On Sums of Practical Numbers and Polygonal Numbers

Sai Teja Somu JustAnswer Bengaluru 560052 India

                  somuteja@gmail.com

Duc Van Khanh Tran Department of Mathematics University of Texas at Austin Austin, TX 78712 USA

                 duc.tranvk@utexas.edu

Abstract

Practical numbers are positive integers n such that every positive integer less than or equal to n can be written as a sum of distinct positive divisors of n. We show that all positive integers can be written as a sum of a practical number and a triangular number, resolving a conjecture by Sun. We also show that all sufficiently large natural numbers can be written as a sum of a practical number and two s-gonal numbers.

1 Introduction

By a natural number, we mean a positive integer. Practical numbers, introduced by Srinivasan in [9], are natural numbers n such that every natural number less than or equal to n can be written as a sum of distinct positive divisors of n. The sequence of practical numbers can also be found on the On-Line Encyclopedia of Integer Sequences (OEIS) (see the sequence \underline{A005153} in the OEIS [7]). Stewart [10] and Sierpiński [6] proved the characterization that a natural number n ≥ 2 with prime factorization n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}}, where p_{1} < p_{2} < ⋯ < p_{k}, is practical if and only if p_{1} = 2 and

p_{j} ≤ σ(p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{j−1}^{a_{j−1}}) + 1

for all 2 ≤ j ≤ k, where σ(·) denotes the sum of divisors.

There have been many works on various additive representations of natural numbers involving practical numbers. Melfi [4] showed that every even natural number is a sum of two practical numbers. Pomerance and Weingartner [5] proved that every sufficiently large odd number can be written as a sum of a practical number and a prime. Somu et al. [8] proved that all natural numbers congruent to 1 modulo 8 are expressible as a sum of a practical number and a square. In this paper, we derive results involving sums of practical and polygonal numbers.

In this article, we consider 0 as a triangular number. In Section 3, we prove that all natural numbers can be written as a sum of a practical number and a triangular number, resolving the conjecture proposed by Sun (see the sequence \underline{A208244} in the OEIS [7]). We prove the following theorem.

Theorem 1. Every natural number can be written as a sum of a practical number and a triangular number.

Section 4 focuses on additive representations of practical and polygonal numbers more generally. We prove that all sufficiently large natural numbers can be written as a sum of a practical number and two s-gonal numbers. We prove the following theorem.

Theorem 2. Let s be a natural number greater than 3. Then there exists a natural number N(s) such that all natural numbers greater than N(s) can be written as a sum of a practical number and two s-gonal numbers.

Finally, in Section 5, we propose two conjectures regarding additive representations of natural numbers involving practical numbers.

2 Notation

We use the following notation:

ℕ: the set of positive integers.

ℕ_{0}: the set of non-negative integers.

σ(·): the sum of positive divisors of a natural number.

≡: We say a ≡ b (mod n) if n divides a − b. gcd(·, ·): the greatest common divisor of two integers.

⌊·⌋: the floor function of a real number, which is the largest integer not exceeding that real number.

P_{s}(n): the n-th s-gonal number given by P_{s}(n) = (s − 2)\frac{n(n−1)}{2} + n = \frac{(s−2)n^{2}−(s−4)n}{2}, for s ≥ 3, n ≥ 0.

O(·): We say f(x) = O(g(x)) if there exists a positive real number M such that |f(x)| ≤ Mg(x) for all sufficiently large x.

3 Proof of Theorem 1

In this section, we prove that all natural numbers can be written as a sum of a practical number and a triangular number, resolving the conjecture proposed by Sun (see the sequence \underline{A208244} in the OEIS [7]). We require three lemmas to prove Theorem 1.

Lemma 3. Let m and n be natural numbers. There exists a natural number 1 ≤ x ≤ 2^{m} − 1 such that

x^{2} ≡ 8n + 1 (mod 2^{m+2}).

Proof. See [8, Lemma 3.2] for proof.

Lemma 4. If m is a practical number and n is a natural number such that n ≤ σ(m) + 1, then mn is a practical number.

Proof. See [3, Corollary 1] for proof.

Lemma 5. If x is an odd natural number, then \frac{x^{2}−1}{8} is a triangular number.

Proof. As x = 2k + 1 for some non-negative integer k, we have \frac{x^{2}−1}{8} = \frac{k(k+1)}{2}. Hence, we can conclude that \frac{x^{2}−1}{8} is a triangular number.

Now we prove Theorem 1.

Proof of Theorem 1. Let n be a natural number and m = ⌊log_{2} \sqrt{8n + 1}⌋. By Lemma 3, there exists a natural number x such that 1 ≤ x ≤ 2^{m} − 1 and x^{2} ≡ 8n + 1 (mod 2^{m+2}). Since x ≤ 2^{m} − 1 and m = ⌊log_{2} \sqrt{8n + 1}⌋, we have x^{2} < 2^{2m} ≤ 8n + 1.

As x^{2} < 8n + 1 and x^{2} ≡ 8n + 1 (mod 2^{m+2}), we have 8n + 1 − x^{2} = 2^{m+2}s for some natural number s. As m = ⌊log_{2} \sqrt{8n + 1}⌋, we have 2^{m+2}s ≤ 8n + 1 ≤ 2^{2m+2}, which implies that s ≤ 2^{m}.

Since 2^{m−1} is a practical number and s ≤ 2^{m} = σ(2^{m−1})+1, by Lemma 4, we can conclude that 2^{m−1}s is a practical number. Notice that \frac{x^{2}−1}{8} is a triangular number, as x is an odd natural number because x^{2} ≡ 8n + 1 (mod 2^{m+2}). Now, we have 8n + 1 − x^{2} = 2^{m+2}s, or equivalently n = 2^{m−1}s + \frac{x^{2}−1}{8}. Since 2^{m−1}s is a practical number and \frac{x^{2}−1}{8} is a triangular number, we can conclude that n is a sum of a practical number and a triangular number. Therefore, all natural numbers can be written as a sum of a practical number and a triangular number.

4 Proof of Theorem 2

In this section, we prove some results regarding additive representations of natural numbers involving practical and polygonal numbers more generally. We require five lemmas to prove Theorem 2.

Lemma 6. Let p be an odd prime, let n be a natural number, and let s be a natural number greater than 3. There exist natural numbers x and y such that

P_{s}(x) + P_{s}(y) ≡ n (mod p).

Proof. If p | (s − 2), then x = n and y = p satisfy

P_{s}(x) + P_{s}(y) ≡ n (mod p).

If p ∤ (s − 2), then

|\{P_{s}(i) mod p : 1 ≤ i ≤ p\}| = |\{(n − P_{s}(j)) mod p : 1 ≤ j ≤ p\}| = \frac{p + 1}{2}.

This implies that

{P_{s}(i) mod p : 1 ≤ i ≤ p} ∩ {(n − P_{s}(j)) mod p : 1 ≤ j ≤ p} ≠ ∅.

Thus, there exist natural numbers x and y such that

P_{s}(x) + P_{s}(y) ≡ n (mod p).

Lemma 7. Let n be a natural number and s be a natural number greater than 3. There exist natural numbers x and y such that

P_{s}(x_{1}) + P_{s}(y_{1}) ≡ n (mod 2)

for all natural numbers x_{1} ≡ x (mod 4) and y_{1} ≡ y (mod 4).

Proof. If n is even, then (x, y) = (4, 4) satisfies the condition above. If n is odd, then (x, y) = (4, 1) satisfies the condition above.

Lemma 8. Let p be a prime congruent to 1 modulo 4. For all natural numbers n and k, there exist natural numbers x and y such that x^{2} + y^{2} ≡ n (mod p^{k}) and p ∤ y.

Proof. We will prove the lemma using mathematical induction on k. Let us first prove the lemma for k = 1. Let n be any natural number. By [1, Theorem 84], we have

|\{i^{2} mod p : 1 ≤ i ≤ p\}| = |\{(n − j^{2}) mod p : 1 ≤ j ≤ p\}| = \frac{p + 1}{2}.

Hence

{i^{2} mod p : 1 ≤ i ≤ p} ∩ {(n − j^{2}) mod p : 1 ≤ j ≤ p} ≠ ∅.

Thus there exist natural numbers x and y such that

x^{2} + y^{2} ≡ n (mod p).

If n ≢ 0 (mod p), then x and y cannot both be multiples of p. Without loss of generality, we can let p ∤ y. Since p ≡ 1 (mod 4), there exists a natural number a such that a^{2} + 1 ≡ 0 (mod p) (see [1, Theorem 86]). So, we can conclude that x = a and y = 1 is a solution to x^{2} + y^{2} ≡ 0 (mod p) with p ∤ y.

Now suppose that there exist natural numbers x and y_{s} such that

x^{2} + y_{s}^{2} ≡ n (mod p^{s})

and p ∤ y_{s}, where s ≥ 1. Let l be any natural number satisfying

l ≡ (\frac{n − x^{2} − y_{s}^{2}}{p^{s}} )(2y_{s})^{−1} (mod p),

and let y_{s+1} = y_{s} + p^{s}l. Now, as y_{s+1} ≡ y_{s} (mod p^{s}) and p ∤ y_{s}, we have p ∤ y_{s+1}. As \frac{x^{2}+y_{s}^{2}−n}{p^{s}} + 2ly_{s} ≡ 0 (mod p), we have

x^{2} + y_{s+1}^{2} = x^{2} + (y_{s} + p^{s}l)^{2}

= x^{2} + y_{s}^{2} + 2p^{s}ly_{s} + p^{2s}l^{2} ≡ x^{2} + y_{s}^{2} + 2p^{s}ly_{s} (mod p^{s+1})

≡ n + p^{s}(\frac{x^{2} + y_{s}^{2} − n}{p^{s}} + 2ly_{s}) (mod p^{s+1})

≡ n (mod p^{s+1}).

Hence, by mathematical induction, for all natural numbers k, there exist natural numbers x and y such that

x^{2} + y^{2} ≡ n (mod p^{k})

and p ∤ y.

Lemma 9. Let s be a natural number greater than 3. There exists an odd prime p not dividing s − 2 such that for all k, n ∈ ℕ, there exist x, y ∈ ℕ such that

P_{s}(x) + P_{s}(y) ≡ n (mod p^{k}).

Proof. Let k and n be any natural numbers, and let p be a prime congruent to 1 modulo 4 such that p ∤ (s − 2). Note that

8(s − 2)P_{s}(x) = (2(s − 2)x − (s − 4))^{2} − (s − 4)^{2}.

Since p ≡ 1 (mod 4), by Lemma 8, there exist natural numbers x_{0} and y_{0} such that

x_{0}^{2} + y_{0}^{2} ≡ 8(s − 2)n + 2(s − 4)^{2} (mod p^{k}).

Let x and y be natural numbers satisfying the congruences

x ≡ 2^{−1}(s − 2)^{−1}(x_{0} + s − 4) (mod p^{k})

and

y ≡ 2^{−1}(s − 2)^{−1}(y_{0} + s − 4) (mod p^{k}).

We have

8(s − 2)P_{s}(x) + 8(s − 2)P_{s}(y) ≡ x_{0}^{2} + y_{0}^{2} − 2(s − 4)^{2} (mod p^{k})
≡ 8(s − 2)n (mod p^{k}).

Since gcd(p^{k}, 8(s − 2)) = 1, we have P_{s}(x) + P_{s}(y) ≡ n (mod p^{k}).

Lemma 10. Let s be a natural number greater than 3, and let p_{i(s)} be the smallest prime for which Lemma 9 holds. There exists a real number A(s) such that for all x ≥ 1, we have

\frac{2P_{s}(2p_{i(s)}x)}{x^{2}} ≤ A(s).

Proof. Since 2P_{s}(2p_{i(s)}x) is a quadratic polynomial, we have

\frac{2P_{s}(2p_{i(s)}x)}{x^{2}} = O(1).

Hence, there exists a real number A(s) such that

\frac{2P_{s}(2p_{i(s)}x)}{x^{2}} ≤ A(s)

for all real numbers x ≥ 1.

Now we are ready to give a proof of Theorem 2.

Proof of Theorem 2. Let p_{i} denote the i-th prime, and let p_{i(s)} be the smallest prime for which Lemma 9 holds. By Lemma 10, there exists a real number A(s) such that

\frac{2P_{s}(2p_{i(s)}x)}{x^{2}} ≤ A(s)

for all real numbers x ≥ 1. Let r be the smallest natural number such that r ≥ i(s) and

\frac{σ(p_{1}p_{2} ⋯ p_{r})}{p_{1}p_{2} ⋯ p_{r}} ≥ A(s).

Such an r is well-defined, as the product ∏_{p prime}(1+ \frac{1}{p}) diverges (see [2, Chapter 7, Theorem 3] and [1, Theorem 19]). Let

N(s) = 2P_{s}(2p_{1}p_{2} ⋯ p_{r}).

Consider any natural number n greater than N(s). Let k be the largest natural number such that

2P_{s}(2p_{1}p_{2} ⋯ p_{i(s)−1}p_{i(s)}^{k}p_{i(s)+1} ⋯ p_{r}) < n.

Let n_{k} = 2p_{1}p_{2} ⋯ p_{i(s)−1}p_{i(s)}^{k}p_{i(s)+1} ⋯ p_{r}. From the definition of k, we have

2P_{s}(n_{k}) < n ≤ 2P_{s}(p_{i(s)}n_{k}).

From Lemma 6, there exists a solution x mod p_{i}, y mod p_{i} to the equation P_{s}(x) + P_{s}(y) ≡ n (mod p_{i}) for 2 ≤ i ≤ r and i ≠ i(s). From Lemma 7, there exists a solution x mod 2p_{1}, y mod 2p_{1} to the equation P_{s}(x) + P_{s}(y) ≡ n (mod p_{1}). From Lemma 9, there exists a solution x mod p_{i(s)}^{k}, y mod p_{i(s)}^{k} to P_{s}(x) + P_{s}(y) ≡ n (mod p_{i(s)}^{k}). Hence, by the Chinese remainder theorem, there exists a solution x mod n_{k}, y mod n_{k} to the equation

P_{s}(x) + P_{s}(y) ≡ n (mod \frac{n_{k}}{2} ).

Thus, there exist natural numbers x, y ≤ n_{k} such that

P_{s}(x) + P_{s}(y) ≡ n (mod \frac{n_{k}}{2} ).

This, together with the fact that n > 2P_{s}(n_{k}), implies that

\frac{2(n − P_{s}(x) − P_{s}(y))}{n_{k}} ∈ ℕ.

Note that

\frac{2(n − P_{s}(x) − P_{s}(y))}{n_{k}} ≤ \frac{2n}{n_{k}} ≤ \frac{4P_{s}(p_{i(s)}n_{k})}{n_{k}}.

By Lemma 10, we have

\frac{2P_{s}(p_{i(s)}n_{k})}{\frac{n_{k}^{2}}{4}} ≤ A(s).

Therefore, we have

\frac{2(n − P_{s}(x) − P_{s}(y))}{n_{k}} ≤ \frac{4P_{s}(p_{i(s)}n_{k})}{n_{k}} ≤ \frac{A(s)n_{k}}{2} .

Also, we have

\frac{σ(\frac{n_{k}}{2} )}{\frac{n_{k}}{2}} = \frac{σ(p_{1}p_{2} ⋯ p_{i(s)−1}p_{i(s)}^{k}p_{i(s)+1} ⋯ p_{r})}{p_{1}p_{2} ⋯ p_{i(s)−1}p_{i(s)}^{k}p_{i(s)+1} ⋯ p_{r}} ≥ \frac{σ(p_{1}p_{2} ⋯ p_{r})}{p_{1}p_{2} ⋯ p_{r}} ≥ A(s).

This implies \frac{A(s)n_{k}}{2} ≤ σ(\frac{n_{k}}{2} ). Therefore, we have

\frac{2(n − P_{s}(x) − P_{s}(y))}{n_{k}} ≤ \frac{A(s)n_{k}}{2} ≤ σ(\frac{n_{k}}{2} ).

Note that \frac{n_{k}}{2} = p_{1}p_{2} ⋯ p_{i(s)−1}p_{i(s)}^{k}p_{i(s)+1} ⋯ p_{r} is a practical number by the characterization of practical numbers (see [10, Section 3]). Thus, by Lemma 4, we can conclude that

\frac{2(n − P_{s}(x) − P_{s}(y))}{n_{k}} (\frac{n_{k}}{2} ) = n − P_{s}(x) − P_{s}(y)

is a practical number. Therefore, the natural number n can be written as a sum of a practical number and two s-gonal numbers.

In Theorem 1, we have proved that all natural numbers can be written as a sum of a triangular number and a practical number. In Theorem 2, we have proved that for all s > 3, all sufficiently large natural numbers can be written as a sum of a practical number and two s-gonal numbers. Now we show that there are infinitely many s > 3 for which we cannot write all sufficiently large natural numbers as a sum of a practical number and an s-gonal number. We also show that as s tends to infinity, the number of natural numbers that cannot be written as a sum of a practical number and two s-gonal numbers tends to infinity. Hence, we cannot drop “sufficiently large” from the statement of Theorem 2. We will require one lemma to prove these claims.

Lemma 11. If q is a practical number such that q is neither divisible by 3 nor by 4, then q = 1 or q = 2.

Proof. For the sake of contradiction, assume that q > 2. Since q > 2 and 4 ∤ q, the natural number q should have at least one odd prime divisor. Let p be the smallest odd prime divisor of q. As 3 ∤ q, we have p ≥ 5. As p ≥ 5 > σ(2) + 1, by the characterization of practical numbers, we can conclude that q is not practical (see [10, Section 3]). This is a contradiction.

Proposition 12.

(a) If s ≡ 0 (mod 12) or s ≡ 4 (mod 12), then infinitely many natural numbers cannot be written as a sum of a practical number and an s-gonal number.

(b) Let E(s) be the number of natural numbers that cannot be written as a sum of a practical number and two s-gonal numbers. Then we have

lim_{s→∞} E(s) = ∞.

Proof.

(a) If s ≡ 0 (mod 12) or s ≡ 4 (mod 12), then

P_{s}(n) = \frac{s − 2}{2}n^{2} + \frac{s − 4}{2}n = an^{2} + bn,

where a = \frac{s−2}{2} and b = \frac{s−4}{2}. Note that a is odd and is not divisible by 3. Also, note that b is even. Hence, as

P_{s}(n) = \frac{1}{a}((an + \frac{b}{2})^{2} − \frac{b^{2}}{4} ),

for all n ∈ ℕ_{0}, we have

P_{s}(n) ≢ a^{−1}(2 − \frac{b^{2}}{4} ) (mod 3),

and

P_{s}(n) ≢ a^{−1}(2 − \frac{b^{2}}{4} ) (mod 4).

This is because 2 is a quadratic non-residue modulo 3 and modulo 4. Let r be a natural number such that

r ≡ a^{−1}(2 − \frac{b^{2}}{4} ) (mod 12),

and let x be any positive real number. Let k be a natural number congruent to r modulo 12 and not exceeding x such that k = P_{s}(m) + q for some m ∈ ℕ_{0} and some practical number q. Since k ≡ r (mod 12), we have 3 ∤ (k −P_{s}(m)) and 4 ∤ (k −P_{s}(m)). Hence, by Lemma 11, we can conclude that q = 1 or q = 2, and thus

k ∈ {P_{s}(n) + 1 : n ∈ ℕ_{0} and P_{s}(n) + 1 ≤ x} ∪ {P_{s}(n) + 2 : n ∈ ℕ_{0} and P_{s}(n) + 2 ≤ x}.

Since

|{P_{s}(n) + 1 : n ∈ ℕ_{0} and P_{s}(n) + 1 ≤ x}| + |\{P_{s}(n) + 2 : n ∈ ℕ_{0} and P_{s}(n) + 2 ≤ x\}| = O(\sqrt{x}),

there are at most O(\sqrt{x}) natural numbers not exceeding x that are congruent to r modulo 12 and expressible as a sum of a practical number and an s-gonal number. Since there are \frac{x}{12} + O(1) natural numbers not exceeding x that are congruent to r modulo 12, there are at least \frac{x}{12} + O(\sqrt{x}) natural numbers not exceeding x that are congruent to r modulo 12 and not expressible as a sum of a practical number and an s-gonal number. Therefore, infinitely many natural numbers cannot be written as a sum of a practical number and an s-gonal number if s ≡ 0 (mod 12) or s ≡ 4 (mod 12).

(b) Let us count the number of natural numbers less than s that can be represented as a sum of a practical number and two s-gonal numbers. Only 0 and 1 are s-gonal numbers less than s. Thus, if n < s is a sum of a practical number and two s-gonal numbers, then

n ∈ {P + i : P is practical, P ≤ s, and i ∈ {0, 1, 2}}.

From a result of Weingartner [12, Theorem 1], it follows that

|\{P + i : P is practical, P ≤ s, and i ∈ \{0, 1, 2\}\}| = O(\frac{s}{log s}).

Hence, at least s + O(\frac{s}{log s}) natural numbers less than s cannot be written as a sum of a practical number and two s-gonal numbers. Therefore, we have

lim_{s→∞} E(s) = ∞.

5 Conjectures on sums of practical and polygonal numbers

In this section, we propose a few conjectures on some additive representations involving practical and polygonal numbers based on numerical computations. All of the code for the conjectures below, written in Python, can be found on Github [11].

For each natural number s > 3, let n_{s} be the number of natural numbers below 10^{8} that cannot be written as a sum of a practical number and an s-gonal number, and let N_{s} be the largest number below 10^{8} that cannot be written as a sum of a practical number and an s-gonal number. We have the following conjecture based on Table 1.

\begin{matrix}s & n_{s} & N_{s} \\ 4 & 17929061 & 99999998 \\ 5 & 13 & 2671 \\ 6 & 101 & 1332329 \\ 7 & 73 & 79445 \\ 8 & 414 & 4005819 \end{matrix}  \begin{matrix}s & n_{s} & N_{s} \\ 9 & 186 & 325808 \\ 10 & 341 & 13613213 \\ 11 & 68 & 105712 \\ 12 & 16663689 & 99999998 \\ 13 & 609 & 1612172 \end{matrix}  \begin{matrix}s & n_{s} & N_{s} \\ 14 & 79 & 106878 \\ 15 & 767 & 1486748 \\ 16 & 16665797 & 99999998 \\ 17 & 106 & 9314 \\ 18 & 1020 & 8541224 \end{matrix}

Table 1: Data regarding the sum of a practical number and an s-gonal number.

Conjecture 13. If s > 3, s ≢ 0 (mod 12), and s ≢ 4 (mod 12), then all sufficiently large natural numbers can be written as a sum of a practical number and an s-gonal number.

For s ∈ {4, 5, 6, 7, 8, 10}, we also computationally verified that all natural numbers below 10^{8} are expressible as a sum of a practical number and two s-gonal numbers. We have the following conjecture, which is a stronger version of Theorem 2, for the cases s ∈ {4, 5, 6, 7, 8, 10}.

Conjecture 14. For s ∈ {4, 5, 6, 7, 8, 10}, all natural numbers can be written as a sum of a practical number and two s-gonal numbers.

The conjecture above does not hold for other values of s. One can verify that 23 cannot be written as a sum of a practical number and two nonagonal numbers and that 11 cannot be written as a sum of a practical number and two s-gonal numbers if s ≥ 11.

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[11] D. V. K. Tran, Github project,

https://github.com/ducvktran/On-Sums-of-Practical-Numbers-and-Polygonal-Numbers,

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2020 Mathematics Subject Classification: Primary 11B83; Secondary 11D85, 11A99. Keywords: practical number, sum, triangular number, polygonal number.

(Concerned with sequence \underline{A000217}, \underline{A005153}, and \underline{A208244}.)

Received March 17 2024; revised versions received April 25 2024; April 30 2024. Published in Journal of Integer Sequences, May 1 2024.

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