An Interesting Lemma for Regular C-fractions

Kwang-Wu Chen Department of International Business Management Ching Yun University No. 229, Jianshing Road, Jungli City Taoyuan, Taiwan 320, R.O.C.

                       kwchen@cyu.edu.tw

Abstract

In this short note we give an interesting lemma for regular C-fractions. Applying this lemma we obtain some congruence properties of some classical numbers such as the Springer numbers of even index, the median Euler numbers, the median Genocchi numbers, and the tangent numbers.

1 The interesting lemma

A regular C-fraction is a continued fraction of the form

a_{0} + K_{n=1}^{∞}(a_{n}z/1) = a_{0} + \frac{a_{1}z}{1} + \frac{a_{2}z}{1} + \frac{a_{3}z}{1} + …
= a_{0} + \frac{a_{1}z}{1 + \frac{a_{2}z}{1 + \frac{a_{3}z}{...}}} ,

where a_{n} ∈ ℂ.

Let f(z) = ∑_{n=0}^{∞} c_{n}z^{n} ∈ ℂ[[z]] be a formal power series. It is known that there exists a one-to-one correspondence between regular C-fractions a_{0} + K_{n=1}^{∞}(a_{n}z/1) and formal power series ∑_{n=0}^{∞} c_{n}z^{n} [6, pp. 252–265].

Now we assume that all coefficients are integral. The lemma we state here gives the division relation between the integral coefficients of the regular C-fraction and the integral coefficients of its corresponding formal power series.

Lemma 1. Let f(z) = ∑_{n=0}^{∞} a_{n}z^{n} ∈ ℤ[[z]] be an integral formal power series. Assume the corresponding uniquely determined regular C-fraction is

∑_{n=0}^{∞} a_{n}z^{n} = \frac{b_{0}}{1} + \frac{bb_{1}z}{1} + \frac{bb_{2}z}{1} + …,  (1)

where b and (b_{n})_{n≥0} are integral. Then a_{n} is divisible by (b_{0}b_{1}b^{n}) for n ≥ 1.

Proof. Setting z = y/b, Equation (1) becomes

∑_{n=0}^{∞} a_{n}(\frac{y}{b})^{n} = \frac{b_{0}}{1} + \frac{b_{1}y}{1} + \frac{b_{2}y}{1} + \frac{b_{3}y}{1} + …
= b_{0} − \frac{b_{0}b_{1}y}{1 + b_{1}y} + \frac{b_{2}y}{1} + \frac{b_{3}y}{1} + ….

Since a_{0} = b_{0}, we have

∑_{n=1}^{∞} \frac{a_{n}}{b_{0}b_{1}b^{n}} y^{n} = \frac{−y}{1 + b_{1}y} + \frac{b_{2}y}{1} + \frac{b_{3}y}{1} + ….

Since the right-hand side of the above identity can be uniquely expressed as a formal power series with integral coefficients, we conclude the proof.

Let f(t) = ∑_{n} a_{n}t^{n} and g(t) = ∑_{n} b_{n}t^{n} (n ≥ 0) be two formal power series with integral coefficients. For a non-negative integer m we write

f(t) ≡ g(t) (mod m) iff a_{n} ≡ b_{n} (mod m) for all n ≥ 0.  (2)

Applying Lemma 1 we can obtain some congruence properties of some classical numbers such as the Springer numbers of even index, the median Euler numbers, the median Genocchi numbers, and the tangent numbers.

2 Applications

The Springer numbers ([1, p. 275]) are defined by

S(x) = e^{x} sech 2x = ∑_{n=0}^{∞} \frac{S_{n}x^{n}}{n!} .  (3)

The even (resp. odd) part of the Springer numbers is what Glaisher ([1, p. 276]) called the numbers P_{n} (resp. Q_{n}). That is to say,

\frac{cosh x}{cosh 2x} = ∑_{n=0}^{∞} \frac{S_{2n}x^{2n}}{(2n)!} ,  \frac{sinh x}{cosh 2x} = ∑_{n=0}^{∞} \frac{S_{2n+1}x^{2n+1}}{(2n + 1)!} .  (4)

Springer introduced these numbers for a problem about root systems, and Arnold showed these numbers as counting various types of snakes ([4, p. 6–p. 7]).

Following the notation and the result in Corollary 3.3 of [1] we put

p(x) = ∑_{n=0}^{∞} S_{2n}x^{2n+1} = x − 3x^{3} + 57x^{5} − …

= \frac{x}{1} + \frac{3x^{2}}{1} + \frac{16x^{2}}{1} + \frac{35x^{2}}{1} + …
+ \frac{16n^{2}x^{2}}{1} + \frac{(4n + 1)(4n + 3)x^{2}}{1} + ….  (5)

Note that our definition of the Springer numbers S_{2n} differs from that in [1]. The unsigned sequence (−1)^{n}S_{2n} : 1, 3, 57, 2763, 250737, …, is the sequence A000281 in [7]. Applying Lemma 1 we have S_{2n} is divisible by 3. Moreover, we have the following theorem.

Theorem 1. For n ≥ 1, the Springer number with even index S_{2n} is divisible by 3 and

\frac{S_{2n}}{3} ≡ (−1)^{n}3^{n−1}  (mod 16).  (6)

Proof. Multiplying x into p(x) and setting t = x^{2}, we have

∑_{n=0}^{∞} S_{2n}t^{n+1} = t − 3t^{2} + 57t^{3} − ⋯ = \frac{t}{1} + \frac{3t}{1} + \frac{16t}{1} + ….

Applying Lemma 1, S_{2n} is divisible by 3 for n ≥ 1. And

∑_{n=1}^{∞} \frac{S_{2n}}{3} t^{n+1} = \frac{−t^{2}}{1 + 3t} + \frac{16t}{1} + \frac{35t}{1} + …  (7)

≡ \frac{−t^{2}}{1 + 3t}  (mod 16)
= ∑_{n=1}^{∞}(−1)^{n}3^{n−1}t^{n+1}.

Comparing the coefficients of t^{n+1}, we have

\frac{S_{2n}}{3} ≡ (−1)^{n}3^{n−1}  (mod 16),  n ≥ 1.

Remark 1. Now we write Equation (7) as

\frac{−t^{2}}{1 + 3t} + \frac{16t}{1} + \frac{35t}{1} + … = \frac{−t^{2}}{1 + 3t} + K_{n=1}^{∞} (\frac{c_{n}t}{1}) ,

where c_{2n−1} = 16n^{2} and c_{n} = (4n + 1)(4n + 3), for n ≥ 1.

If we take the modulus c_{2} = 35 instead of c_{1} = 16 for Equation (7) in the above proof. Then we have

∑_{n=1}^{∞} \frac{S_{2n}}{3} t^{n+1} ≡ \frac{−t^{2}}{1 + 19t}  (mod 35)

≡ \frac{−t^{2}}{1 − 16t}  (mod 35)
= ∑_{n=1}^{∞}(−16^{n−1})t^{n+1}.

Comparing the coefficients of t^{n+1}, we have

\frac{S_{2n}}{3} ≡ −16^{n−1}  (mod 35),  n ≥ 1.  (8)

Since 16^{3} ≡ 1 (mod 35), we can also write Equation (8) as follows: for k ≥ 1,

\frac{S_{2n}}{3} ≡ \{ \begin{matrix}34 (mod 35), & if n = 3k − 2; \\ 19 (mod 35), & if n = 3k − 1; \\ 24 (mod 35), & if n = 3k. \end{matrix}  (9)

Similarly, we take another c_{n} as the modulus for Equation (7), then we can get the congruences for S_{2n}/3 under the modulus c_{n}.

Let us define the Euler numbers E_{n} through the exponential generating function E(x):

E(x) = sech x + tanh x = ∑_{n=0}^{∞} \frac{E_{n}x^{n}}{n!} .

We construct the Seidel matrix (a_{n,m})_{n,m≥0} associated with the sequence (0, E_{1}, E_{2}, E_{3}, …) as follows:

1. The first row (a_{0,n})_{n≥0} of the matrix is the initial sequence (0, E_{1}, E_{2}, E_{3}, …).

2. Each entry a_{n,m} of the n-th row is the sum of the entry immediately above and of the entry above and to the right of it:

a_{n,m} = a_{n−1,m} + a_{n−1,m+1}.

The resulting Seidel matrix is

0 1 −1 −2  5 16 −61 ⋯
1 0 −3 3 21 −45 ⋯
1 −3 0 24 −24 ⋯
−2 −3 24 0 ⋯
−5 21 24 ⋯
16 45 ⋯
61 ⋯
⋯

The absolute values of the upper diagonal sequence 1, 3, 24, 402, … are called the median Euler numbers R_{n} (see [1, Section 4] or [7, Sequence A002832]). Using the same method as above, we have

Theorem 2. For n ≥ 1, the median Euler number R_{n} is divisible by 3 and

\frac{R_{n}}{3} ≡ 3^{n−1}  (mod 5).  (10)

Proof. Since the ordinary generating function of the median Euler numbers R_{n} satisfies the continued fraction representation [1, Proposition 7]:

r(x) = ∑_{n=0}^{∞}(−1)^{n}R_{n}x^{n+1} = x − 3x^{2} + 24x^{3} − 402x^{4} + 11616x^{5} − ⋯
= \frac{x}{1} + \frac{3x}{1} + \frac{5x}{1} + \frac{2 · 7x}{1} + \frac{2 · 9x}{1} + ….  (11)

Applying Lemma 1, R_{n} is divisible by 3 for n ≥ 1. And

∑_{n=1}^{∞}(−1)^{n} \frac{R_{n}}{3} x^{n+1} = \frac{−x^{2}}{1 + 3x} + \frac{5x}{1} + \frac{14x}{1} + …

≡ \frac{−x^{2}}{1 + 3x}  (mod 5)
= ∑_{n=1}^{∞}(−1)^{n}3^{n−1}x^{n+1}.

Comparing the coefficients of x^{n+1}, we complete the proof.

The Genocchi numbers G_{n} [7, Sequence A036968] are defined by

\frac{2x}{e^{x} + 1} = ∑_{n=0}^{∞} \frac{G_{n}x^{n}}{n!} .

The median Genocchi numbers H_{2n+1} (see [1, 2], or [7, Sequence A005439]) can be defined by H_{1} = 1 and

H_{2n+1} = ∑_{k=0}^{⌊ \frac{n−1}{2} ⌋} \binom{n}{2k + 1}G_{2n−2k}, n ≥ 1,

where ⌊x⌋ denotes the greatest integer not exceeding x.

Theorem 3. For n ≥ 1, the median Genocchi number H_{2n+3} is divisible by 2^{n} and

\frac{H_{2n+3}}{2^{n}} ≡ \{ \begin{matrix}1 (mod 6), & if n is odd; \\ 4 (mod 6), & if n is even. \end{matrix}  (12)

Proof. Since the ordinary generating function of the median Genocchi numbers H_{2n+1} satisfies the continued fraction representation [1, p. 295]

h(x) = ∑_{n=0}^{∞} H_{2n+1}x^{n+1} = x − x^{2} + 2x^{3} − 8x^{4} + 56x^{5} − ⋯

= \frac{x}{1} + \frac{x}{1} + \frac{x}{1} + \frac{2^{2}x}{1} + \frac{2^{2}x}{1} + \frac{3^{2}x}{1} + \frac{3^{2}x}{1} + ….  (13)

From [1, Lemma 1] we have

\frac{x}{1} + \frac{c_{1}x}{1} + \frac{c_{2}x}{1} + \frac{c_{3}x}{1} + …
= x − \frac{c_{1}x^{2}}{1 + (c_{1} + c_{2})x} − \frac{c_{2}c_{3}x^{2}}{1 + (c_{3} + c_{4})x} − \frac{c_{4}c_{5}x^{2}}{1 + (c_{5} + c_{6})x} − …  (14)
= \frac{x}{1 + c_{1}x} − \frac{c_{1}c_{2}x^{2}}{1 + (c_{2} + c_{3})x} − \frac{c_{3}c_{4}x^{2}}{1 + (c_{4} + c_{5})x} − ….  (15)

Then we can rewrite the continued fraction representation of h(x) as

h(x) = x − \frac{x^{2}}{1 + 2x} − \frac{2^{2}x^{2}}{1 + 2 · 2^{2}x} − \frac{2^{2} · 3^{2} · x^{2}}{1 + 2 · 3^{2}x} − … − \frac{n^{2}(n + 1)^{2}x^{2}}{1 + 2 · (n + 1)^{2}x} − ….

Hence

− ∑_{n=1}^{∞} H_{2n+1}x^{n} = \frac{x}{1 + 2x} − \frac{2^{2} · x^{2}}{1 + 2 · 2^{2}x} − ….

Now we apply Equation (15), and transform the above equation to

− ∑_{n=0}^{∞} H_{2n+3}x^{n+1} = K_{n=0}^{∞} (\frac{c_{n}x}{1}) ,

where c_{0} = 1, c_{2n−1} = c_{2n} = n(n + 1), for n ≥ 1.

Applying Lemma 1, H_{2n+3} is divisible by 2^{n} for n ≥ 1, and

∑_{n=1}^{∞} \frac{H_{2n+3}}{2^{n}} x^{n} = \frac{x}{1 + x} + \frac{x}{1} + \frac{3x}{1} + \frac{3x}{1} + \frac{6x}{1} + \frac{6x}{1} + …  (16)
≡ \frac{x}{1 + x} + \frac{x}{1} + \frac{3x}{1 + 3x}  (mod 6)
≡ \frac{x}{3x^{2} + 2x + 1}  (mod 6)
≡ \frac{x}{3x^{2} − 4x + 1}  (mod 6)
= \frac{x}{(3x − 1)(x − 1)} = \frac{1}{2} · \frac{1}{1 − 3x} − \frac{1}{2} · \frac{1}{1 − x}
= ∑_{n=0}^{∞} (\frac{3^{n} − 1}{2}) x^{n}.

Comparing the coefficients of x^{n}, we have

\frac{H_{2n+3}}{2^{n}} ≡ \frac{3^{n} − 1}{2}  (mod 6)
= 3^{n−1} + 3^{n−2} + ⋯ + 3 + 3^{0}.

Since 3^{n} ≡ 3 (mod 6), for n ≥ 1, we have

\frac{H_{2n+3}}{2^{n}} ≡ (n − 1) · 3 + 1 ≡ 3n − 2  (mod 6).  (17)

If n = 2k − 1, for k ≥ 1, then

\frac{H_{2n+3}}{2^{n}} ≡ 3(2k − 1) − 2 ≡ 1 (mod 6).

If n = 2k, for k ≥ 1, then

\frac{H_{2n+3}}{2^{n}} ≡ 3(2k) − 2 ≡ 4 (mod 6).

Hence we complete our proof.

Using the similar method, we could get Barsky’s result ([2, Theorem 1]): for n ≥ 1,

\frac{H_{2n+3}}{2^{n}} ≡ \{ \begin{matrix}3 (mod 4), & if n is odd; \\ 2 (mod 4), & if n is even. \end{matrix}  (18)

The tangent numbers T_{n} are defined by

1 + tanh x = ∑_{n=0}^{∞} \frac{T_{n}x^{n}}{n!} .

The unsign tangent numbers are the sequence [7, Sequence A009006]. The tangent numbers T_{n} are closely related to the Bernoulli numbers:

T_{2n−1} = 2^{2n}(2^{2n} − 1)B_{2n}/2n.  (19)

Theorem 4. For n ≥ 1, the tangent number T_{2n+1} is divisible by 2^{n} and

\frac{T_{2n+1}}{2^{n}} ≡ (−1)^{n}4^{n−1}  (mod 6).  (20)

Proof. We use the classical continued fraction representation for the ordinary generating function of the tangent numbers T_{n} [1, Corollary 3.1]

∑_{n=0}^{∞} T_{n}x^{n+1} = x + x^{2} − 2x^{4} + 16x^{6} − 272x^{8} + …

= x + \frac{x^{2}}{1} + \frac{2x^{2}}{1} + \frac{6x^{2}}{1} + … + \frac{n(n + 1)x^{2}}{1} + ….  (21)

Changing the variable x^{2} as t we have

t + ∑_{n=1}^{∞} T_{2n+1}t^{n+1} = \frac{t}{1} + \frac{2t}{1} + \frac{6t}{1} + … + \frac{n(n + 1)t}{1} + ….

Applying Lemma 1, T_{2n+1} is divisible by 2^{n} for n ≥ 1. And

∑_{n=1}^{∞} \frac{T_{2n+1}}{2^{n}} t^{n+1} = \frac{−t^{2}}{1 + t} + \frac{3t}{1} + \frac{6t}{1} + …

≡ \frac{−t^{2}}{1 + t + 3t}  (mod 6)
= ∑_{n=1}^{∞}(−1)^{n}4^{n−1}t^{n+1}.

Comparing the coefficients of t^{n+1}, we complete the proof.

The result that T_{2n+1} is divisible by 2^{n} is not new. Howard [5, Theorem 8] proved in an elementary way that for every n ≥ 1 the number (2^{n+1}(1−2^{2n})/2n)B_{2n} is an integer. That is to say, T_{2n−1} is divisible by 2^{n−1}. Ramanujan (see [3, p. 7]) proved some similar congruence properties, such as

\frac{2(2^{4n+2} − 1)}{2n + 1}B_{4n+2}, and \frac{−2(2^{8n+4} − 1)}{2n + 1}B_{8n+4}

are integers of the form 30k + 1, for n ≥ 0. And it means that T_{4n+1}, T_{8n+3} are divisible by 2^{4n}, 2^{8n+1}, respectively, and

\frac{T_{4n+1}}{2^{4n}} ≡ \frac{−T_{8n+3}}{2^{8n+1}} ≡ 1 (mod 30).

3 Acknowledgements

The author would like to thank the referee for some useful comments and suggestions.

References

[1] D. Dumont, Further triangles of Seidel-Arnold type and continued fractions related to Euler and Springer numbers, Adv. in Appl. Math. 16, No. 3 (1995), 275–296.

[2] G.-N. Han, J. Zeng, On a q-sequence that generalizes the median Genocchi numbers, Ann. Sci. Math. Québec 23, No. 1 (1999), 63–72.

[3] G. H. Hardy, P. V. Seshu Aiyar, and B. M. Wilson, Collected Papers of Srinivasa Ramanujan, Chelsea Pub. Co., 1962.

[4] M. E. Hoffman, Derivative polynomials, Euler polynomials, and associated integer sequence, Electron. J. Combin. 6 (1999), #R21, 13 pp.

[5] F. T. Howard, Applications of a recurrence for the Bernoulli Numbers, J. Number Theory 52, No. 1 (1995), 157–172.

[6] L. Lorentzen, H. Waadeland, Continued Fractions with Applications, North-Holland, Netherlands, 1992.

[7] N. J. A. Sloane, editor (2003), The On-Line Encyclopedia of Integer Sequences,

    http://www.research.att.com/˜ njas/sequences/.

2000 Mathematics Subject Classification: Primary 11A55; Secondary 11B68. Keywords: Continued fractions, Springer numbers, Euler numbers, Genocchi numbers, Tangent numbers.

(Concerned with sequences \underline{A000281}, \underline{A002832}, \underline{A005439}, \underline{A009006}, and \underline{A036968}.)

Received October 20 2003; revised version received November 8 2003. Published in Journal of Integer Sequences, January 16 2004.

Return to Journal of Integer Sequences home page.