Polygonal, Sierpiński, and Riesel numbers

Daniel Baczkowski and Justin Eitner Department of Mathematics The University of Findlay

Findlay, OH 45840 USA

 baczkowski@findlay.edu
  eitnerj@findlay.edu

Carrie E. Finch^{1} and Braedon Suminski^{2} Department of Mathematics Washington and Lee University

Lexington, VA 24450

USA

     finchc@wlu.edu
suminskib14@mail.wlu.edu

Mark Kozek Department of Mathematics

Whittier College Whittier, CA 90608 USA

  mkozek@whittier.edu

Abstract

In this paper, we show that there are infinitely many Sierpiński numbers in the sequence of triangular numbers, hexagonal numbers, and pentagonal numbers. We also show that there are infinitely many Riesel numbers in the same sequences. Furthermore, we show that there are infinitely many n-gonal numbers that are simultaneously Sierpiński and Riesel.

1Partially supported by a Lenfest Grant.

2Supported by the R. E. Lee Summer Scholar program.

1 Introduction

Polygonal numbers are those that can be expressed geometrically by an arrangement of equally spaced points. For example, a positive integer n is a triangular number if n dots can be arranged in the form of an equilateral triangle. Similarly, n is a square number if n dots can be arranged in the form of a square. The diagram below represents the first four hexagonal numbers, which are 1, 6, 15, and 28.

In 1960, Sierpiński [11] showed that there are infinitely many odd positive integers k with the property that k ·2^{n} +1 is composite for all positive integers n. Such an integer k is called a Sierpiński number in honor of Sierpiński’s work. Two years later, Selfridge (unpublished) showed that 78557 is a Sierpiński number. To this day, this is the smallest known Sierpiński number. As of this writing, there are six candidates smaller than 78557 to consider: 10223, 21181, 22699, 24737, 55459, 67607. See http://www.seventeenorbust.com for the most up-to-date information.

Riesel numbers are defined in a similar way: an odd positive integer k is Riesel if k·2^{n}−1 is composite for all positive integers n. These were first investigated by Riesel in 1956 [10]. The smallest known Riesel number is 509203. As of this writing there are 50 remaining candidates

smaller that 509203 to consider. See http://www.prothsearch.net/rieselprob.html for

the most recent information.

The tool used to construct these numbers is a covering system. A collection of congruences

r_{1} (mod m_{1})
r_{2} (mod m_{2})
⋮
r_{t} (mod m_{t})

is called a covering system of congruences, also called a covering system, if each integer n satisfies n ≡ r_{i} (mod m_{i}) for some 1 ≤ i ≤ t. This technique was first introduced by Erdős who later used the idea to show that there are infinitely many odd numbers that are not of the form 2^{k} + p, where p is a prime [4].

Previous work has been done to show an intersection between Sierpiński or Riesel numbers with familiar integer sequences such as the Fibonacci numbers [8, 9] and the Lucas numbers [1]. Perfect power Sierpiński numbers and Riesel numbers have been studied in depth; in particular, there are infinitely many Sierpiński numbers of the form k^{r} for any fixed positive integer r [2, 5]. For Riesel numbers, there are infinitely many k such that k^{r} is Riesel for values of r with gcd(r, 12) ≤ 3 [2] and for gcd(r, 105) = 1 [7]. In this paper we expand on these findings by considering the intersection of sequences of polygonal numbers with Sierpiński and Riesel numbers. As the kth polygonal number for an n-sided polygon is given by \frac{1}{2}(k^{2}(n − 2) − k(n − 4)), which is quadratic in k, we build on techniques for constructing perfect power Sierpiński numbers and binomial Sierpiński and Riesel numbers (cf. [5, 6]). Through the use of coverings, we construct polygonal numbers that are also Riesel numbers and Sierpiński numbers.

Cohen and Selfridge showed that there are infinitely many numbers that are simultaneously Sierpiński and Riesel [3]. The smallest Sierpiński-Riesel number that came from their construction has 26 digits. Several others also produced Sierpiński-Riesel numbers; for example, Brier (unpublished) produced an example with 41 digits in 1998, and Gallot (unpublished) produced an example with 27 digits in 2000. In 2008, an example with 24 digits was produced [5]. Recently, a Sierpiński-Riesel number that is 22 digits long was discovered by Clavier (unpublished). Entry A180247 in the Online Encyclopedia of Integer Sequences has more information about these results. (See http://oeis.org/A180247.) We use similar techniques in this paper to find polygonal numbers that are Sierpiński-Riesel.

All computations were performed using the computer algebra system Maple.

2 Triangular and hexagonal numbers

2.1 Triangular-Sierpiński numbers

Let T_{k} denote the k^{th} triangular number. That is, T_{k} = \frac{k(k+1)}{2}. Now, consider the implications in Table 1 below.

n ≡ 1 (mod 2)  & k ≡ 1 (mod 3)  ⟹  3 | (T_{k} · 2^{n} + 1)
n ≡ 0 (mod 3)  & k ≡ 3 (mod 7)  ⟹  7 | (T_{k} · 2^{n} + 1)
n ≡ 2 (mod 4)  & k ≡ 1 or 3 (mod 5)  ⟹  5 | (T_{k} · 2^{n} + 1)
n ≡ 4 (mod 8)  & k ≡ 1 or 15 (mod 17)  ⟹  17 | (T_{k} · 2^{n} + 1)
n ≡ 8 (mod 12)  & k ≡ 4 or 8 (mod 13)  ⟹  13 | (T_{k} · 2^{n} + 1)
n ≡ 16 (mod 24) & k ≡ 53 or 187 (mod 241) ⟹ 241 | (T_{k} · 2^{n} + 1)

Table 1

The congruences for n in Table 1 form a covering. In each row, the congruence for k results in T_{k} · 2^{n} + 1 being divisible by one of the primes in P = {3, 5, 7, 17, 13, 241} for every positive integer in the congruence class for n. Assume that our values of k in each row will be chosen large enough from the congruence for k such that T_{k} · 2^{n} + 1 is larger than its prime divisor in that row. It follows that each of these T_{k} · 2^{n} + 1 must be composite for every positive integer in this congruence class. To ensure that T_{k} is odd (in order to satisfy the definition of a Sierpiński number), we also include the congruence k ≡ 1 or 2 (mod 4). If k = 4ℓ + 1, we have

T_{k} = \frac{1}{2}k(k + 1) = \frac{1}{2}(4ℓ + 1)(4ℓ + 2) = (4ℓ + 1)(2ℓ + 1),

and if k = 4ℓ + 2, we have

T_{k} = \frac{1}{2}k(k + 1) = \frac{1}{2}(4ℓ + 2)(4ℓ + 3) = (2ℓ + 1)(4ℓ + 3),

both of which are clearly odd.

Now, we find the intersection of all congruences for k to find a T_{k} that is a Sierpiński number. Using the Chinese remainder theorem, we have the following result.

Theorem 1. There are infinitely many Sierpiński numbers in the sequence of triangular numbers.

The smallest solution to the congruences for k that we find using the Chinese remainder theorem is 698953, and if k is a natural number with k ≡ 698953 (mod 4·3·5·7·13·17·241), then T_{k} is a Sierpiński number. Thus, the smallest Triangular-Sierpiński number from this construction is 244267997581.

2.2 Triangular-Riesel numbers

Consider the implications in Table 2 below.

n ≡ 0 (mod 2)  & k ≡ 1 (mod 3)  ⟹  3 | (T_{k} · 2^{n} − 1)
n ≡ 0 (mod 3)  & k ≡ 1 or 5 (mod 7)  ⟹  7 | (T_{k} · 2^{n} − 1)
n ≡ 1 (mod 4)  & k ≡ 2 (mod 5)  ⟹  5 | (T_{k} · 2^{n} − 1)
n ≡ 7 (mod 8)  & k ≡ 8 (mod 17)  ⟹  17 | (T_{k} · 2^{n} − 1)
n ≡ 11 (mod 12) & k ≡ 5 or 7 (mod 13)  ⟹  13 | (T_{k} · 2^{n} − 1)
n ≡ 16 (mod 24) & k ≡ 5 or 235 (mod 241) ⟹ 241 | (T_{k} · 2^{n} − 1)

Table 2

We also include k ≡ 1 or 2 (mod 4) to ensure T_{k} is odd. Once again, the congruences for n in Table 2 form a covering. In each row, the congruences for k result in T_{k} · 2^{n} − 1 being divisible by one of the primes in the set

P = {3, 7, 5, 17, 13, 241}

for every positive integer n in the same row. As before, assume that our values of k will be chosen large enough from those in the congruence for k so that T_{k} is larger than any of the primes in the set P. It follows that each of these T_{k} · 2^{n} − 1 must be composite for every positive integer n. We then find the intersection of all congruences for k to find a T_{k} that is a Riesel number. By use of the Chinese remainder theorem, we see that the smallest residue class that is in the intersection of all congruences for k is k ≡ 888802 (mod 4 · 3 · 7 · 5 · 17 · 13 · 241). Thus, for every such value of k, the triangular number T_{k} is Riesel. Hence, we have the following theorem.

Theorem 2. There are infinitely many Riesel numbers in the sequence of triangular numbers.

2.3 Hexagonal-Sierpiński and Riesel numbers

Let H_{k} denote the k^{th} hexagonal number. That is, H_{k} = 2k^{2} − k. Notice that if k = 2ℓ + 1, we have T_{k} = \frac{1}{2}k(k + 1) = 2ℓ^{2} + 3ℓ + 1 = H_{ℓ+1}. That is, when k is odd, the triangular number T_{k} is also a hexagonal number. Thus, if we include the congruences from Table 1 and k ≡ 1 (mod 2), then we will have a subset of the triangular numbers that are also Sierpiński numbers in addition to hexagonal numbers. The smallest such k is 698953, and all positive integers k congruent to 698953 modulo 4 · 3 · 7 · 5 · 17 · 13 · 241 also give T_{k} which are also Sierpiński and hexagonal. Similarly, the positive integers k that are congruent to 2916817 modulo 4 · 3 · 7 · 5 · 17 · 13 · 241 yield T_{k} which are Riesel and also hexagonal. Thus, we have the following corollaries.

Corollary 3. There are infinitely many Sierpiński numbers in the sequence of hexagonal numbers.

Corollary 4. There are infinitely many Riesel numbers in the sequence of hexagonal numbers.

2.4 Triangular-Sierpiński-Riesel numbers

Table 3 below gives congruences for k to construct triangular numbers T_{k} that are simultaneously Sierpiński and Riesel.

n ≡ 1 (mod 2)  & k ≡ 1 (mod 3)  ⟹  3 | (T_{k} · 2^{n} + 1)
n ≡ 1 (mod 3)  & k ≡ 2 or 4 (mod 7)  ⟹  7 | (T_{k} · 2^{n} + 1)
n ≡ 5 (mod 9)  & k ≡ 23 or 49 (mod 73)  ⟹  73 | (T_{k} · 2^{n} + 1)
n ≡ 6 (mod 12)  & k ≡ 1 or 11 (mod 13)  ⟹  13 | (T_{k} · 2^{n} + 1)
n ≡ 8 (mod 18)  & k ≡ 6 or 12 (mod 19)  ⟹  19 | (T_{k} · 2^{n} + 1)
n ≡ 2 (mod 36)  & k ≡ 15 or 21 (mod 37)  ⟹  37 | (T_{k} · 2^{n} + 1)
n ≡ 20 (mod 36) & k ≡ 24 or 84 (mod 109)  ⟹ 109 | (T_{k} · 2^{n} + 1)
n ≡ 4 (mod 5)  & k ≡ 13 or 17 (mod 31)  ⟹  31 | (T_{k} · 2^{n} + 1)
n ≡ 6 (mod 10)  & k ≡ 3 or 7 (mod 11)  ⟹  11 | (T_{k} · 2^{n} + 1)
n ≡ 8 (mod 20)  & k ≡ 9 or 31 (mod 41)  ⟹  41 | (T_{k} · 2^{n} + 1)
n ≡ 0 (mod 15)  & k ≡ 69 or 81 (mod 151)  ⟹ 151 | (T_{k} · 2^{n} + 1)
n ≡ 12 (mod 60) & k ≡ 20 or 40 (mod 61)  ⟹  61 | (T_{k} · 2^{n} + 1)
n ≡ 0 (mod 2)  & k ≡ 1 (mod 3)  ⟹  3 | (T_{k} · 2^{n} − 1)
n ≡ 1 (mod 4)  & k ≡ 2 (mod 5)  ⟹  5 | (T_{k} · 2^{n} − 1)
n ≡ 7 (mod 8)  & k ≡ 8 (mod 17)  ⟹  17 | (T_{k} · 2^{n} − 1)
n ≡ 11 (mod 16) & k ≡ 128 (mod 257)  ⟹ 257 | (T_{k} · 2^{n} − 1)
n ≡ 11 (mod 24) & k ≡ 90 or 150 (mod 241)  ⟹ 241 | (T_{k} · 2^{n} − 1)
n ≡ 3 (mod 48) & k ≡ 41 or 55 (mod 97) ⟹ 97 | (T_{k} · 2^{n} − 1)
n ≡ 19 (mod 48) & k ≡ 315 or 357 (mod 673) ⟹ 673 | (T_{k} · 2^{n} − 1)

Table 3

In Table 3, the congruences for n above the horizontal line form a covering. This part of the table ensures that the congruences for k yield a Sierpiński number T_{k}. In addition, the congruences for n below the horizontal line also form a covering. Thus, the bottom part of the table ensures that the congruences for k yield a Riesel number. Notice that the congruences for k above the line and those below the line are compatible; the only modulus that is repeated in the two parts of the table is 3, and in both instances, we have k ≡ 1 (mod 3).

Now we include the congruence k ≡ 1 or 2 (mod 4) to ensure that T_{k} is odd, and then use the Chinese remainder theorem to combine all of the congruences for k. The smallest solution to this set of congruences is

k ≡ 92290397124858700233022 (mod 270351155161021554764103899940).

We conclude there are infinitely many Sierpiński-Riesel numbers in the sequence of triangular numbers, and the smallest example resulting from this construction is

4258758700732063521204486546872386447899742753.

We state this result as a theorem below.

Theorem 5. There are infinitely many triangular numbers that are simultaneously Sierpiński numbers and Riesel numbers.

2.5 Hexagonal-Sierpiński-Riesel numbers

If we again include the congruence k ≡ 1 (mod 2) with the congruences in the previous subsection, we then have triangular numbers that are also hexagonal, in addition to being both Sierpiński and Riesel. Combining these congruences using the Chinese remainder theorem, we find the smallest solution to this set of congruences is

k ≡ 24743267730877977274574137 (mod 270351155161021554764103899940),

then T_{k} is hexagonal, Sierpiński, and Riesel. We conclude with the following:

Theorem 6. There are infinitely many hexagonal numbers that are simultaneously Sierpiński and Riesel numbers.

3 Pentagonal numbers

Let P_{k} denote the k^{th} pentagonal number. We then have P_{k} = \frac{1}{2}k(3k − 1). In this section, we show that there are infinitely many pentagonal-Sierpiński numbers, infinitely many pentagonal-Riesel numbers, and infinitely many pentagonal numbers that are simultaneously Sierpiński and Riesel.

3.1 Pentagonal-Sierpiński numbers

Consider the implications in Table 4 below.

n ≡ 1 (mod 2)  & k ≡ 1 (mod 3)  ⟹  3 | (P_{k} · 2^{n} + 1)
n ≡ 2 (mod 3)  & k ≡ 2 or 3 (mod 7)  ⟹  7 | (P_{k} · 2^{n} + 1)
n ≡ 2 (mod 4)  & k ≡ 1 (mod 5)  ⟹  5 | (P_{k} · 2^{n} + 1)
n ≡ 4 (mod 8)  & k ≡ 1 or 5 (mod 17)  ⟹  17 | (P_{k} · 2^{n} + 1)
n ≡ 0 (mod 12)  & k ≡ 3 or 6 (mod 13)  ⟹  13 | (P_{k} · 2^{n} + 1)
n ≡ 16 (mod 24) & k ≡ 189 or 213 (mod 241) ⟹ 241 | (P_{k} · 2^{n} + 1)

Table 4

Observe that if k ≡ 1 or 2 (mod 4), then P_{k} is odd. To see this, notice that if k = 4ℓ + 1, we have

P_{k} = \frac{1}{2}k(3k − 1) = \frac{1}{2}(4ℓ + 1)(12ℓ + 3 − 1) = (4ℓ + 1)(6ℓ + 1),

and if k = 4ℓ + 2 we have

P_{k} = \frac{1}{2}k(3k − 1) = \frac{1}{2}(4ℓ + 2)(12ℓ + 6 − 1) = (2ℓ + 1)(12ℓ + 5),

which are both clearly odd. Thus, we also include k ≡ 1 or 2 (mod 4) in order to construct Sierpiński numbers in this sequence.

Once again, the congruences for n in Table 4 form a covering. In each row, the congruences for k and n result in P_{k} · 2^{n} + 1 being divisible by one of the primes in the set P = {3, 5, 7, 13, 17, 241}. As before, assume that our values of k will be chosen large enough from those in the congruence classes for k so that P_{k} is larger than any of the primes in the set P. It follows that each of these P_{k} · 2^{n} − 1 must be composite for every positive integer n. We deduce the intersection of all congruences for k to find a P_{k} that is a Sierpiński number. Using the Chinese remainder theorem for the congruences for k, we find that there are infinitely many such k, and the smallest solution that arises out of these congruences is

k ≡ 56101 (mod 22369620).

We conclude the following:

Corollary 7. There are infinitely many pentagonal numbers that are Sierpiński numbers.

3.2 Pentagonal-Riesel numbers

In this section, we demonstrate the following result:

Theorem 8. There are infinitely many pentagonal numbers that are Riesel numbers.

We again prove this statement using a covering of the integers, shown in Table 5 below.

n ≡ 0 (mod 2)  & k ≡ 1 (mod 3)  ⟹  3 | (P_{k} · 2^{n} − 1)
n ≡ 2 (mod 3)  & k ≡ 6 (mod 7)  ⟹  7 | (P_{k} · 2^{n} − 1)
n ≡ 3 (mod 4)  & k ≡ 3 or 4 (mod 5)  ⟹  5 | (P_{k} · 2^{n} − 1)
n ≡ 1 (mod 8)  & k ≡ 10 or 13 (mod 17)  ⟹  17 | (P_{k} · 2^{n} − 1)
n ≡ 1 (mod 12)  & k ≡ 11 (mod 13)  ⟹  13 | (P_{k} · 2^{n} − 1)
n ≡ 21 (mod 24) & k ≡ 61 or 100 (mod 241) ⟹ 241 | (P_{k} · 2^{n} − 1)

Table 5

The congruences for n form a covering of the integers, so we again use the Chinese remainder theorem to combine the congruences for k (including k ≡ 1 or 2 (mod 4)); the smallest solution for k that satisfies all of the congruences in the table is

k ≡ 590029 (mod 22369620).

For any of these solutions for k, the expression P_{k} · 2^{n} − 1 is divisible by one of the primes in the set P = {3, 5, 7, 13, 17, 241}.

3.3 Pentagonal-Sierpiński-Riesel

We show now that there are infinitely many pentagonal numbers that are simultaneously Sierpiński and Riesel. Consider the congruences in Table 6.

In Table 6, the congruences for n above the horizontal line form a covering of the integers. Thus, the congruences for k above this line yield pentagonal numbers P_{k} that are Sierpiński. Similarly, the congruences for n below the horizontal line also form a covering of the integers. Thus, the corresponding congruences for k in the bottom part of the table yield pentagonal numbers P_{k} that are also Riesel. Again, the congruences for k above and below the line are compatible; the only repeated modulus is 3, and in both parts of the table, we have k ≡ 1 (mod 3). When we also include k ≡ 1 or 2 (mod 4) to make sure that the resulting P_{k} is also an odd integer, we find that there are 2^{17} solutions for k modulo

M = 4 · 3 · 5 · 7 · 11 · 13 · 17 · 19 · 31 · 37 · 41 · 61 · 73 · 97 · 109 · 151 · 241 · 257 · 673.

The smallest such solution is

k ≡ 180972518141277924651218 (mod M),

yielding the smallest pentagonal-Sierpiński-Riesel from this construction:

49126578483592751315774667185145775331460999677.

Thus, we have shown the following result.

Theorem 9. There are infinitely many pentagonal numbers that are simultaneously Sierpiński and Riesel numbers.

4 Polygonal numbers

Theorem 10. For infinitely many values of s, there are infinitely many s-gonal numbers that are Sierpiński numbers.

n ≡ 0 (mod 2)  & k ≡ 2 (mod 3)  ⟹  3 | (P_{k} · 2^{n} + 1)
n ≡ 1 (mod 4)  & k ≡ 3 or 4 (mod 5)  ⟹  5 | (P_{k} · 2^{n} + 1)
n ≡ 1 (mod 10)  & k ≡ 2 (mod 11)  ⟹  11 | (P_{k} · 2^{n} + 1)
n ≡ 7 (mod 8)  & k ≡ 9 or 14 (mod 17)  ⟹  17 | (P_{k} · 2^{n} + 1)
n ≡ 3 (mod 18)  & k ≡ 15 or 17 (mod 19)  ⟹  19 | (P_{k} · 2^{n} + 1)
n ≡ 11 (mod 24) & k ≡ 162 or 240 (mod 241) ⟹ 241 | (P_{k} · 2^{n} + 1)
n ≡ 3 (mod 16)  & k ≡ 140 or 203 (mod 257) ⟹ 257 | (P_{k} · 2^{n} + 1)
n ≡ 43 (mod 48) & k ≡ 32 or 33 (mod 97) ⟹ 97 | (P_{k} · 2^{n} + 1)
n ≡ 27 (mod 48) & k ≡ 112 or 337 (mod 673) ⟹ 673 | (P_{k} · 2^{n} + 1)
n ≡ 1 (mod 2)  & k ≡ 2 (mod 3)  ⟹  3 | (P_{k} · 2^{n} − 1)
n ≡ 6 (mod 10)  & k ≡ 2 (mod 11)  ⟹  11 | (P_{k} · 2^{n} − 1)
n ≡ 4 (mod 12)  & k ≡ 4 or 5 (mod 13)  ⟹  13 | (P_{k} · 2^{n} − 1)
n ≡ 12 (mod 18) & k ≡ 15 or 17 (mod 19)  ⟹  19 | (P_{k} · 2^{n} − 1)
n ≡ 24 (mod 36) & k ≡ 29 or 33 (mod 37)  ⟹  37 | (P_{k} · 2^{n} − 1)
n ≡ 10 (mod 20) & k ≡ 19 or 36 (mod 41)  ⟹  41 | (P_{k} · 2^{n} − 1)
n ≡ 58 (mod 60) & k ≡ 50 or 52 (mod 61)  ⟹  61 | (P_{k} · 2^{n} − 1)
n ≡ 6 (mod 36)  & k ≡ 83 or 99 (mod 109)  ⟹ 109 | (P_{k} · 2^{n} − 1)
n ≡ 2 (mod 3)  & k ≡ 6 (mod 7)  ⟹  7 | (P_{k} · 2^{n} − 1)
n ≡ 0 (mod 9)  & k ≡ 1 or 48 (mod 73)  ⟹  73 | (P_{k} · 2^{n} − 1)
n ≡ 4 (mod 5)  & k ≡ 22 or 30 (mod 31)  ⟹  31 | (P_{k} · 2^{n} − 1)
n ≡ 7 (mod 15)  & k ≡ 28 or 73 (mod 151)  ⟹ 151 | (P_{k} · 2^{n} − 1)

Table 6

Proof. Observe that the k^{th} s-gonal number is given by

S_{k} = S_{k}(s) = \frac{1}{2}k((s − 2)k − (s − 4)).

Using the congruences in Table 1, if s ≡ 3 (mod p) for each p in the set

P := {3, 5, 7, 13, 17, 241},

then we have

T_{k} ≡ S_{k} (mod ∏_{p∈P} p)

and

n ≡ 1 (mod 2)  & k ≡ 1 (mod 3)  ⟹  3 | (S_{k} · 2^{n} + 1)
n ≡ 0 (mod 3)  & k ≡ 3 (mod 7)  ⟹  7 | (S_{k} · 2^{n} + 1)
n ≡ 2 (mod 4)  & k ≡ 1 or 3 (mod 5)  ⟹  5 | (S_{k} · 2^{n} + 1)
n ≡ 4 (mod 8)  & k ≡ 1 or 15 (mod 17)  ⟹  17 | (S_{k} · 2^{n} + 1)
n ≡ 8 (mod 12)  & k ≡ 4 or 8 (mod 13)  ⟹  13 | (S_{k} · 2^{n} + 1)
n ≡ 16 (mod 24) & k ≡ 53 or 187 (mod 241) ⟹ 241 | (S_{k} · 2^{n} + 1)

Table 7

This implies that the expression S_{k} · 2^{n} + 1 is composite for all positive integers n if k lies in the intersection of the congruence classes listed in the table above since the congruences for n form a covering of the integers. If we also include the congruence k ≡ 1 (mod 4), then the resulting polygonal number S_{k} is odd since k = 4ℓ + 1 implies

S_{k} = (4ℓ + 1)(2ℓs − 4ℓ + 1).

The conclusion follows.

Using the same technique, we also deduce the following results.

Theorem 11. For infinitely many values of s, there are infinitely many s-gonal numbers that are Riesel numbers.

Theorem 12. For infinitely many values of s, there are infinitely many s-gonal numbers that are Sierpiński-Riesel numbers.

5 Acknowledgments

The authors thank the referee for the valuable suggestions. The third author wishes to thank the Department of Mathematics at The University of Findlay for the hospitality during her visit. The fourth author is grateful for the hospitality of the Mathematics Department at Washington and Lee University during his visit. The authors gratefully acknowledge the computational support from Washington and Lee University student Elliot Emadian (’17) and Professor Gregg Whitworth (Biology).

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2010 Mathematics Subject Classification: Primary 11A07; Secondary 11Y55. Keywords: Sierpiński number, Riesel number, triangular number, covering congruence.

(Concerned with sequences \underline{A000217}, \underline{A000290}, \underline{A000326}, \underline{A000384}, and \underline{A180247}.)

Received November 25 2014; revised version received June 19 2015; July 9 2015. Published in Journal of Integer Sequences, July 16 2015.

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