Complementary Bell Numbers and p-adic Series

Deepak Subedi Faculty of Science and Technology ICFAI University Himachal Pradesh Kalujhinda, Solan, HP-174103 India

                     deepak12321@gmail.com

Abstract In this article, we generalize a result of Murty on the non-vanishing of complementary Bell numbers and irrationality of a p-adic series. This generalization leads to a sequence of polynomials. We partially answer the question of existence of an integral zero of those polynomials.

1 Introduction

Murty and Sumner [5] have shown that there is a sequence of integers a_{k}, b_{k} such that the following equality

∑_{n=0}^{∞} n^{k}n! = a_{k} ∑_{n=0}^{∞} n! + b_{k}  (1)

holds in ℚ_{p}. Alexander [6] has shown that a_{k} vanishes at most twice. In Proposition 1, we generalize Eq. (1) to show that for non-negative integers k, j there exist two sequences of integers a_{k}(j), b_{k}(j) such that the following equality

∑_{n=0}^{∞} n^{k}(n + j)! = a_{k}(j)α + b_{k}(j)  (2)

holds in ℚ_{p}, α being the p-adic sum ∑_{n=0}^{∞} n!.

It is obvious that we would like to identify non-negative integers k and j such that

a_{k}(j) = 0.

This would mean that the infinite sum on left-hand side of Eq. (2) is just an integer b_{k}(j). In fact by Proposition 1 it would follow that

a_{1}(0) = 0, a_{2}(1) = 0

and

b_{1}(0) = −1, b_{2}(1) = 2.

Again by Proposition 1, it follows that

∑_{n=0}^{∞} n · n! = −1

and

∑_{n=0}^{∞} n^{2}(n + 1)! = 2.

On the other hand, Dragovich [2] has shown that if the series

∑_{n=0}^{∞} n!

converges to a rational number in ℚ_{p} for every prime p, then the series cannot converge to the same rational number. Furthermore, the fact a_{2}(j) > 0 for every integer j ≥ 2 leads us to conclude that for a fixed integer j ≥ 2 if the series

∑_{n=0}^{∞} n^{2}(n + j)!

converges to a rational number in ℚ_{p}, then it cannot converge to a fixed rational number in ℚ_{p} for every prime p. In order to show that a_{k}(j) is non-zero for selected values of k, j we need certain identities for a_{k}(j). We derive a few of those identities in the next section.

2 Recurrence for the polynomial

We begin this section by considering the series ∑_{n=0}^{∞}(n + j)! for a fixed non-negative integer j. Observe that

∑_{n=0}^{∞}(n + j)! = ∑_{n=0}^{∞} n! − (0! + 1! + 2! + ⋯ + (j − 1)!).

Kurepa’s left factorial, K(m) for a non-negative integer m is given by

K(m) = \begin{cases}0, & if m = 0; \\ 0! + 1! + 2! + ⋯ + (m − 1)!, & if m is a positive integer. \end{cases}

It then follows that

∑_{n=0}^{∞}(n + j)! = α − K(j).

For reasons which will be clear in a moment, we define

a_{0}(x) = 1
b_{0}(j) = −K(j).

Therefore, it follows that

∑_{n=0}^{∞}(n + j)! = a_{0}(j)α + b_{0}(j).  (3)

We hereby present the main result of this section.

Proposition 1. Let k ≥ 0 and j ≥ 0 be fixed integers. Then there is a polynomial a_{k}(x) and an integer b_{k}(j) such that

∑_{n=0}^{∞} n^{k}(n + j)! = a_{k}(j)α + b_{k}(j)

where a_{k}(x) and b_{k}(j) are defined inductively on k as follows:

a_{k}(x) = a_{k−1}(x + 1) − (x + 1)a_{k−1}(x), k ≥ 1
b_{k}(j) = b_{k−1}(j + 1) − (j + 1)b_{k−1}(j), k ≥ 1

and

a_{0}(x) = 1
b_{0}(j) = −K(j).

Proof. The proof is by induction on k. The case k = 0 has already been worked out. So we may assume k ≥ 0 and that the proposition holds for k.

Observe that

∑_{n=0}^{∞} n^{k}(n + j + 1)! = ∑_{n=0}^{∞} n^{k+1}(n + j)! + (j + 1) ∑_{n=0}^{∞} n^{k}(n + j)!.

Thus using ∑_{n=0}^{∞} n^{k}(n + j)! = a_{k}(j)α + b_{k}(j) and then comparing the coefficient of α, term without α, we have

a_{k}(j + 1) − (j + 1)a_{k}(j) = a_{k+1}(j)  (4)

and

b_{k}(j + 1) − (j + 1)b_{k}(j) = b_{k+1}(j).

Corollary 2. The series ∑_{n=0}^{∞} n^{k}(n + j)! converges to an integer whenever a_{k}(j) vanishes.

Corollary 3.

a_{k}(0) = a_{k+1}(−1).

The next proposition may remind us about a similar kind of property exhibited by Bernoulli polynomials.

Proposition 4. The derivative of a_{k}(x) is given by

\frac{d}{dx}a_{k}(x) = −ka_{k−1}(x), k ≥ 1.

Proof. The proposition is easily seen to be true for k = 1 and we prove the proposition by induction on k. We differentiate the expression

a_{K}(x + 1) − (x + 1)a_{K}(x) = a_{K+1}(x)

given in Proposition 1 to obtain

a′_{K}(x + 1) − a_{K}(x) − a′_{K}(x)(x + 1) = a′_{K+1}(x).

We assume the proposition holds for k = K to obtain

−Ka_{K−1}(x + 1) − a_{K}(x) + Ka_{K−1}(x)(x + 1) = a′_{K+1}(x).

Again we consider Proposition 1 to obtain the desired result.

Proposition 5. If c_{i,k} denotes the coefficients of x^{i} in a_{k}(x) then

−kc_{i,k−1} = (i + 1)c_{i+1,k}

for non-negative integers i, k and i ≤ k − 1.

Proof. The proof follows by comparing constant term and the coefficient of powers of x in Proposition 4.

We note that if k is a prime then for i ≤ k − 2

gcd(i + 1, k) = 1.

Hence i + 1 must divide c_{i,k−1} and k must divide c_{i+1,k} but a_{k}(0) = a_{k} and so we can write

a_{p}(x) ≡ a_{p} − x^{p} (mod p).

We proceed for a few more congruences for a_{k}(x). We start with a proposition that states a_{k}(x) can be determined using a_{k} and the binomial coefficients.

Proposition 6. The polynomial a_{k}(x) is given by

a_{k}(x) = ∑_{i=0}^{k}^{k}C_{i}a_{i}(−1)^{k−i}x^{k−i}.

Proof. Applying induction on Proposition 4, it follows that

\frac{d^{i}}{dx^{i}} a_{k}(x) = (−1)^{i}k(k − 1)(k − 2) ⋯ (k − i + 1)a_{k−i}(x).

We write a_{k}(x) as

a_{k}(x) = ∑_{i=0}^{k} b_{i}x^{i}.

Then \frac{d^{i}}{dx^{i}} a_{k}(x) at x equal to 0 must be b_{i}i!. Hence b_{i} must be

(−1)^{i} ^{k}C_{i}a_{k−i}(0).

Using the fact that a_{k−i}(0) = a_{k−i} the result follows.

Now, we include a table containing first few polynomials a_{k}(x).

k  a_{k}(x)
0  1
1  −x
2  −1 + x^{2}
3  1 + 3x − x^{3}
4  +2 − 4x − 6x^{2} + x^{4}
5  −9 − 10x + 10x^{2} + 10x^{3} − x^{5}
6  9 + 54 x + 30 x^{2} − 20 x^{3} − 15 x^{4} + x^{6}
7  50 − 63 x − 189 x^{2} − 70 x^{3} + 35 x^{4} + 21 x^{5} − x^{7}
8  −267 − 400x + 252x^{2} + 504 x^{3}+ 140 x^{4} − 56x^{5} − 28x^{6} + x^{8}
9  413 + 2403 x + 1800 x^{2} − 756 x^{3} − 1134 x^{4} − 252 x^{5} + 84 x^{6}
+36 x^{7} − x^{9}
10  2180 − 4130 x − 12015 x^{2} − 6000 x^{3} + 1890 x^{4} + 2268 x^{5}+
420 x^{6} − 120 x^{7} − 45 x^{8} + x^{10}
11 −17731 − 23980 x + 22715 x^{2} + 44055 x^{3} + 16500 x^{4} − 4158 x^{5}
−4158x^{6} − 660x^{7} + 165x^{8} + 55x^{9} − x^{11}
12  50533 + 212772x + 143880x^{2} − 90860x^{3} − 132165x^{4} − 39600x^{5}
+8316x^{6} + 7128x^{7} +990x^{8} − 220 x^{9} − 66x^{10} + x^{12}
13 110176 − 656929x − 1383018x^{2} − 623480x^{3} + 295295x^{4} + 343629 x^{5}
+85800x^{6} − 15444x^{7} − 11583x^{8} − 1430x^{9} + 286x^{10} + 78x^{11} − x^{13}
14  −1966797 − 1542464x + 4598503x^{2} + 6454084x^{3} + 2182180x^{4}
−826826x^{5} − 801801x^{6} − 171600x^{7} + 27027x^{8} + 18018x^{9} + 2002x^{10}
−364x^{11} − 91x^{12} + x^{14}

It is easy to see from the table that

a_{1}(0) = 0, a_{2}(1) = 0

and so we would like to identify integers k and j such that a_{k}(j) is zero/nonzero. We partially identify such k and j in the next section.

3 On non-vanishing of the polynomials

The next proposition helps us in concluding non-vanishing of a_{k}(x) whenever k is a prime.

Proposition 7. If p is a prime then a_{p}(j) does not vanish for every j in ℤ with j incongruent to 1 modulo p.

Proof. The proposition can be easily verified for p = 2. For p ≥ 3, Proposition 6 and the congruence

\binom{p}{i} ≡ 0 (mod p) for 1 ≤ i ≤ p − 1

leads us to

a_{p}(j) ≡ a_{p} − j^{p} (mod p).

Murty [5] has shown that

a_{p} ≡ 1 (mod p).

Considering Fermat’s theorem the proposition follows.

Observe that a_{1}(x) = −x > 0 for x < 0. More generally, we have the following proposition.

Proposition 8. a_{k}(x) > (k − 1)! for k ≥ 1 and x ≤ −k.

Proof. We prove this proposition by induction on k.

Assume a_{k}(x) > (k − 1)! for x ≤ −k holds for some fixed k ≥ 1 then the recurrence relation

a_{k+1}(x) = a_{k}(x + 1) − (x + 1)a_{k}(x)

for a_{k}(x) gives us

a_{k+1}(x) > (k − 1)! + (k − 1)(k − 1)!

for x ≤ −k − 1.

Hence the proposition follows.

With this proposition it is clear that a_{k}(x) does not vanish for k ≥ 1 and x ≤ −k.

To analyse a_{k}(x) further, we start with the following result.

Proposition 9. For a non-negative integer m and an integer j

a_{p+m}(j) ≡ a_{m+1}(j) + a_{m}(j) (mod p).

Proof. For an integer j, by Proposition 6 it follows that

a_{p}(j) ≡ 1 − j (mod p).

Applying the recurrence given in Proposition 1 and the fact

a_{2}(j) + a_{1}(j) = j^{2} − j − 1,

it follows that

a_{p+1}(j) ≡ a_{2}(j) + a_{1}(j) (mod p).

Again, applying the recurrence in Proposition 1 repeatedly we obtain the desired result.

Proposition 10. For a prime p such that

p ≡ 2, 3 (mod 5),

a_{p+1}(j) does not vanish for any integer j. Proof. By previous proposition

a_{p+1}(j) ≡ j^{2} − j − 1 (mod p).

However, for a prime p ≠ 2, 5 considering

4(j^{2} − j − 1) = (2j − 1)^{2} − 5,

it is clear that a_{p+1}(j) is not congruent to 0 modulo p whenever the Legendre symbol

\binom{5}{p} = −1.

Now, \binom{5}{p} = −1 if and only if

p ≡ 2, 3 (mod 5).

Hence the proposition follows. Corollary 11. a_{8}(j), a_{14}(j), a_{18}(j) and a_{24}(j) does not vanish for any integer j. Proposition 12. For a prime p ≡ 2, 5, 6, 7, 8, 11 (mod 13) and an integer j not divisible by p, a_{p+2}(j) does not vanish. Proof. By Proposition 9, for m = 2, one has

a_{p+2}(j) ≡ −j(j^{2} − j − 3) (mod p).

Hence the proposition follows. Proposition 13. If a_{p}(1) is not divisible by p^{2}, then a_{p}(x) is an irreducible polynomial over ℚ. Proof. By Proposition 6 we have

a_{p}(x + 1) ≡ a_{p} − (x + 1)^{p} (mod p).

Considering the congruence for a_{p} given by Murty [5] again it follows that

a_{p}(x + 1) ≡ −x^{p} (mod p).

Hence by Eisenstein’s criterion, the result follows.

As a consequence of Proposition 13 it is clear that if a_{p}(1) is not divisible by p^{2}, then there does not exist an integer j such that a_{p}(j) = 0. The next proposition gives us a conditional statement for deciding whether a_{p}(j) is different from 1.

Proposition 14. For an odd prime p, if a_{p} − 1 is not divisible by p^{2}, then a_{p}(x) − 1 is an irreducible polynomial.

Proof. Following the steps of Proposition 13 we have

a_{p}(x) − 1 ≡ −x^{p} (mod p).

Hence by Eisenstein’s criterion for the irreducibility of a polynomial, the result follows.

Proposition 15. For non-negative integers m, t and an integer j the following congruence holds

a_{tp+m}(j) ≡ ∑_{i=o}^{t}^{t}C_{i}a_{m+i}(j) (mod p).  (5)

Proof. The case t = 0 is obviously true. As our induction hypothesis we assume that the congruence in Eq. (5) is true for some t ≥ 0 and by Proposition 9 it follows that

a_{(t+1)p+m}(j) ≡ a_{tp+m+1}(j) + a_{tp+m}(j) (mod p).  (6)

Hence by our induction hypothesis

a_{(t+1)p+m}(j) ≡ ∑_{i=o}^{t}^{t}C_{i}a_{m+1+i}(j) + ∑_{i=o}^{t}^{t}C_{i}a_{m+i}(j) (mod p)  (7)

≡ ∑_{i=o}^{t+1}^{t+1}C_{i}a_{m+i}(j) (mod p).  (8)

Hence the result follows by induction.

Proposition 16. For non-negative integers m, i and an integer j the following congruence holds

a_{p^{i}+m}(j) ≡ a_{m+1}(j) + ia_{m}(j) (mod p).

Proof. The case i = 0 is a trivial case and the case i = 1 follows from Proposition 9. So we assume i ≥ 2.

For 1 ≤ j ≤ p^{i−1} − 1, considering the congruence

\binom{p^{i−1}}{j} ≡ 0 (mod p)

and t = p^{i−1} in the above proposition it follows that

a_{p^{i}+m}(j) ≡ a_{m}(j) + a_{p^{i−1}+m}(j) (mod p).

Repeating the previous step r number of times where r ≤ i − 1 we have

a_{p^{i}+m}(j) ≡ ra_{m}(j) + a_{p^{i−r}+m}(j) (mod p).

Choosing r = i − 1 we have

a_{p^{i}+m}(j) ≡ (i − 1)a_{m}(j) + a_{p+m}(j) (mod p).

The result follows from the above congruence and Proposition 9. Corollary 17. For a positive integer i,

a_{p^{i}} ≡ i (mod p).

Proof. Choosing m, j equal to 0 the corollary follows. Proposition 18. a_{p^{zp}}(j) does not vanish for any integer j not divisible by p. Proof. We consider i = zp for some non-negative integer z in the previous proposition to obtain

a_{p^{zp}+m}(j) ≡ a_{m+1}(j) (mod p).

Choosing m = 0, the result follows. Proposition 19. For a non-negative integer t and an integer j

a_{3t}(j) ≠ 0.

Proof. Considering i = 2, p = 2 in Proposition 16, it follows that

a_{4+m}(j) ≡ a_{1+m}(j) (mod 2).

Choosing m = 2, 5, 8, ⋯ it is easy to see that for a positive integer t

a_{3t} ≡ a_{3}(j) (mod 2).

The fact

a_{3}(j) ≢ 0 (mod 2)

leads to the desired result.

Proposition 20. For a non-negative integer t a_{pt} ≡ a_{t−1} (mod p).

Proof. Through Proposition 6 it is easy to see that

a_{k}(−1) = ∑_{i=0}^{k}^{k}C_{i}a_{i}

However, by the recurrence 1

a_{k}(−1) = a_{k−1}(0).

By congruence (5)

a_{pt+m}(j) ≡ ∑_{i=0}^{t}^{t}C_{i}a_{m+i}(j) (mod p).

For m = 0, j = 0 above congruence reduces to

a_{pt}(0) ≡ ∑_{i=0}^{t}^{t}C_{i}a_{i}(0) (mod p)

and so

a_{pt}(0) ≡ a_{t−1} (mod p).

Proposition 21. For a positive integer i and an integer j if

j ≢ i (mod p)

then

a_{p^{i}}(j) ≠ 0.

Proof. Choosing m = 0 in Proposition 16 we have

a_{p^{i}}(j) ≡ a_{1}(j) + ia_{0}(j) (mod p).

The result follows.

The next result gives us a much stronger congruence of a_{k}(j). Theorem 22. For non-negative integers t, m, a positive integer n, an odd prime p and an integer j such that

j ≡ 0, 1, 2 (mod p)

the following congruence

a_{\frac{p^{p}−1}{p−1} ·p^{n−1}t+m}(j) ≡ a_{m}(j) (mod p^{n})

holds.

Proof. We consider three cases: j ≡ 0 (mod p), j ≡ 1 (mod p) and j ≡ 2 (mod p) Case 1. For an integer j ≡ 0 (mod p) and a positive integer r, it is easy to see that

\binom{\frac{p^{p}−1}{p−1} · p^{n−1}}{r}(−j)^{r} = \frac{\frac{p^{p}−1}{p−1}(−j)^{r} · p^{n−1}}{r} \binom{\frac{p^{p}−1}{p−1} · p^{n−1} − 1}{r − 1} ≡ 0 (mod p^{n}).

For an odd prime p, following a slightly different notation, Alexander [6] has proved that

a_{\frac{p^{p}−1}{p−1} ·p^{n−1}t+m} ≡ a_{m} (mod p^{n}), t, m being non-negative integers.

Hence by Proposition (6), it follows that for an integer j ≡ 0 (mod p)

a_{\frac{p^{p}−1}{p−1}} (j) ≡ 1 (mod p^{n}).  (9)

For simplicity we denote \frac{p^{p}−1}{p−1} · p^{n−1} by k. Case 2. In this case we are expressing a_{k}(j) in terms of a_{k+1}(j − 1) and a_{k}(j − 1) and then deriving the required congruence.

Replacing j by j − 1, Eq. (4) can be written as

a_{k}(j) = a_{k+1}(j − 1) + ja_{k}(j − 1).  (10)

For an integer j ≡ 1 (mod p) and k = \frac{p^{p}−1}{p−1} · p^{n−1}, n ≥ 1

a_{k+1}(j − 1) ≡ a_{k+1} − (j − 1)(k + 1)a_{k} (mod p^{n}).

Again using the result of Alexander [6], it follows that

a_{k+1}(j − 1) ≡ −(j − 1) ≡ a_{1}(j − 1) (mod p^{n}).

Hence, it follows that

a_{k}(j) ≡ 1 (mod p^{n}) for j ≡ 1 (mod p)  (11)

Case 3. In this case, we express a_{k}(j) in term of a_{k+2}(j − 2) and other similar terms. Replacing k by k + 1 and j by j − 1, Eq. (10) can be written as

a_{k+1}(j − 1) = a_{k+2}(j − 2) + (j − 1)a_{k+1}(j − 2) and  (12)

replacing j by j − 1, Eq. (10) can be written as

a_{k}(j − 1) = a_{k+1}(j − 2) + (j − 1)a_{k}(j − 2)  (13)

Eliminating a_{k}(j − 1), a_{k+1}(j − 1) from Eqs. (10), (12), and (13), it follows that

a_{k}(j) = a_{k+2}(j − 2) + (j − 1)a_{k+1}(j − 2) + j{a_{k+1}(j − 2) + (j − 1)a_{k}(j − 2)}.  (14)

But for an integer j ≡ 2 (mod p), by Proposition 6 it is easy to see that

a_{k+2}(j − 2) ≡ a_{k+2} + (k + 2)(2 − j)a_{k+1} + \frac{(k + 2)(k + 1)}{2}(j − 2)^{2}a_{k} (mod p^{n})

Again, considering the result of Alexander it would follow that

a_{k+2}(j − 2) ≡ −1 + (j − 2)^{2} ≡ a_{2}(j − 2) (mod p^{n}).  (15)

Also it is easy to obtain that

a_{k+1}(j − 2) ≡ a_{0}(j − 2) (mod p^{n}).  (16)

Applying Eqs. (14),(15),(16) it would follow that for an integer j ≡ 2 (mod p),

a_{k}(j) ≡ 1 (mod p^{n}).  (17)

Therefore, using Eq. (4) the result follows.

Corollary 23. For non-negative integers t, m and a positive integer n, the following congruence

a_{13·3^{n−1}t+m}(j) ≡ a_{m}(j) (mod 3^{n})

holds.

Proof. The proof follows by considering p = 3 in previous proposition.

Remark: The polynomials a_{k}(x) were previously analyzed by Wannemacker [8]. He has verified numerically that a_{k}(x) is irreducible over ℤ for all 6 ≤ k ≤ 200. He conjectured that a_{k}(x) is irreducible over ℤ for all k ≥ 6.

References

[1] B. Dragovich, On some p-adic series with factorials, in p-adic Functional Analysis, Lect. Notes Pure Appl. Math., Vol. 192, Dekker, 1997, pp. 95–105.

[2] B. Dragovich, On p-adic power series, preprint,

   http://arxiv.org/abs/math-ph/0402051.

[3] B. Dragovich, On some finite sums with factorials, Facta Universitasis (Nis) Ser. Math. Inform. 14 (1999), 1–10.

[4] L. Comtet, Advanced Combinatorics, D. Reidel Publishing Company, 1974.

[5] M. Ram Murty and S. Sumner, On the p-adic series ∑_{n=1}^{∞} n^{k} · n!, in Number Theory, CRM Proc. Lecture Notes, 36, Amer. Math. Soc., 2004, pp. 219–227.

[6] N. C. Alexander, Non-vanishing of Uppuluri-Carpenter numbers, preprint,

   http://tinyurl.com/oo36das.

[7] N. J. A. Sloane, The On-line Encyclopedia of Integer Sequences, http://oeis.org.

[8] S. D. Wannemacker, Annihilating polynomials and Stirling numbers of the second kind, Ph. D. thesis, University College Dublin, Ireland, 2006.

[9] Y. Yang, On a multiplicative partition function, Electron. J. Combin. 8 (2001) #R19.

2001 Mathematics Subject Classification: Primary 11A07; Secondary 40A30. Keywords: p-adic series, complementary Bell number.

(Concerned with sequence \underline{A000587}.)

Received June 24 2013; revised version received February 3 2014. Published in Journal of Integer Sequences, February 15 2014.

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