogq_{n})

and inductively

logz(e^{A−1} logq_{n}) = log( \frac{e^{A−1} logq_{n}}{logz(e^{A−1} logq_{n})})

= log( \frac{e^{A−1} logq_{n}}{logz_{k−1} (e^{A−1} logq_{n}) + O( \frac{log log log q_{n}}{(log log q_{n} )^{k−1}} )} )

= A − 1 + loglogq_{n} − loglogz_{k−1} (e^{A−1} logq_{n}) + O( \frac{logloglogq_{n}}{(loglogq_{n})^{k}})

= logz_{k}(e^{A−1} logq_{n}) + O( \frac{logloglogq_{n}}{(loglogq_{n})^{k}})

for any k. □ Proof of Equation (2.6). From the asymptotic arithmetic mean

\frac{1}{n} ∑_{j=1}^{n} φ(j) = \frac{3}{π^{2}} n + O( (logn)^{\frac{2}{3}} (loglogn)^{\frac{3}{4}} )

(see [10]) we get

logq_{n} = \frac{3 log2}{π^{2}}n^{2} + O( n(logn)^{\frac{2}{3}} (loglogn)^{\frac{3}{4}} ) .

Solving for n yields

n = π\sqrt{\frac{logq_{n}}{3 log2}} + O( (loglogq_{n})^{\frac{2}{3}} (logloglogq_{n})^{\frac{3}{4}} ) ,

and so the claim follows by using (1.2), since φ(n) ⩽ n − 1, with equality whenever n is a prime. □

Proof of Equation (2.7). The asymptotic arithmetic mean

(2.20)  \frac{1}{n} ∑_{j=1}^{n} ω(j) = loglogn + O(1)

(Theorem 430 on page 472 of [6]) implies logq_{n} ⩽ n logloglogn+O(n), and trivially logq_{n} ⩾ bn with some positive constant b. Solving for n gives

\frac{logq_{n}}{loglogloglogq_{n} + O(1)} ⩽ n ⩽ \frac{logq_{n}}{b}.

To see that the lower bound of the claim always holds we use a result

(2.21)  ω(n) ⩽ \frac{logn}{loglogn}( 1 + \frac{1}{loglogn} + \frac{2.89726 …}{(loglogn)^{2}})

of Nicolas (Proposition 5 in [8]) and inequality (1.2). For the upper bound we use the fact that when n is a product of first primes, we have

(2.22)  ω(n) = \frac{logn}{loglogn − 1 + O( \frac{log log log n}{log log n} )} .

The claims follow from this and (1.2). □

Proof of Equation (2.8). Now the asymptotic arithmetic mean (2.20) directly gives us the estimate

\frac{logq_{n}}{log2} = n loglogn + O(n),

from which we solve

n = \frac{logq_{n}}{log2 logloglogq_{n} + O(1)} .

The claims follow from inequalities (2.21), (2.22) and (1.2). □

Proof of Equation (2.9). Again we want to be sharper than usual because we know the exact worst-case behavior of Ω(n). We shall use a theorem of Hardy and Ramanujan (Theorem C′ in [5]), stating that whenever f(n) is a function tending to infinity, we have

(2.23)  loglogn − f(n)\sqrt{loglogn} ⩽ Ω(n) ⩽ loglogn + f(n)\sqrt{loglogn}

for almost all n. In particular, by choosing f(n) = (loglogn)^{\frac{1}{4}} and using Abel’s summation formula (2.13) we get an estimate

(2.24) ∏_{j=1}^{n−1} ( 1 + \frac{q_{j−1}}{a_{j+1} q_{j}} ) ⩽ ∏_{j=1}^{n(1+o(1))} ( 1 + \frac{1}{loglogj − (loglogj)^{\frac{3}{4}}} ) e^{o(n)} = e^{o(n)} .

Now this and the asymptotic arithmetic mean

(2.25)  \frac{1}{n} ∑_{j=1}^{n} Ω(j) = loglogn + B + o(1),

where

B = 1 − loglog2 + ∫_{2}^{∞} \frac{∑ _{p⩽t} \frac{1}{p} logp − logt dt}{t(logt)^{2}} + ∑_{p prime} \frac{1}{p(p − 1)}

is a known constant (Theorem 430 on page 472 of [6]) imply

logq_{n} ⩽ n logloglogn + o(n).

As a lower bound we only get, by using (2.23), that

logq_{n} ⩾ n logloglogn(1 + o(1)).

Solving for n yields

\frac{logq_{n}}{loglogloglogq_{n} + o(1)} ⩽ n ⩽ \frac{logq_{n}}{loglogloglogq_{n}(1 + o(1))} .

Because obviously Ω(n) ⩽ logn/ log2, with equality whenever n is a power of 2, both claims now follow by using (1.2). □

Proof of Equation (2.10). We may use the upper bound (2.24) and the asymptotic arithmetic mean (2.25) to get the estimate

\frac{logq_{n}}{log2} = n loglogn + Bn + o(n).

Now

n = \frac{logq_{n}}{log2(logloglogq_{n} + B + o(1))} ,

and so the claim follows by using (1.2), since Ω(n) ⩽ logn/ log2, with equality whenever n is a power of 2. □

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Department of Mathematics and (Received 25 04 2014) Centre of Excellence IT4Innovation, division of UO, Institute for Research and Applications of Fuzzy Modeling University of Ostrava, Ostrava, Czech Republic

 hancl@osu.cz

Department of Mathematical Sciences University of Oulu, Oulu, Finland

 kalle.leppala@oulu.fi