Infinite Sets of b-Additive and b-Multiplicative Ramanujan-Hardy Numbers

Viorel Niţică Department of Mathematics West Chester University of Pennsylvania West Chester, PA 19383 USA

                   vnitica@wcupa.edu

Abstract

Let b a numeration base. A b-additive Ramanujan-Hardy number N is an integer for which there exists at least one integer M, called the additive multiplier, such that the product of M and the sum of base-b digits of N, added to the reversal of the product, gives N. We show that for any b there exist infinitely many b-additive Ramanujan-Hardy numbers and infinitely many additive multipliers. A b-multiplicative Ramanujan-Hardy number N is an integer for which there exists at least an integer M, called the multiplicative multiplier, such that the product of M and the sum of base-b digits of N, multiplied by the reversal of the product, gives N. We show that for b ≡ 4 (mod 6), and for b = 2, there exist infinitely many b-multiplicative Ramanujan-Hardy numbers and infinitely many multiplicative multipliers. If b even, b ≡ 0 (mod 3) or b ≡ 2 (mod 3), we show there exist infinitely many numeration bases for which there exist infinitely many b-multiplicative Ramanujan-Hardy numbers and infinitely many multiplicative multipliers.

These results completely answer two questions and partially answer two other questions asked in a previous paper of the author.

1 Introduction

Let b ≥ 2 be a numeration base. In Niţică [6], motivated by some properties of the taxicab number, 1729, we introduce the classes of b-additive Ramanujan-Hardy (or b-ARH) numbers and b-multiplicative Ramanujan-Hardy (or b-MRH) numbers. The first class consists of numbers N for which there exists at least an integer M, called the additive multiplier, such that the product of M and the sum of base-b digits of N, added to the reversal of the product, gives N. The second class consists of numbers N for which there exists at least an integer M, called the multiplicative multiplier, such that the product of M and the sum of base-b digits of N, multiplied by the reversal of the product, gives N.

It is asked [6, Question 6] if the set of b-ARH numbers is infinite and it is asked [6, Question 8] if the set of additive multipliers is infinite. It is shown [6, Theorems 12 and 15] that the answer is positive if b is even. The case b odd is left open. It is asked [6, Question 7] if the set of b-MRH numbers is infinite and it is asked [6, Question 9] if the set of multiplicative multipliers is infinite. It is shown [6, Theorem 30] that the answer is positive if b is odd. The case b even is left open.

We recall that Niven (or Harshad) numbers are numbers divisible by the sum of their decimal digits. Niven numbers have been extensively studied. See, for instance, Cai [1], Cooper and Kennedy [2], De Koninck and Doyon [3], and Grundman [4]. Of interest are also b-Niven numbers, which are numbers divisible by the sum of their base-b digits. See, for example, Fredricksen, Ionaşcu, Luca, and Stănică [5]. A b-MRH-number is a b-Niven number. High degree b-Niven numbers are introduced in [7].

The goal of this paper is to show that, for any numeration base, there exist infinitely many b-ARH numbers and infinitely many distinct additive multipliers. We also show that, for b ≡ 4 (mod 6), and for b = 2, there exist infinitely many b-MRH numbers, and infinitely many distinct multiplicative multipliers. If b even, b ≡ 0 (mod 3) or b ≡ 2 (mod 3), we show there are infinitely many numeration bases for which there exist infinitely many bmultiplicative Ramanujan-Hardy numbers and infinitely many multiplicative multipliers. These results completely answer the first two questions from [6] revisited above, and partially answer the other two. We observe that a trivial example of infinitely many b-MRH numbers is given by the powers of 10. Our examples have at least two digits different from zero. Finding infinitely many b-MRH numbers with all digits different from zero remains an open question.

Our results about b-ARH numbers also give solutions to the Diophantine equation N·M = reversal(N ·M). Motivated by this link, we show that the Diophantine equation has solution for all integers N not divisible by the numeration base b. We do not know how to answer the following related question:

Question 1. Does there exist, for any integer N, an integer M such that N · M is a b-ARH number (or a b-MRH number, or a b-Niven number)?

Our final result shows that for any string of digits I there exist infinitely many b-Niven numbers that contain I in their base-b representation. We do not know a similar result for the classes of b-ARH and b-MRH numbers.

2 Statements of the main results

Let s_{b}(N) denote the sum of base-b digits of integer N. If x is a string of digits, let (x)^{∧k} denote the base-10 integer obtained by repeating x k-times. Let [x]_{b} denote the value of the string x in base b.If N is an integer, let N^{R} denote the reversal of N, that is, the number obtained from N writing its digits in reverse order. The operation of taking the reversal is dependent on the base. In the definition of a b-ARH-number/b-MRH number N we take the reversal of the base-b representation of s_{b}(N)M.

Theorem 2. Let α ≥ 1 integer, b ≥ α+1 integer, and k = (1+α)^{ℓ}, ℓ ≥ 0. Assume b ≡ 2+α (mod 2 + 2α). Define

N_{k} = [(1α)^{∧k}]_{b}.

Then there exists M ≥ 0 integer such that

s_{b}(N_{k}) · M = (s_{b}(N_{k}) · M)^{R} = \frac{N_{k}}{2} .

In particular, the numbers N_{k}, k ≥ 1, are b-ARH numbers and b-Niven numbers.

The proof of Theorem 2 is done in Section 3. Remark 3. The particular case b = 10, α = 2, of Theorem 2, which gives N_{k} = (12)^{3^{ℓ}}, is covered by [6, Example 10]. Theorem 2 does not give any information if b = 2.

The following proposition gives positive answers to [6, Questions 5 and 6].

Proposition 4. For any b ≥ 2, there exist infinitely many b-ARH numbers and infinitely many additive multipliers. The b-ARH numbers are also b-Niven numbers.

The proof of Proposition 4 is done in Section 4. Remark 5. Note that [6, Theorems 12 and 15] show, for all even bases, infinitely many b-ARH numbers that are not b-Niven numbers. The case of odd base is open. The question of finding infinitely many b-Niven numbers that are not b-ARH numbers is also open. It is shown in [6, Theorem 28] that for any base there exist infinitely many numbers that are not b-ARH numbers.

The result in Theorem 2 gives many base-10 solutions for the equation:

N · M = (N · M)^{R}.  (1)

One can try to solve the equation (1) ,where (N · M)^{R} is the reversal of N · M written in base b, for any numeration base b.

Observe that if N is divisible by b, then (N · M)^{R} has less digits then N · M, therefore N is not a solution of (1). Note also that if N = N^{R} and N has k digits then (1) always has an infinite set of solutions with

M = [(1(0)^{∧ℓ})^{∧p}1]_{b}, ℓ ≥ k − 1, p ≥ 0.

Consequently, if (N_{0}, M_{0}) is a solution of (1), then (1) has infinite sets of solutions of types (N_{0}, M) and (N, M_{0}).

Theorem 6. Let b ≥ 2 and N ≥ 1 integer such that b |̸N. Then N is a solution of (1).

The proof of Theorem 6 is done in Section 5. For base 10, a proof belonging to David Radcliffe can be found at [8]. We learned about this reference from J. Shallit. We generalize the proof for an arbitrary numeration base. After our paper was written, we learned from J. Shallit [9] that he also has a proof of Theorem 6.

A b-numeric palindrome is a base-b integer N such that N = N^{R}.

Corollary 7. All integers, not divisible by b, are factors of b-numeric palindromes.

Definition 8. The multiplicity of a multiplicative multiplier M is the number of (N, M) solutions of (1).

It was observed above that for any solution (N, M) of (1), M has infinite multiplicity. The following theorem shows infinitely many solutions of (1) independent of above.

Theorem 9. Let b ≥ 2 a numeration base. Then, for all k ≥ 0, we have

[1(b − 1)]_{b} · [(b − 1)^{∧k}]_{b} = [1(b − 2)(b − 1)^{∧k−2}(b − 2)1]_{b}.

The proof of Theorem 9 is done in Section 6.

Our next results show, for b even, more examples of infinite sets of of b-ARH.

Theorem 10. Let b ≥ 2 even. Let a ∈ {1, 2, …, b − 1} and let k ≥ 0 be an integer.

(a) Let

N_{k} = [a(0)^{∧k}a]_{b}.

Then N_{k} is a b-ARH number, but not a b-Niven number.

(b) Let

N_{k} = [(1(0)^{∧k})^{∧b} 0 ((0)^{∧k}1)^{∧b}]_{b}.

Then N_{k} is a b-ARH number, but not a b-Niven number.

(c) Let

N_{k} = [((0)^{∧k}1)^{∧b} 0 (1(0)^{∧k})^{∧b}]_{b}.

Then N_{k} is a b-ARH number and a b-Niven number.

The proof of Theorem 10 is done in Section 7.

The following theorem gives partial answers to [6, Questions 7 and 8].

Theorem 11.

(a) Let b ≡ 4 (mod 6). Let k ≥ 1 integer such that k ≡ 1 (mod 3). Define

α_{k} = [1(0)^{∧k}(b − 2)]_{b}.

Then N_{k} = α_{k} · (α_{k})^{R} is a b-MRH number.

(b) Let b = 2 and let k ≥ 1 be an even integer. Define

α_{k} = [1(0)^{∧k}1]_{2}.

Then N_{k} = α_{k} · (α_{k})^{R} is a b-MRH number.

In particular, for any numeration base b, b ≡ 4 (mod 6), and for b = 2, there exist infinitely many b-MRH numbers and infinitely many multipliers.

The proof of Theorem 11 is done in Section 8.

Our next result lists several infinite sequences of 10-MRH-numbers.

Proposition 12. Assume k ≥ 1 integer and define N_{k} = α_{k} · (α_{k})^{R}, where α_{k} is one of the following numbers:

• [1(0)^{∧k}8]_{10}, k ≡ 1 (mod 3),

• [7(0)^{∧k}2]_{10},

• [5(0)^{∧k}4]_{10},

• [4(0)^{∧k}5]_{10}

Then N_{k} is a 10-MRH number.

The first item in Proposition 12 follows as a corollary of Theorem 11. The other items can be proved using the same approach as in the proof of Theorem 11.

If b even, b ≡ 0 (mod 3) or b ≡ 2 (mod 3), the next theorem shows there are infinitely many numeration bases for which there exist infinitely many b-MRH numbers and infinitely many multipliers.

Theorem 13.

(a) Let b ≥ 18, b = 6a, and a ≡ 1 (mod 25). Let α_{k} = [1(0)^{∧k}4]_{b} with k ≡ 4 (mod 5). Then N_{k} = α_{k} · α_{k}^{R} is a b-MRH number. The corresponding multipliers are distinct.

(b) Let b ≥ 18, b = 8a, a ≡ 1 (mod 25), and a ≡ 1 (mod 3). Let α_{k} = [1(0)^{∧k}4]_{b} with k ≡ 4 (mod 20). Then N_{k} = α_{k} · α_{k}^{R} is a b-MRH number. The corresponding multipliers are distinct.

The proof of Theorem 13 is done in section 9.

Theorem 14. For any base b and for any string of base b digits I there exist infinitely many b-Niven numbers that contain the string I in their base-b representation.

Proof. Let I be a string of base-b digits. There exist infinitely many base-b strings J such that s_{b}([IJ]_{b}) is a power of b, say b^{k}, k ≥ 1. Then the number N_{J} = [IJ(0)^{∧k}]_{b} is a b-Niven number.

3 Proof of Theorem 2

Proof. The condition b ≡ 2 + α (mod 2 + 2α) implies that b + α is even. The base-b representation for N_{k}/2 is N_{k}/2 = [(0\frac{b+α}{2})^{∧k}]_{b}. One has that:

s_{b}(N_{k}) = k · (1 + α) = (1 + α)^{ℓ+1}.  (2)

The value of N_{k}/2 in base 10 is obtained summing a geometric series.

\frac{N_{k}}{2} = \frac{b + α}{2} · b^{2k−2} + \frac{b + α}{2} · b^{2k−4} + ⋯ + \frac{b + α}{2} · b^{2} + \frac{b + α}{2} = \frac{b + α}{2} · \frac{b^{2k} − 1}{b^{2} − 1}
= \frac{b + α}{2} · \frac{b^{2(1+α)^{ℓ}} − 1}{b^{2} − 1}.  (3)

Note that N_{k}/2 = (N_{k}/2)^{R}. We finish the proof of the theorem if we show that:

(1 + α)^{ℓ+1}|\frac{b + α}{2} · \frac{b^{2(1+α)^{ℓ}} − 1}{b^{2} − 1}.  (4)

We prove (4) by induction on ℓ. For ℓ = 0 equation (4) becomes 1 + α|\frac{b+α}{2}, which is true because b ≡ 2 + α (mod 2 + 2α).

Now we assume that (4) is true for ℓ and show that it is true for ℓ + 1.

\begin{matrix}\frac{b + α}{2} · \frac{b^{2(1+α)^{ℓ+1}} − 1}{b^{2} − 1} = \frac{b + α}{2} · \frac{(b^{2(1+α)^{ℓ}})^{1+α} − 1}{b^{2} − 1} \\ = \frac{b + α}{2} · \frac{b^{2(1+α)^{ℓ}} − 1}{b^{2} − 1} (B^{α} + B^{α−1} + ⋯ + B^{2} + B + 1) , \end{matrix}  (5)

where

B = b^{2(1+α)^{ℓ}}.  (6)

The congruence b ≡ 2 + α (mod 2 + 2α) implies that

b^{2} ≡ (2 + α)^{2} ≡ α^{2} + 4α + 4 ≡ α^{2} ≡ 1 (mod 1 + α),

which implies that

b^{m} ≡ 1 (mod 1 + α), m even.  (7)

From (6) and (7) follows that B^{p} ≡ 1 (mod 1 + α), 1 ≤ p ≤ α, so

1 + α|B^{α} + B^{α−1} + ⋯ + B^{2} + B + 1.  (8)

Combining (4) (for ℓ) and (8), and using (5), it follows that (4) is true for ℓ + 1.

4 Proof of Proposition 4

Proof. The case b = 2 is covered by [6, Theorem 12]. If b ≥ 3, choose α = b − 2 and apply Theorem 2. We show that, for a fixed b, the multipliers appearing in the proof of Theorem 2 are all distinct. It follows from (2) and (3) that the multiplier for N_{k} is given by:

M = \frac{\frac{N_{k}}{2}}{s_{b}(N_{k})} = \frac{\frac{b+α}{2} · \frac{b^{2(1+α)^{ℓ}}−1}{b^{2}−1}}{(1 + α)^{ℓ+1}} .  (9)

Note that α = b − 2. After algebraic manipulations, equation (9) becomes

M = \frac{b^{2(1+α)^{ℓ}} − 1}{(b − 1)^{ℓ}(b^{2} − 1)}.

In order to show that the multipliers are distinct it is enough to show that the sequence of multipliers is strictly increasing as a function of ℓ That is, we need to show that:

\frac{b^{2(1+α)^{ℓ}} − 1}{(b − 1)^{ℓ}(b^{2} − 1)} < \frac{b^{2(1+α)^{ℓ+1}} − 1}{(b − 1)^{ℓ+1}(b^{2} − 1)}.  (10)

After algebraic manipulations (10) becomes

(b − 1)(b^{2(1+α)^{ℓ}} − 1) < b^{2(1+α)^{ℓ+1}} − 1.  (11)

After denoting

B = b^{2(1+α)^{ℓ}} = b^{2(b−1)^{ℓ}},

right hand side of (11) factors as follows:

b^{2(1+α)^{ℓ+1}} − 1 = (b^{2(1+α)^{ℓ}} − 1)(B^{α} + B^{α−1} + ⋯ + B + 1).  (12)

Now (11) follows from (12) and the following inequality:

b − 1 < b^{2(b−1)^{ℓ}}, ℓ ≥ 0, ℓ ≥ 0, b ≥ 3.

5 Proof of Theorem 6

Proof. Let b = p_{1}^{α_{1}}p_{2}^{α_{2}} ⋯ p_{k}^{α_{k}}, α_{i} ≥ 1, p_{i} prime, 1 ≤ i ≤ k. We recall that a base-b integer N is divisible by p_{i}^{γ} if the last γ digits of N form a base-b integer divisible by p_{i}^{γ}. Let N = p_{1}^{β_{1}}p_{2}^{β_{2}} ⋯ p_{k}^{β_{k}}w, where gcd(w, b) = 1. Let m = max(β_{1}, β_{2}, ⋯, β_{k}). Let L be the base-b integer equal to p_{1}^{β_{1}}p_{2}^{β_{2}} ⋯ p_{k}^{β_{k}}. As b |̸N, the last digit of L is not 0. Let ℓ be the length of L. Consider the base-b palindrome P = [L^{R}(0)^{∧m−ℓ}L]_{b}, where L^{R} is the reversal of base-b representation of L. As P is divisible by p_{1}^{β_{1}}p_{2}^{β_{2}} ⋯ p_{k}^{β_{k}}, this is the end of the proof if w = 1.

Assume w > 1. Let ϕ be Euler’s totient function which counts the positive integers up to a given integer n that are relatively prime to n. As gcd(w, b) = 1 Euler’s theorem implies that b^{ϕ(w)} − 1 ≡ 0 (mod w).

Let r be an multiple of ϕ(w) which is greater than l + m, the length of P. Let q ≥ 1 a multiple of b^{ϕ(w)} − 1. Consider the infinite family of integers given by

Q_{r,q} = [1((0)^{∧r−1}1)^{∧q}]_{b} = 1 + b^{r} + b^{2r} + ⋯ + b^{qr}
= 1 + b^{r} + b^{2r} + ⋯ + b^{qr} + q − q  (13)
= (b^{r} − 1) + (b^{2r} − 1) + (b^{3r} − 1) + ⋯ + (b^{qr} − 1) + q.

All terms in the last part of (13) are divisible by b^{ϕ(w)} −1, so Q_{r,q} is divisible by b^{ϕ(w)} −1 and by w. We finish the proof observing that P · Q_{r,q} is a base-b palindrome divisible by N.

6 Proof of Theorem 9

Proof. Observe that:

\begin{matrix}(b − 1) · (b − 1) = b(b − 2) + 1 = [(b − 2)1]_{b} \\ (b − 1)b^{k} + (b − 1)b^{k} = b^{k} + (b − 2)b^{k−1} = [1(b − 2)0^{∧k}]_{b}. \end{matrix}  (14)

Using (14) we get

[1(b − 1)]_{b} · [(b − 1)^{∧k}]_{b} = (b + b − 1) · (∑_{i=0}^{k−1}(b − 1)b^{i})

= ∑_{i=0}^{k−1} ((b − 1)b^{i+1} + (b(b − 2) + 1) b^{i})

= ∑_{i=1}^{k}(b − 1)b^{i} + ∑_{i=0}^{k−1} (b(b − 2) + 1) b^{i}

= (b − 1)b^{k} + ∑_{i=1}^{k−1} ((b − 1) + b(b − 2) + 1)b^{i} + b(b − 2) + 1

= (b − 1)b^{k} + ∑_{i=1}^{k−1}(b − 1)b^{i+1} + b(b − 2) + 1

= (b − 1)b^{k} + (b − 1)b^{k} + ∑_{i=1}^{k−2}(b − 1)b^{i+1} + b(b − 2) + 1

= b^{k} + (b − 2)b^{k−1} + ∑_{i=1}^{k−2}(b − 1)b^{i+1} + b(b − 2) + 1

= [1(b − 2)(b − 1)^{∧k−2}(b − 2)1]_{b}.

7 Proof of Theorem 10

Proof. (a) Note that s_{b}(N_{k}) = 2a. As b is even, there exists an integer M such that:

2a · M = [a(0)^{∧k+1}]_{b}.

The following computation shows that N_{k} is a b-ARH number:

s_{b}(N_{k}) · M + (s_{b}(N_{k}) · M)^{R} = [a(0)^{∧k+1}]_{b} + [a]_{b} = [a(0)^{∧k}a]_{b} = N_{k}.

To show that N_{k} is not b-Niven observe that N_{k}/a = [1(0)^{∧k}1]_{b} is odd.

(b) Note that s_{b}(N_{k}) = 2b. As b is even, the multiplier M = [(1(0)^{∧k})^{∧b}(0)^{∧kb+b−1}]_{b}/2 is an integer.

The following computation shows that N_{k} is a b-ARH number:

\begin{matrix}s_{b}(N_{k}) · M + (s_{b}(N_{k}) · M)^{R} \\ = [(1(0)^{∧k})^{∧b}(0)^{∧kb+b}]_{b} + [((0)^{∧k}1)^{∧b}]_{b} = [(1(0)^{∧k})^{∧b}0((0)^{∧k}1)^{∧b}]_{b} = N_{k}. \end{matrix}

To show that N_{k} is not b-Niven observe that N_{k} is not divisible by b.

(c) The proof is similar to that of b).

8 Proof of Theorem 11

Proof. (a) Using the fact that

(b − 2)^{2} = b^{2} − 4b + 4 = b(b − 4) + 4 = [(b − 4)4]_{b},

an equivalent base-b representation for N_{k} is given by

N_{k} = \begin{cases}[(b − 2)(0)^{∧k−1}(b − 4)5(0)^{∧k}(b − 2)]_{b}, & if b ≠ 4; \\ [2(0)^{∧k−1}11(0)^{∧k}2]_{4}, & if b = 4. \end{cases}  (15)

If b ≠ 4 one has s_{b}(N_{k}) = 3(b − 1) and if b = 4 one has s_{4}(N_{k}) = 6. To finish the proof of case a) it is enough to show that α_{k} is divisible by s_{b}(N_{k}).

If b ≠ 4 we get

α_{k} = b^{k+1} + b − 2 = b^{k+1} − 1 + b − 1 = (b − 1)(b^{k} + b^{k−1} + ⋯ + b^{2} + b + 2)

and

b^{k} + b^{k−1} + ⋯ + b^{2} + b + 2 ≡ k + 2 ≡ 0 (mod 3).

For the first congruence we used b ≡ 1 (mod 3) and for the second we used k ≡ 1 (mod 3). If b = 4, then clearly α_{k} is divisible by 2. Moreover

α_{k} = 4^{k+1} + 2 = (3 + 1)^{k+1} + 2 ≡ 0 (mod 3),

which shows that α_{k} is divisible by 6.

(b) Now assume that b = 2. Then an equivalent base-2 representation for N_{k} is given by

N_{k} = [1(0)^{∧k−1}10(0)^{∧k}1]_{2},

so s_{2}(N_{k}) = 3. To finish the proof, we use the fact that k is even to show that α_{k} is divisible by 3:

α_{k} = 2^{k+1} + 1 = (3 − 1)^{k+1} + 1 ≡ 0 (mod 3).

To prove the last claim in the theorem, we show that the multipliers corresponding to various values of k are distinct. This follows from the explicit formulas below. All sequences of multipliers are strictly increasing as functions of k.

If b = 2 the sequence of multipliers is given by M_{k} = \frac{2^{k+1}+1}{3}.

If b = 4 the sequence of multipliers is given by M_{k} = \frac{4^{k+1}+2}{6}.

If b > 4 the sequence of multipliers is given by M_{k} = \frac{b^{k+1}+b−2}{3(b−1)}.

9 Proof of Theorem 13

Proof. (a) The base-b representation for N_{k} is

N_{k} = [4(0)^{∧k−1}(17)(0)^{∧k}4]_{b}.

Therefore s_{b}(N_{k}) = 25. If k = 5ℓ + 4. one has that:

α_{k} = 6^{k}a^{k} + 4 ≡ (6^{5})^{ℓ}6^{4} + 4 ≡ (7776)^{ℓ} · 296 + 4 ≡ 0 (mod 25).

Hence N_{k} is a b-MRH number with multiplier \frac{α_{k}}{25} = \frac{(6a)^{k}+4}{25}.

(b) As above, s_{b}(N_{k}) = 25. If k = 20ℓ + 4, one has that:

α_{k} = 8^{k}a^{k} + 4 ≡ (8^{20})^{ℓ}8^{4} + 4 ≡ (76)^{ℓ} · 96 + 4 ≡ 0 (mod 25).

Hence N_{k} is a b-MRH number with multiplier \frac{α_{k}}{25} = \frac{(8a)^{k}+4}{25}.

10 Acknowledgments

The author would like to thank the editor and the referee for valuable comments that helped him write a better paper.

References

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

[2] C. N. Cooper and R. E. Kennedy, On consecutive Niven numbers, Fibonacci Quart., 21 (1993), 146–151.

[3] J. M. De Koninck and N. Doyon, Large and small gaps between consecutive Niven numbers, J. Integer Sequences 6 (2003), Article 03.2.5.

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

[5] H. Fredricksen, E. J. Ionaşcu, F. Luca, and P. Stănică, Remarks on a sequence of minimal Niven numbers, in S. W. Golomb, G. Gong, T. Helleseth and H.-Y. Song, eds., Sequences, Subsequences, and Consequences, Lecture Notes in Comp. Sci., Vol. 4893, Springer, 2007, pp. 162–168.

[6] V. Niţică, About some relatives of the taxicab number, J. Integer Sequences, 21 (2018), Article 18.9.4.

[7] V. Niţică, High degree b-Niven numbers, to appear in Integers. Preprint available at

   http://arxiv.org/abs/1807.02573, July 6 2018.
[8] D. Radcliffe, World of numbers, http://www.worldofnumbers.com/em36.htm.

[9] J. Shallit, Palindromic multiples, preprint, 2018.

2010 Mathematics Subject Classification: Primary 11B83; Secondary 11B99. Keywords: numeration base, b-Niven number, reversal, additive b-Ramanujan-Hardy number, multiplicative b-Ramanujan-Hardy number, palindrome.

(Concerned with sequences \underline{A005349}, \underline{A067030}, \underline{A305130}, and \underline{A305131}.)

Received December 26 2018; revised versions received March 14 2019; March 30 2019; April 4 2019. Published in Journal of Integer Sequences, May 24 2019.

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