Arithmetic Progressions of b-Prodigious Numbers

Michael Gohn

Dept. of Mathematics and Computer Sci. DeSales University 2755 Station Ave.

Center Valley, PA 18034

USA

mg7785@desales.edu

Joshua Harrington Department of Mathematics Cedar Crest College 100 College Dr. Allentown, PA 18104 USA

                        joshua.harrington@cedarcrest.edu

Sophia Lebiere Department of Mathematics

Tufts University 419 Boston Ave. Medford, MA 02155

USA

slebie01@tufts.edu

Hani Samamah Department of Mathematics University of Florida Gainesville, FL 32611 USA

                               hanisamamah@ufl.edu

Kyla Shappell Department of Mathematics

Spring Hill College 4000 Dauphin St. Mobile, AL 36608

USA

kyla.m.shappell@email.shc.edu

Tony W. H. Wong Department of Mathematics Kutztown University of Pennsylvania 15200 Kutztown Road Kutztown, PA 19530 USA

                               wong@kutztown.edu

Abstract

A positive integer n is called a b-prodigious number if n is divisible by the product of its non-zero base-b digits. In this article, we investigate the maximum length of an arithmetic progression of b-prodigious numbers and the maximal length of consecutive sequences of b-prodigious numbers.

1 Introduction

In the On-Line Encyclopedia of Integer Sequences, sequence \underline{A055471} [8] is the sequence of positive integers that are divisible by the product of their nonzero digits. On the French website Diophante [4], this sequence is called nombres prodigieux, which translates to prodigious numbers. In recognition of this translation, for an integer b ≥ 2, we say that a positive integer n is a b-prodigious number if it is divisible by p_{b}(n), which is defined as the product of the nonzero digits of n in its base-b representation. Clearly, every positive integer is a 2-prodigious number. For b = 10, De Koninck and Luca [3] showed that for large enough values of x, the number of b-prodigious numbers less than x, denoted by N_{0}(x), satisfies x^{0.495} < N_{0}(x) < x^{0.901}. This result was later improved and generalized to arbitrary base by Sanna [6].

In a similar manner, a positive integer n is called a Niven number if it is divisible by the sum of its digits. More generally, for an integer b ≥ 2, a b-Niven number is a positive integer n that is divisible by s_{b}(n), which is defined as the sum of the digits of n in its base-b representation. In 1992, Cooper and Kennedy [2] showed that there exist sequences of 20 consecutive integers that are all Niven numbers and further proved that no sequences of 21 consecutive Niven numbers exist. Their result was partially generalized in 1994 by Grundman [5], who showed that every sequence of consecutive integers of length at least 2b + 1 must contain a term that is not b-Niven. Grundman’s bound on the maximum length of consecutive Niven numbers was shown to be tight for b ∈ {2, 3} by Cai [1] and when b ≥ 4 by Wilson [9]. In this paper, we establish analogous results for b-prodigious numbers.

Let b ≥ 3 be a positive integer and let ℓ be the smallest positive integer that does not divide b. In Section 2 of this paper, we show that there exist sequences of b + ℓ consecutive integers that are all b-prodigious and we further show that no sequence of b+ℓ+1 consecutive b-prodigious numbers exist. In fact, we obtain this result by considering consecutive terms in an arithmetic progression. In Section 3, we investigate maximal length of consecutive integers whose terms are all b-prodigious (here, maximal stands for sequences that cannot be extended on either ends).

2 The maximum length of arithmetic progressions of b-prodigious numbers

We begin this section with the following lemma without proof.

Lemma 1. Let b ≥ 2. Then for all integers c > 0, 1 ≤ j ≤ b − 1, and 0 ≤ β < α, we have

p_{b}(cb^{α} + jb^{β}) = p_{b}(c) · j.

Furthermore, if cb^{α} + jb^{β} is a b-prodigious number, then

j | (cb^{α} + jb^{β}) and p_{b}(c) | (cb^{α} + jb^{β}).

In view of Lemma 1, the relation p_{b}(c) | (cb^{α} + jb^{β}) is a necessary condition for cb^{α} + jb^{β} to be a b-prodigious number. For this reason, it will be useful to force p_{b}(c) = 1 while also guaranteeing that c satisfies certain congruence constraints.

Lemma 2. Let {x ≡ r_{v} (mod m_{v}) : 1 ≤ v ≤ w} be a consistent system of congruences such that r_{v} = 0 if gcd(m_{v}, b) ≠ 1. Then there exists a positive integer c such that c satisfies this system of congruences and p_{b}(c) = 1.

Proof. For each 1 ≤ v ≤ w, let m_{v} = m_{v,1}m_{v,2} such that m_{v,1} divides a nonnegative power of b and gcd(m_{v,2}, b) = 1. Let M_{1} = lcm{m_{v,1} : 1 ≤ v ≤ w} and M_{2} = lcm{m_{v,2} : 1 ≤ v ≤ w}. By the conditions on r_{v}, the system of congruences is equivalent to

{x ≡ 0 (mod M_{1}), x ≡ R (mod M_{2})}

for some 1 ≤ R ≤ M_{2}. Then c = ∑_{u=t}^{R+t−1} b^{uφ(M_{2})} satisfies the desired properties, where φ is the Euler’s totient function and t is a nonnegative integer such that M_{1} | b^{tφ(M_{2})}.

For a positive integer d, we say that a sequence S of integers is a d-AP if all consecutive terms of S have a common difference d. When d | b and d < b/2, the following theorem establishes the maximum length of a d-AP whose terms are all b-prodigious numbers.

Theorem 3. Let b > 2 and d be positive integers such that d | b and d < b/2. Define D_{b} = {0} ∪ {1 ≤ j ≤ b − 1 : j | b}, and for each j ∈ D_{b}, let k_{j} = min{κ ∈ ℕ : (j + κd) ∤ b}. Then the maximum length of a b-prodigious d-AP is b/d + k, where k = max{k_{j} : j ∈ D_{b}}.

Proof. Suppose S is a b-prodigious d-AP of length at least b/d + k. Let the first term of S be ab + j for some nonnegative integers a and j with 0 ≤ j ≤ b − 1. Then S contains ab + j, ab + (j + d), …, (a + 1)b + (j + (k − 1)d).

If there exists 1 ≤ i ≤ k−1 such that j+(i−1)d < b ≤ j+id, then j+(i−1)d ≥ b−d > b/2. Thus, (j + (i − 1)d) ∤ b. Since j + (i − 1)d is a factor of both p_{b}(ab + (j + (i − 1)d)) and p_{b}((a + 1)b + (j + (i − 1)d)) by Lemma 1, we deduce that ab + (j + (i − 1)d) and (a + 1)b + (j + (i − 1)d) cannot be both b-prodigious, a contradiction. Hence j + id ≤ b − 1 for all 0 ≤ i ≤ k − 1. Furthermore, since ab + (j + id) and (a + 1)b + (j + id) are both b-prodigious, we have j, j + d, …, j + (k − 1)d ∈ D_{b}. This implies that j + (k − 1)d ≤ b/2, so j + kd < b. Therefore, ab + (j + kd) and (a + 1)b + (j + kd) cannot be both prodigious since (j + kd) ∤ b by the definition of k. In other words, S cannot be of length greater than b/d + k.

Finally, note that there exists a positive integer c such that c satisfies the congruence x ≡ 0 (mod (b − 1)!) and p_{b}(c) = 1 by Lemma 2. As a result, if j ∈ D_{b} satisfies k_{j} = k, then

cb^{2} + j, cb^{2} + (j + d), …, cb^{2} + b + (j + (k − 1)d)

is a b-prodigious d-AP of length b/d + k. This is because p_{b}(cb^{2} + (j + id)) = j + id by Lemma 1 when 0 < j + id < b; together with (b − 1)! | c, we have (j + id) | (cb^{2} + (j + id)). Likewise, p_{b}(cb^{2} + b + (j + id)) = j + id when 0 < j + id < j + kd; together with (j + id) | b, we have (j + id) | (cb^{2} + b + (j + id)).

As an example of how to use Theorem 3, consider the case b = 12 and d = 4. Then D_{b} = {0, 1, 2, 3, 4, 6}, k_{1} = k_{3} = k_{4} = k_{6} = 1, and k_{0} = k_{2} = 2. Therefore, k = 2 and the maximum length of a 12-prodigious 4-AP is 12/4 + 2 = 5. Following the definitions in the proof of Lemma 2, we have m_{1} = 11!, M_{1} = 2^{8} · 3^{4}, and M_{2} = 5^{2} · 7 · 11. Then R = M_{2}, φ(M_{2}) = 1200, and t = 1. Hence we have c = ∑_{u=1}^{1925} 12^{1200u}, and

144c, 144c + 4, 144c + 8, 144c + 12, 144c + 16

forms a 12-prodigious 4-AP of length 5. Note that although our theorem provides a method to construct a b-prodigious d-AP of maximum length, the sequence constructed using this method may not be the simplest. For example, the sequence 2, 6, 10, 14, 18 forms a 12prodigious 4-AP of length 5.

When d = 1, the statement of Theorem 3 simplifies to the following corollary.

Corollary 4. Let b > 2. The maximum length of a sequence of consecutive b-prodigious numbers is b + ℓ, where ℓ is the smallest positive integer that does not divide b.

Proof. From Theorem 3 we have k_{0} = ℓ ≤ k. On the other hand, for all j ∈ D_{b}, there always exists a multiple of ℓ in {j + κ : 1 ≤ κ ≤ ℓ}; thus k_{j} ≤ ℓ. Combining both directions, our result follows from Theorem 3, since k = ℓ.

Since Theorem 3 only applies when d | b and d < b/2, we end this section by addressing the maximum length of a b-prodigious (b/2)-AP.

Theorem 5. Let b > 2 be even. Then the maximum length of a b-prodigious (b/2)-AP is 6.

Proof. Suppose S is a b-prodigious (b/2)-AP of length at least 6. Then S contains ab + (j + b/2) and (a + 1)b + (j + b/2) for some nonnegative integers a and j with 0 ≤ j < b/2. By Lemma 1, j + b/2 divides both ab + (j + b/2) and (a + 1)b + (j + b/2). Hence (j + b/2) | b, which implies that j = 0.

If S contains both cb^{2} + ib and cb^{2} + ib + b/2 for some nonnegative integers n and 1 ≤ i ≤ b − 1, then i | (cb^{2} + ib) and (i · b/2) | (cb^{2} + ib + b/2) by Lemma 1. Hence i divides (cb^{2} + ib + b/2) − (cb^{2} + ib) = b/2. Moreover, cb^{2} + ib + b/2 = mi · b/2 for some integer m. Dividing both sides of the last equation by b/2, we have 2cb+2i+1 = mi, so i | (2cb+2i+1).

Combining the established observations that i | (b/2) and i | (2cb+2i+1), we have i = 1. Therefore, S is of the form

(c − 1)b^{2} + (b − 1)b + b/2, cb^{2}, cb^{2} + b/2, cb^{2} + b, cb^{2} + b + b/2, cb^{2} + 2b

for some positive integer c. Note that this is a b-prodigious (b/2)-AP of length 6 if c = ∑_{u=0}^{b/2−1} b^{u}.

3 Maximal lengths of consecutive b-prodigious numbers

We say that the sequence n, n + 1, n + 2, …, n + t − 1 of consecutive b-prodigious numbers is maximal of length t if n − 1 and n + t are not b-prodigious numbers. Recall from Corollary 4 that b+ℓ is the maximum length of consecutive b-prodigious numbers, where ℓ is the smallest positive integer that does not divide b. In this section, we determine the existence of maximal sequences of consecutive b-prodigious numbers of lengths b ≤ t ≤ b + ℓ − 1.

We begin this section with the following three lemmas.

Lemma 6. Let b ≥ 2. If ab + j and ab + (j + 1) are both b-prodigious numbers, where a is a positive integer and 0 ≤ j ≤ b − 2, then p_{b}(a) = 1.

Proof. If ab+j and ab+(j +1) are b-prodigious, then p_{b}(a) | (ab+j) and p_{b}(a) | (ab+(j +1)) by Lemma 1. It follows that p_{b}(a) | 1. Thus p_{b}(a) = 1.

Lemma 7. Let b > 2, let ℓ be the smallest positive integer that does not divide b, and let S be a maximal sequence of consecutive b-prodigious numbers of length at least b. If the first term of S is of the form cb^{2} for some positive integer c, then S is of length b + ℓ. If the first term of S is of the form cb^{2} +1 for some nonnegative integer c, then c = 0 and S is of length at b + ℓ − 1.

Proof. If the first term of S is of the form cb^{2} for some positive integer c, then since S has at least b terms, p_{b}(c) = 1 by Lemma 6. Thus, by Lemma 1, we have p_{b}(cb^{2} + b + j) = j for 1 ≤ j ≤ b − 1. Hence, if ℓ is the smallest positive integer that does not divide b, then cb^{2}, cb^{2} + 1, …, cb^{2} + b + (ℓ − 1) are each b-prodigious.

If the first term of S is of the form cb^{2} + 1 for some nonnegative integer c, then c = 0 since cb^{2} is not b-prodigious. Noting that b + ℓ is not b-prodigious since p_{b}(b + ℓ) = ℓ by Lemma 1 and ℓ ∤ (b + ℓ), our result follows.

Lemma 8. Let b > 2 and let S be a maximal sequence of consecutive b-prodigious numbers of length at least b + 1. Then the first term of S is either 1, of the form cb^{2}, or of the form cb^{2} + b for some positive integer c.

Proof. Let ab + j be the first term of S for some nonnegative integers a and j with j ≤ b − 1. Then the last term of S is at least (a + 1)b + j. This implies that j = 0 or j | b; thus j ≤ b/2. If 1 ≤ j ≤ b/2, then Lemma 6 implies that p_{b}(a) = p_{b}(a + 1) = 1, and if j = 0, then p_{b}(a) = 1. Therefore, the first term of S is of the forms cb^{2} + j or cb^{2} + b for some nonnegative integers c and j with 0 ≤ j ≤ b/2.

If c = 0, then since every integer 1 ≤ j ≤ b − 1 is b-prodigious, the first term of the maximal sequence S of consecutive b-prodigious numbers is 1. Hence it remains to consider c > 0, and we have p_{b}(c) = 1. If 2 ≤ j ≤ b/2, then since cb^{2} + 2(j − 1) ∈ S is b-prodigious, we know that j − 1 divides cb^{2} + 2(j − 1) − (j − 1) = cb^{2} + (j − 1). As a result, cb^{2} + (j − 1) is also b-prodigious, contradicting that cb^{2} + j is the first term of the maximal sequence S. Finally, j ≠ 1 by Lemma 7.

Theorem 9. Let b > 2 and let S be a maximal sequence of consecutive b-prodigious numbers of length m for some b + 1 ≤ m ≤ b + ℓ − 1. Then

(a) m cannot be strictly between b + 1 and b + ℓ − 1;

(b) m = b + ℓ − 1 if and only if the first term of S is 1; and

(c) m = b + 1 if and only if either b is odd and the first term of S is 1 or b is a power of 2.

Proof. By Lemmas 7 and 8, the first term of S is either 1 or of the form cb^{2} + b for some positive integer c. If the first term of S is 1, then m = b + ℓ − 1 by Lemma 7. In the remainder of this proof, assume that the first term of S is cb^{2} + b.

If m > b + 1, then S contains both cb^{2} + 2b and cb^{2} + 2b + 1, contradicting Lemma 6. This implies that m = b + 1 and establishes part (a). Hence m = b + ℓ − 1 if and only if ℓ = 2, which is equivalent to b being odd. However, b cannot be odd since 2 | (cb^{2} + b + 2) and 2 | (cb^{2} + 2b). Therefore, if m = b + ℓ − 1, the first term of S cannot be cb^{2} + b. This establishes part (b).

If b = 2^{α}m for some integer α ≥ 1 and odd integer m > 1, then note that 2^{α+1} ≤ b − 1 and 2^{α+1} ∤ (cb^{2} + b + 2^{α+1}) since 2^{α+1} | b^{2} but 2^{α+1} ∤ b. This contradicts that S contains cb^{2} + b + 2^{α+1}. Hence it remains to show that such a maximal sequence S exists when b = 2^{α} for some integer α ≥ 2. This can be achieved by finding a positive integer c such that j | (cb^{2} + b + j) for each 1 ≤ j ≤ b − 1 and p_{b}(c) = 1.

For each 1 ≤ j ≤ b − 1, let j = 2^{β_{j}}j′ for some nonnegative integer β_{j} and odd integer j′. Note that β_{j} < α, so j | (cb^{2}+b+j) is equivalent to cb^{2}+b ≡ 0 (mod j′). Since gcd(b, j′) = 1, we have c ≡ −b^{−1} (mod j′). Such an integer c exists by Lemma 2, which completes the proof of part (c).

Lemma 10. Let b > 2 and let

ab + (b − 1), (a + 1)b, (a + 1)b + 1, …, (a + 1)b + (b − 2)

be a b-prodigious sequence. Then b − 1 is a prime.

Proof. Suppose by way of contradiction that b − 1 is not a prime. Then there exists 2 ≤ j ≤ b − 2 such that j | b − 1. We deduce that j divides ab + (b − 1) since (b − 1) | ab + (b − 1)) by Lemma 1. Since j also divides (a + 1)b + j by Lemma 1, it follows that j divides (a + 1)b + j − (ab + (b − 1)) = j + 1, a contradiction.

Theorem 11. Let b > 2. Then there exists a maximal sequence S of consecutive b-prodigious numbers of length b if and only if either b − 1 is a prime or b is a power of an odd prime.

Proof. Let S be a maximal sequence of consecutive b-prodigious numbers of length b. By Lemma 6, S is a subsequence of

(c − 1)b^{2} + (b − 1)b + (b − 1), cb^{2}, cb^{2} + 1, …, cb^{2} + 2b

for some nonnegative integer c such that c = 0 or p_{b}(c) = 1. Suppose by way of contradiction that b − 1 is not a prime and b is not a power of an odd prime. Then Lemma 10 implies that the first term of S cannot be (c − 1)b^{2} + (b − 1)b + (b − 1) nor cb^{2} + (b − 1). Also, the first term of S cannot be cb^{2} nor cb^{2} + 1 since cb^{2} + b and cb^{2} + b + 1 are b-prodigious, and the first term of S cannot be cb^{2} + b + 1 since cb^{2} + b is b-prodigious.

If the first term of S is cb^{2} + b, then b cannot be even; otherwise, cb^{2} + 2b is b-prodigious. Since b is not a power of an odd prime, there exist distinct primes p < q such that pq | b. Let α be the greatest integer such that p^{α} | b. Then p^{α+1} ≤ b − 1, but cb^{2} + b + p^{α+1} is not b-prodigious, a contradiction. Hence S is given by

cb^{2} + j, cb^{2} + (j + 1), …, cb^{2} + b + (j − 1)

for some 2 ≤ j ≤ b − 2. Since j − 1 divides both cb^{2} + b + (j − 1) and cb^{2} + (j − 1) by Lemma 1, this implies that (j − 1) ∤ b since cb^{2} + b + (j − 1) is b-prodigious while cb^{2} + (j − 1) is not.

Since cb^{2} + (b − 1) is b-prodigious, we have by Lemma 1 that (b − 1) | c. If there exists 2 ≤ j̃ ≤ j such that j̃ | b − 1, then j̃ ∤ b; thus j̃ ∤ (cb^{2} + b + j̃), a contradiction. Hence j is less than every prime factor of b − 1, implying that j < \sqrt{b − 1} since b − 1 is not a prime. Note that b − 1 ≥ 4, so j < \sqrt{b − 1} ≤ \frac{b−1}{2}. As a result, j ≤ 2(j − 1) < b − 1, so cb^{2} + 2(j − 1) is b-prodigious. Together with cb^{2} + b + (j − 1) being b-prodigious, we deduce from Lemma 1 that (j − 1) | b, a contradiction.

Conversely, we assume that b − 1 is a prime or b is a power of an odd prime. If b − 1 is a prime, then let

B = max{β ∈ ℤ : p^{β} | j for some prime p | b and 1 ≤ j ≤ b − 2}

and

N = max{n ∈ ℕ : n | lcm{j ∈ ℕ : 1 ≤ j ≤ b − 2} and gcd(n, b) = 1}.

This construction ensures that j | b^{B}N for all 1 ≤ j ≤ b − 2. By Lemma 2, there exists a positive integer \overline{c} such that \overline{c} ≡ b^{−2}(−b^{2} +1)−b^{B−2} +1 (mod (b−1)^{B}), \overline{c} ≡ −b^{B−2} (mod N), \overline{c} ≡ 0 (mod b^{B−1}), and p_{b}(\overline{c}) = 1. Define c = \overline{c} + b^{B−2}; thus p_{b}(c) = 1. Note that

p_{b}((c − 1)b^{2} + (b − 1)b + (b − 1)) = p_{b}((\overline{c} + b^{B−2} − 1)b^{2} + (b − 1)b + (b − 1))

= p_{b}(\overline{c}b^{2} + b^{B} − 1)

= p_{b} (\overline{c}b^{2} + ∑_{t=0}^{B−1}(b − 1)b^{t})

= (b − 1)^{B} and p_{b}(cb^{2} + j) = j for all 1 ≤ j ≤ b − 2 by Lemma 1. Furthermore,

(c − 1)b^{2} + (b − 1)b + (b − 1) = (\overline{c} + b^{B−2} − 1)b^{2} + (b − 1)b + (b − 1)
= \overline{c}b^{2} + b^{B} − 1
≡ (b^{−2}(−b^{2} + 1) − b^{B−2} + 1)b^{2} + b^{B} − 1 (mod (b − 1)^{B})
≡ (−b^{2} + 1) − b^{B} + b^{2} + b^{B} − 1  (mod (b − 1)^{B})
≡ 0  (mod (b − 1)^{B})

and cb^{2} + j = (\overline{c} + b^{B−2})b^{2} + j = \overline{c}b^{2} + b^{B} + j is divisible by j since \overline{c}b^{2} + b^{B} ≡ 0 (mod N) and \overline{c}b^{2} + b^{B} ≡ 0 (mod b^{B}). Hence

(c − 1)b^{2} + (b − 1)b + (b − 1), cb^{2}, cb^{2} + 1, …, cb^{2} + (b − 2)

forms a sequence of consecutive b-prodigious numbers of length b. This sequence is maximal since (c−1)b^{2} +(b−1)b+(b−2) and (c−1)b^{2} +(b−1)b+(b−1) cannot both be b-prodigious by Lemma 6 and cb^{2} + (b − 1) = ((c − 1)b^{2} + (b − 1)b + (b − 1)) + b is not divisible by b − 1.

Lastly, if b = p^{α} for some odd prime p, then for each 1 ≤ j ≤ b−1, let β_{j} be the maximum integer such that p^{β_{j}} | j and j = p^{β_{j}}j′. By Lemma 2, there exists a positive integer c such that c ≡ −p^{−α} (mod j′) for all 1 ≤ j ≤ b − 1 and p_{b}(c) = 1. Then

cb^{2} + b, cb^{2} + b + 1, …, cb^{2} + b + (b − 1)

forms a maximal sequence of consecutive b-prodigious numbers of length b, since cb^{2} + b + (b − 1) is even, and hence cb^{2} + 2b is not b-prodigious.

4 Concluding remarks

Although the main focus of our paper is on consecutive b-prodigious numbers, there are many other questions about b-prodigious numbers that could be investigated. One such question is to consider the intersection of b-prodigious numbers with other integer sequences. In this direction, we end our paper with a theorem that considers the b-prodigious-Niven numbers, i.e., the intersection of the set of b-prodigious numbers and the set of b-Niven numbers.

Lemma 12 ([7]). Let b ≥ 2. If x and y are two b-Niven numbers such that s_{b}(x) = s_{b}(y), then s_{b}(x) | (x − y).

Theorem 13. The maximum length of a sequence of consecutive b-prodigious-Niven numbers is exactly

(a) 4 if b = 2;

(b) b if b > 2 is even; and (c) b + 1 if b > 2 is odd.

Proof. Every positive integer is 2-prodigious, so part (a) follows from the result by Cai [1] on 2-Niven numbers as mentioned in the introduction. In the rest of the proof, we consider b > 2.

To prove parts (b) and (c), we first note that 1, 2, …, b and 1, 2, …, b+1 are maximal sequences of consecutive b-prodigious-Niven numbers when b is even and b is odd, respectively. We call these the “trivial” sequences. To search for nontrivial sequences, let a and j be positive integers with 1 ≤ j ≤ b − 1 such that the sequence ab + j, ab + j + 1, …, (a + 1)b + j − 1 is b-prodigious-Niven. Since ab + (b − 1) is b-prodigious, we have by Lemma 1 that (b − 1) | a and thus (b − 1) | s_{b}(a). If j = b − 1, then (a + 1)b + b − 1 is not b-prodigious since (b−1) ∤ (a+1). If 1 ≤ j ≤ b−2, then by Lemma 6 we see that p_{b}(a) = 1, and we deduce that s_{b}(ab+j) = s_{b}((a+1)b+(j−1)) = s_{b}(a)+j. Thereforewe have (s_{b}(a)+j) | (b−1) by Lemma 12, contradicting that (b − 1) | s_{b}(a). In other words, every nontrivial sequence of consecutive bprodigious-Niven numbers with length at least b is a subsequence of ab, ab+1, …, ab+(b−1) or ab + (b − 1), (a + 1)b, …, (a + 1)b + (b − 2) with maximum length b.

5 Acknowledgments

These results are based on work supported by the National Science Foundation under grant numbered DMS-1852378.

References

[1] T. Cai, On 2-Niven numbers and 3-Niven numbers, Fibonacci Quart. 34 (1996), 118– 120.

[2] C. Cooper and R. E. Kennedy, On consecutive Niven numbers, Fibonacci Quart. 31 (1993), 146–151.

[3] J. De Koninck and F. Luca, Positive integers divisible by the product of their nonzero digits, Port. Math. 64 (2007), 75–85. Corrigendum, Port. Math. 74 (2017), 169–170.

[4] Diophante.fr, A365. Les Nombres prodigieux, 2022, http://www.diophante.fr/

    problemes-par-themes/arithmetique-et-algebre/a3-nombres-remarquables/

    3578-a365-les-nombres-prodigieux.

[5] H. Grundman, Sequences of consecutive n-Niven numbers, Fibonacci Quart. 32 (1994), 174–175.

[6] C. Sanna, On numbers divisible by the product of their nonzero base b digits, Quaest. Math. 43 (2020), 1563–1571.

[7] H. Grundman, J. Harrington, and T. W. H. Wong, Arithmetic progressions of b-Niven numbers, submitted.

[8] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, https://oeis.org, 2022.

[9] B. Wilson, Construction of 2n consecutive n-Niven numbers, Fibonacci Quart. 35 (1997), 122–128.

2020 Mathematics Subject Classification: Primary 11A63; Secondary 11B25. Keywords: prodigious number, Niven number, arithmetic progression.

(Concerned with sequence \underline{A055471}.)

Received June 16 2022; revised version received October 18 2022. Published in Journal of Integer Sequences, October 25 2022.

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