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n 1 there are infinitely many very triangular numbers, by noting that |(t2 +3)2| =6 ∈ T for all n ∈ N. One can readily extend his argument as follows. Theorem 2.1 (Infinitude of VT -II). Suppose that 2(ℓ + 1) is an even triangular number, and let n> 2ℓ−1. Then t2n+2ℓ−1 ∈ VT . In particular, there are infinitely many very triangular numbers. Proof. Note that (2n + ℓ−1 2i)(2n +2ℓ) ℓ−1 i=0 = (2n + 2i)(2n−1 +2ℓ−1) 2 i=0 P X ℓ−1 ℓ−1 = 22n−1 +2n+ℓ−1 + 2n−1+i + 2ℓ−1+i. i=0 i=0 X X The restriction on n ensures that each power of 2 appearing in the last expression is distinct (2n − 1 > n + ℓ − 1 > n + ℓ − 2 > ··· >n − 1 > 2ℓ − 2 > ··· > ℓ − 1). Since there are exactly 2(ℓ + 1)-many distinct powers, we conclude that 1 n ℓ (t2 +2 −1)2 = 2(ℓ + 1) ∈ T, and consequently, t2n+2ℓ−1 ∈ VT as claimed.
Theorem 2.1 shows that given any even triangular number k, there are infinitely many very triangular numbers with k many 1s in their binary representation. Our next result shows that given any triangular number k, there are very triangular numbers with k many 1s in their binary representation. Theorem 2.2 (Infinitude of VT -III). Let k ∈ T be a triangular number. Then
k ℓ k k ℓ 1 k t2 −2 ∈ VT for all 0 ≤ ℓ ≤ 2 . Moreover, |(t2 −2 )2| = k for all 0 ≤ ℓ ≤ 2 . Proof. Let k ∈ T be given. The cases i =0, 1 are treated in Theorem 3.4. Suppose k now that 2 ≤ ℓ ≤ 2 . We compute k ℓ k ℓ (2 −2 )(2 − 2 + 1) 1 k+ℓ+1 k−ℓ−1 ℓ k−ℓ 2ℓ t k ℓ = = 2 (2 − 1)+2 (2 − 1)+2 2 −2 2 2 k−ℓ−2 k−ℓ−1 = 2k+ℓ+j + 2i+ℓ−1 +22ℓ−1 j=0 i=0 X X k−ℓ−2 ℓ−1 k−ℓ−1 = 2k+ℓ+j + 2i+ℓ−1 + 2i+ℓ−1 +22ℓ−1 j=0 i=0 X X Xi=ℓ k−ℓ−2 ℓ−1 k+ℓ+j i+ℓ−1 k−1 = 2 + 2 +| 2{z . } j=0 i=0 X X k Given the restriction 2 ≤ ℓ ≤ 2 we see that all powers of 2 in the last expression are distinct. Since there are (k − ℓ − 1)+ ℓ +1= k many of these, we conclude that 1 |(t2k−2ℓ )2| = k, as claimed. One may wonder whether given an odd triangular number, are there very tri- angular numbers with that many 1s in their binary representation, and if so, how 1 many? Setting ℓ = 0 in Theorem 2.2 shows that |(t2k−1)2| = (2k−2)−(k−1)+1 = k, hence the answer to the first question in in the positive. The answer to the second GUIDED BY THE PRIMES - AN EXPLORATION OF VERY TRIANGULAR NUMBERS 3 question – as presented in the following two propositions – is perhaps somewhat surprising. While there are infinitely many very triangular numbers vt ∈ VT with 1 |(vt)2| = 15, 21, 45, 55,...(all odd triangular numbers greater than 3), we are rea- sonably certain that there are only four very triangular numbers with three 1s in their binary representations. Proposition 2.3. Let ℓ > 1 so that 2ℓ +1 is a triangular number. There are 1 infinitely many very triangular numbers vt ∈ VT with |(vt)2| =2ℓ +1.
Proof. Given ℓ as in the statement, consider the triangular numbers t22ℓ−2ℓ+1. We compute
1 2ℓ ℓ 2ℓ ℓ t 2ℓ ℓ = 2 − 2 +1 2 − 2 +2 2 −2 +1 2 = 24 ℓ−1 − 23ℓ +22ℓ +1 − 2ℓ − 2ℓ−1 +1 ℓ−2 ℓ = 23ℓ+i + 2ℓ+i − 2ℓ−1 +1 i=0 i=0 X X ℓ−2 ℓ = 23ℓ+i + 2ℓ+i +2ℓ−1 +1. i=0 i=1 X X Given the restriction on ℓ, the the powers of 2 in the above summands are all distinct, and there are (ℓ − 1)+ ℓ +1+1=2ℓ + 1 many of them. We complete this section with a discussion of triangular numbers with three 1s in their binary representations. N 1 1 Conjecture 2.4. If n ∈ is such that |(n)2|≥ 6, then |(tn)2|≥ 4. 1 1 Proposition 2.5. The only very triangular numbers tn with |(n)2|≤ 5 and |(tn)2| = 3 are 21, 28, 276 and 1540.
1 1 1 Proof. The case of |(n)2| = 1 trivially yields |(tn)2| = 2 6= 3. The cases |(n)2| = 2, 3, 4, 5 are all handled by case analysis. Lest we fall victim of paralysis by (sub)- 1 case analysis, we only present here the case |(n)2| = 3, as this case is ‘big enough’ to be representative of the others, and to indicate the general method of reasoning; yet small enough not to cause the reader any undue discomfort. Suppose thus 1 σ(1) σ(2) σ(3) that |(n)2| = 3, and write n = 2 +2 +2 for some increasing function σ : {1, 2, 3} ֒→ N∪{0}. We will need to distinguish between the cases when σ(1) = 0 and when σ(1) > 0. Consider first the case σ(1) ≥ 1, and the corresponding diagram (see Figure 1) representing the summands in n(n + 1), with solid arrows in the diagram pointing towards known larger quantities, and dashed ones between quantities we cannot compare without further assumptions. If σ(2) > 2σ(1), then the sum looks like 2 + 22σ(1) +2σ(2)+ terms whose powers 1 are all greater than σ(2). As such, we conclude that |(n(n + 1))2|≥ 4. If σ(2) = 2σ(1), then (2.1) n(n +1)=2σ(1) +22σ(1)+1 +23σ(1)+1 +24σ(1) + terms with powers involving σ(3). We now consider the subcases σ(1) = 1 and σ(1) > 1. In the former, we obtain n(n +1)=2+23 +25 + terms with powers involving σ(3). 4 AUDREYBAUMHECKELANDTAMAS´ FORGACS´
2σ(1) + 2σ(2) + 2σ(3)
22σ(1) + 2σ(1)+σ(2) + 2σ(1)+σ(3)
2σ(1)+σ(2) + 22σ(2) + 2σ(2)+σ(3)
2σ(1)+σ(3) + 2σ(2)+σ(3) + 22σ(3)
Figure 1. 1 The case |(n)2| = 3 and σ(1) ≥ 1.
The smallest of the remaining powers is σ(3) ≥ 3. If σ(3) = 4 or σ(3) ≥ 6, we 1 1 trivially obtain that |(n(n + 1))2| ≥ 4. If σ(3) = 3, we get |(n(n + 1))2| = 4, 1 while if σ(3) = 5, we obtain |(n(n + 1))2| = 6. If on the other hand σ(1) > 1, then the four displayed powers of 2 in (2.1) are all distinct. We must therefore have σ(3) = 2σ(1)+ 1, or σ(3) = 3σ(1) + 1, as all other cases immediately yield 1 |(n(n + 1))2|≥ 4. Suppose first that σ(3) = 2σ(1)+1. Then n(n +1)=2σ(1) +22σ(1)+2 +23σ(1)+1 +23σ(1)+2 +24σ(1)+3 1 with all terms distinct, since σ(1) > 1. Thus |(n(n + 1))2| = 5. Finally, if σ(3) = 3σ(1) + 1, then n(n +1)=2σ(1) +22σ(1)+1 +23σ(1)+2 +24σ(1) +24σ(1)+2 +25σ(1)+2 +26σ(1)+2, 1 from which |(n(n + 1))2|≥ 6 readily follows. We now consider the case σ(1) = 0. The corresponding diagram is given in Figure 2:
1 + 2σ(2) + 2σ(3)
1 + 2σ(2) + 2σ(3)
2σ(2) + 22σ(2) + 2σ(2)+σ(3)
2σ(3) + 2σ(2)+σ(3) + 22σ(3)
Figure 2. 1 The case |(n)2| = 3 and σ(1) = 0. GUIDED BY THE PRIMES - AN EXPLORATION OF VERY TRIANGULAR NUMBERS 5
We see that in this case (†) n(n +1)=2+2σ(2) +2σ(2)+1 +22σ(2) + terms with powers involving σ(3). The only way for the first four summands not all to be distinct is if σ(2)+1 = 2σ(2), or σ(2) = 1. We rewrite n(n +1)=22 +23 + terms with powers involving σ(3). 1 2 7 If σ(3) > 3, we obtain |(n(n + 1))2| ≥ 4. If σ(3) = 3, we get n(n +1)=2 +2 , 1 3 4 5 hence |(n(n + 1))2| = 2. Finally, if σ(3) = 2, we arrive at n(n +1)=2 +2 +2 , 1 2 3 4 corresponding to the very triangular number tn = 2 n(n +1)=2 +2 +2 = 28. If the first four summands in (†) are distinct, we must have σ(3) = σ(2) + 1 if we want to reduce the number of summands. In this case however n(n +1)=2+2σ(2) +2σ(2)+3 +22σ(2) +22σ(2)+3, 1 which means that |(n(n+1))2|≥ 4 no matter what σ(2) is. Since we have exhausted 1 all the cases, we conclude that the only very triangular number tn with |(n)2| =3 1 and |(tn)2| = 3 is t7 = 28. For the sake of completeness we present (without proof) the remaining very triangular numbers with three 1s in their binary representation: 1 1 1 |(n)2| = 2 gives t6 = 21; |(n)2| = 4 gives t23 = 276, and |(n)2| = 5 yields t55 = 1540.
3. The twin very triangular number theorem and arithmetic progressions
The twin prime conjecture has occupied the minds of many great mathematicians since its formulation, and has seen some great progress towards a positive resolution in recent years. The analogous conjecture with respect to very triangular numbers states that there are infinitely many consecutive triangular numbers which are also very triangular. Fortunately, along with being easy to state, this result is also reasonably easy to prove. We begin with two preliminary results. Lemma 3.1. Let k ∈ N. Then 22k−2 +22k−3 + ··· +2k−1 is triangular number. Corollary 3.2. Any number of the form 22k−2 +22k−3 + ··· +2k−1, k ∈ T is very triangular. Remark 3.3. (i) Corollary 3.2 provides an alternative proof of the Theorem on page 217 of [6] establishing the existence of infinitely many very triangular numbers. (ii) We also see that there is at least one very triangular number with k many 1s in its binary representation for every k ∈ T . (iii) Not every very triangular number is of this form. For example, (21)2 = 10101. Theorem 3.4 (The twin very triangular number theorem). There are infinitely many integers n ∈ N such that tn,tn+1 ∈ VT . Proof. Let k ∈ T , k> 1 be given. By Corollary 3.2
2k−2 11 ··· 1 00 ··· 0 = 2i ∈ VT. k k−1 i=Xk−1 | {z } | {z } 6 AUDREYBAUMHECKELANDTAMAS´ FORGACS´
k−1 i k Let n + 1 := 11 ··· 1 = i=0 2 =2 − 1. Then k P k−1 k−1 2k−2 (n +| 1)({zn}+ 2) (2k − 1)2k t = = = 2i 2k−1 = 2k−1+i = 2i. n+1 2 2 i=0 ! i=0 X X i=Xk−1 Since tn+1 = tn + (n + 1), we see that tn+1 −(n+1) = (11 ··· 1 00 ··· 0)−(11 ··· 1) ∈ k k−1 k T . The reader will note that | {z } | {z } | {z } 2k−2 k−1 11 ··· 1 00 ··· 0 − 11 ··· 1 = 2i − 2i i=0 k k−1 k i=Xk−1 X 2k−2 | {z } | {z } | {z } = 2i − (2k − 1) i=Xk−1 2k−2 = 2i +2k−1 +1, i=Xk+1 and that 2k−2 1 2i +2k−1 +1 = (k − 2)+2= k ∈ T.
i=k+1 ! X 2 We conclude that tn ∈ VT . Since tn+1 ∈ VT as well, the proof is complete.
Remark 3.5. The following observations are in order:
(i) The proof of Theorem 3.4 shows that if k ∈ T , then t2k −2,t2k−1 ∈ VT . Not all twin pairs of very triangular numbers are of this form, however, as demonstrated by the pair (903, 946). (ii) While the theorem exhibits twin very triangular numbers tn,tn+1 with the 1 1 property that |(tn)2| = |(tn+1)2| = k for some k ∈ T , this property does not 1 1 hold in general for twin pairs. For instance, |(t175)2| = 10, while |(t176)2| = 6. (ii) These pairs cannot be ‘continued’ into triples of consecutive very triangular k N numbers, since t2 ∈/ VT for any k ∈ . Theorem 3.4 raises the question of just how long strings of consecutive triangular numbers one can have that are also very triangular. The reader can easily check that 1 1 1 1 1 1 |(t581)2| = |(t582)2| = |(t583)2| = 10 and that|(t1702)2| = |(t1703)2| = |(t1704)2| = 1 |(t1705)2| = 10 (there are many more such examples) which would indicate that perhaps one can find arbitrarily long strings of consecutive triangular numbers that are also very triangular. More generally, one may look for very triangular numbers in arithmetic progressions. In the ‘prime-realm’ we are thus led to the celebrated Green-Tao theorem concerning such progressions. Theorem 3.6 (Green-Tao [3]). Let π(N) denote the number of primes less than or equal to N. If A is a subset of the prime numbers such that |A ∩ [1,N]| lim sup > 0, N→∞ π(N) GUIDED BY THE PRIMES - AN EXPLORATION OF VERY TRIANGULAR NUMBERS 7 then for all positive k, the set A contains infinitely many arithmetic progressions of length k. In particular, the entire set of prime numbers contains arbitrarily long arithmetic progressions. We formulate the analogous conjecture regarding very triangular numbers as follows. Conjecture 3.7. For any positive integer k, there exists an arithmetic progression AP (k) of length k so that tj ∈ VT for all j ∈ AP (k). A plausible approach to proving this conjecture could start with the following theorem of Szemr´edi. Theorem 3.8 (Szemer´edi, [5]). If A ⊂ N is such that |A ∩{1, 2,...,N}| lim sup > 0, N→∞ N then for all k ∈ N, A contains infinitely many arithmetic progressions of length k.
If πV T (x) and πT (x) denote the counting functions of very triangular and trian- gular numbers less than or equal to x, then π (t ) |VT ∩{t ,t ,...,t }| |σ(N) ∩{1, 2,...,N}| V T N = 1 2 N = , πT (tN ) N N ∞ where {tσ(n)}n=1 is the sequence of very triangular numbers. Consequently, by πVT (tN ) N applying Theorem 3.8 we see that if lim supN→∞ πT (tN ) > 0, then σ( ) contains infinitely many arithmetic progressions of length k for every k ∈ N. For any such k−1 progression a · j + b, 0 ≤ j ≤ k − 1, the set {ta·j+b}j=0 is a subset of VT . Our CAS computations – limited by available memory – are inconclusive in terms of πVT (tN ) the potential value of the critical quantity lim supN→∞ πT (tN ) . Undeterred, and following historical precedence1, we next demonstrate the existence of infinitely many arithmetic progressions AP (k) or length six or less, such that {tj }j∈AP (k) ⊂ VT 2. In fact, we establish the existence of strings of consecutive very triangular numbers which are also consecutive triangular numbers. The only result we need in order to be able to demonstrate our claim is the following proposition. Proposition 3.9. Suppose that n> 5. Then for all m ≥ n and 0 ≤ k< 2⌊(n−1)/2⌋, the following equality holds:
m 1 n 1 (t2 +3+k)2 = (t2 +3+k)2 .
Proof. Note first that for any n and 0 ≤ k, 2n−1 n+1 n n−1 t2n+3+k =2 +2 + (k + 1)2 +2 +4+2+ tk +3k.
1Before proving the full result, Szemer´edi first established the existence of infinitely many arithmetic progressions with lengths k = 4 (see [4]) 2When it comes to prime numbers, the terminology primes in arithmetic progression refers to consecutive terms in an arithmetic progression which are all prime. Consecutive primes in arithmetic progression on the other hand refers to consecutive terms in an arithmetic progression which are prime, with only composite numbers between them. 8 AUDREYBAUMHECKELANDTAMAS´ FORGACS´
Given the restriction k< ⌊(n − 1)/2⌋, we see that k(k + 1) 2⌊(n−1)/2⌋(2⌊(n−1)/2⌋ + 1) t = < ≤ 2n−2 +2⌊(n−1)/2⌋−1 < 2n−2 +2n−4, k 2 2 and 3k< 3 · 2⌊(n−1)/2⌋ =2⌊(n−1)/2⌋+1 +2⌊(n−1)/2⌋ < 2n−2 +2n−3. n−1 It follows that tk +3k< 2 , and hence
m 1 2n−1 n+1 n n−1 1 1 (t2 +3+k)2 = (2 +2 + (k + 1)2 +2 )2 + (tk +3k +4+2)2 2m−1 m+1 m m−1 1 1 = (2 +2 + (k + 1)2 +2 )2 + (tk +3k +4+2) 2
n 1 = (t2 +3+k)2 .
The proof is complete.
N 5 Corollary 3.10. There are infinitely many j ∈ such that {tj+ℓ}ℓ=0 ⊂ VT . Proof. In light of Proposition 3.9, it suffices to find an n> 5 and integers 0 ≤ k< ⌊(n−1)/2⌋ n 5 k +1 < k +2 < k +3 < k +4 < k +5 < 2 so that {t2 +3+k+ℓ}ℓ=0 ⊂ VT . Using a CAS, we find that n = 30, and k = 1870 < 1871 < 1872 < 1873 < 1874 < 1875 < 2⌊(30−1)/2⌋ =214 is such a pair. In fact 1 1 1 (t230+1873)2 = (t230+1874)2 = (t230+1875)2 = 1 1 1 (t230+1876)2 = (t230+1877)2 = (t230+1878)2 = 21.
We remark that the first instance of six consecutive triangular numbers that are all very triangular occurs much sooner: {t30301,t30302,t30303,t30304,t30305,t30306}⊂ VT . Corollary 3.10 generalizes the twin very triangular number theorem, and could be viewed as the sextuplet very triangular number theorem, although we suspect that such a moniker is not likely to catch on.
4. Gaps and consecutive strings Results concerning strings of consecutive natural numbers that either must con- tain primes, or do not contain primes have a simple elegance about them. Bertrand’s postulate (see [1]) states that for all integers n> 3 there exists a prime p satisfying n
Table 1. Bertrand’s postulate for very triangular numbers (small n)
n tn m ∈ VT t2n 4 10 21,28 36 5 15 21,28 55 6 21 28 78 7 28 N/A 105 8 26 N/A 136 9 45 N/A 171 10 55 190 210 ...... 19 210 231, 276, 378, 435, 630 741 ......
Next, we turn our attention to the gaps between consecutive very triangular numbers.
k k k 2 k 2 k Proposition 4.2. Suppose that k ∈ T and that 4 | k. If 2 −2 k Proof. Let n, k =4ℓ be as in the statement, and write n =2k − 2 2 + m. Suppose p first that m =2 for some 0 ≤ p ≤⌊log2 ℓ⌋. In this case tn can be written as 4ℓ−1 4ℓ−1 1 t = 2j +2p 2j +2p +1 n 2 jX=2ℓ jX=2ℓ 2 4ℓ−1 4ℓ+p 4ℓ−1 1 = 2j + 2j + 2j +22p +2p 2 jX=2ℓ j=2Xℓ+p+1 jX=2ℓ 8ℓ−1 4ℓ+p 4ℓ−1 1 = 2j +24ℓ + 2j + 2j +22p +2p 2 j=6Xℓ+1 j=2Xℓ+p+1 jX=2ℓ 8ℓ−1 4ℓ−1 4ℓ−1 1 = 2j +24ℓ+p+1 + 2j + 2j +22p +2p 2 j=6Xℓ+1 j=2Xℓ+p+1 jX=2ℓ 2ℓ+p 1 8ℓ−1 4ℓ = 2j +24ℓ+p+1 + 2j + 2j +22p +2p . 2 j=6Xℓ+1 j=2Xℓ+p+2 jX=2ℓ Given that 0 ≤ p ≤⌊log2 ℓ⌋, we see that p ≤ 2p< 2ℓ ≤ 2ℓ + p< 2ℓ + p +2 < 4ℓ< 4ℓ + p +1 < 6ℓ +1 10 AUDREYBAUMHECKELANDTAMAS´ FORGACS´ as soon as ℓ ≥ 2, and hence 1 4ℓ +1= k + 1 if p =0 |(tn)2| = ∈/ T for any k ∈ T with k ≥ 3. (4ℓ +2= k + 2 if p ≥ 1 1 Finally, if ℓ = 1, then n = 13 and |(t13)2| =5 ∈/ T . k Assume now that n, k = 4ℓ are as in the statement, and write n = 2k − 2 2 + m, with r σ(j) m = 2 for some 2 ≤ r ≤⌊log2 ℓ⌋, j=1 X and .{⌊σ : {1, 2,...,r} ֒→{0, 1, 2,..., ⌊log2 ℓ A simple calculation shows that 1 k r r 2k−1 t = 2k/2 + 2j + 2σ(j) 2σ(j) +1 + 2j . n 2 j=1 j=1 j=Xk/2+1 X X j=3Xk/2+1 The reader will note that r k 2 ≤ 2σ(j) +1 ≤ 2σ(r)+1 ≤ 2r+1 ≤ 2log2(ℓ)+1 ≤ < 2k/2, 2 j=1 X and that k 2σ(r)+1 ≤ 2r +1 ≤ 2 log (ℓ)+1 < 2ℓ = . 2 2 Consequently, the largest power of 2 in the expression k r 2j 2σ(j) +1 j=1 j=Xk/2+1 X is less than or equal to 3k/2, whereas the largest power of 2 in the expression r r 2σ(j) 2σ(j) +1 j=1 j=1 X X is less than k/2. Therefore, 1 k r r k |(t )1| = 2j + 2σ(j) 2σ(j) +1 + n 2 2 j=k/2+1 j=1 j=1 2 X X X 1 k r (4.1) = 2j 2σ(j) +1 j=k/2+1 j=1 2 X X 1 r r k + 2σ(j) 2σ(j) +1 + 2 j=1 j=1 2 X X GUIDED BY THE PRIMES - AN EXPLORATION OF VERY TRIANGULAR NUMBERS 11 We compute the first summand in (4.1). To this end, note that k r k r 2j 2σ(j) +1 = 2j + 2σ(j)+k+1 − 2σ(j)+k/2+1 j=1 j=1 j=Xk/2+1 X j=Xk/2+1 X k r k+σ(i) = 2j + 2j i=1 j=Xk/2+1 X j=σ(iX)+k/2+1 k+σ(r) k r−1 k+σ(i) = 2j + 2j + 2j i=1 j=σ(rX)+k/2+1 j=Xk/2+1 X j=σ(iX)+k/2+1 σ(r)+k/2 k+σ(r) r−1 k+σ(i) = 2k+1 + 2j + 2j + 2j i=1 j=Xk/2+1 j=σ(rX)+k/2+2 X j=σ(iX)+k/2+1 σ(r−1)+k/2 k+σ(r) k+σ(r−1) = 2k+1 +2k/2+σ(r)+1 + 2j + 2j + 2j j=Xk/2+1 j=σ(rX)+k/2+2 j=σ(r−X1)+k/2+2 r−2 k+σ(i) + 2j i=1 X j=σ(iX)+k/2+1 σ(r−2)+k/2 = 2k+1 +2k/2+σ(r)+1 +2k/2+σ(r−1)+1 + 2j j=Xk/2+1 k+σ(r) k+σ(r−1) k+σ(r−2) + 2j + 2j + 2j j=σ(rX)+k/2+2 j=σ(r−X1)+k/2+2 j=σ(r−X2)+k/2+2 r−3 k+σ(i) + 2j i=1 X j=σ(iX)+k/2+1 σ(1)+k/2 = 2k+1 +2k/2+σ(r)+1 +2k/2+σ(r−1)+1 + ··· +2k/2+σ(2)+1 + 2j j=Xk/2+1 k+σ(r) k+σ(r−1) k+σ(r−2) k+σ(1) + 2j + 2j + 2j + ··· + 2j j=σ(rX)+k/2+2 j=σ(r−X1)+k/2+2 j=σ(r−X2)+k/2+2 j=σ(1)+Xk/2+2 σ(1)+k/2 = 2k+σ(1)+1 +2k/2+σ(r)+1 +2k/2+σ(r−1)+1 + ··· +2k/2+σ(2)+1 + 2j j=Xk/2+1 k+σ(r) k+σ(3) k+σ(2) k + 2j + ··· + 2j + 2j + 2j j=σ(rX)+k/2+2 j=σ(3)+Xk/2+2 j=σ(2)+Xk/2+2 j=σ(1)+Xk/2+2 σ(1)+k/2 = 2k+σ(2)+1 +2k/2+σ(r)+1 +2k/2+σ(r−1)+1 + ··· +2k/2+σ(2)+1 + 2j j=Xk/2+1 k+σ(r) k+σ(3) k+σ(1) k + 2j + ··· + 2j + 2j + 2j j=σ(rX)+k/2+2 j=σ(3)+Xk/2+2 j=σ(2)+Xk/2+2 j=σ(1)+Xk/2+2 12 AUDREYBAUMHECKELANDTAMAS´ FORGACS´ σ(1)+k/2 = 2k+σ(r)+1 + 2k/2+σ(r)+1 +2k/2+σ(r−1)+1 + ··· + 2k/2+σ(2)+1 + 2j j=Xk/2+1 k+σ(r−1) k+σ(r−2) k+σ(1) k + 2j + 2j + ··· + 2j + 2j j=σ(rX)+k/2+2 j=σ(r−X1)+k/2+2 j=σ(2)+Xk/2+2 j=σ(1)+Xk/2+2 r σ(1)+k/2 r k+σ(j−1) k = 2k+σ(r)+1 + 2k/2+σ(j)+1 + 2j + 2i + 2j j=2 j=2 X j=Xk/2+1 X i=σ(jX)+k/2+2 j=σ(1)+Xk/2+2 k+σ(r−1) r r−1 k+σ(j−1) = 2k+σ(r)+1 + 2j + 2k/2+σ(j)+1 + 2i j=3 j=2 j=σ(rX)+k/2+2 X X i=σ(jX)+k/2+2 σ(1)+k/2 k + 2k/2+σ(2)+1 + 2j + 2j j=Xk/2+1 j=σ(1)+Xk/2+2 k+σ(r−1) r r−1 k+σ(j−1) = 2k+σ(r)+1 + 2j + 2k/2+σ(j)+1 + 2i j=3 j=2 j=σ(rX)+k/2+2 X X i=σ(jX)+k/2+2 σ(1)+k/2 k/2+σ(2) + 2k+1 + 2j + 2j j=Xk/2+1 j=σ(1)+Xk/2+2 k r r−1 k+σ(j−1) = 2k+σ(r)+1 +2k+σ(r−1)+1 + 2j + 2k/2+σ(j)+1 + 2i j=3 j=2 j=σ(rX)+k/2+2 X X i=σ(jX)+k/2+2 σ(1)+k/2 k/2+σ(2) + 2j + 2j . j=Xk/2+1 j=σ(1)+Xk/2+2 Now, r r−1 k+σ(j−1) 2k/2+σ(j)+1 + 2i j=3 j=2 X X i=σ(jX)+k/2+2 r r k+σ(j−2) = 2k/2+σ(j)+1 + 2i j=3 j=3 X X i=σ(j−X1)+k/2+2 r k/2+σ(j) = 2k+σ(j−2)+1 + 2i , j=3 X i=σ(j−X1)+k/2+2 hence, we arrive at the expression r k r k/2+σ(j) σ(1)+k/2 2k+σ(j)+1 + 2j + 2i + 2j . j=1 j=2 X j=σ(rX)+k/2+2 X i=σ(j−X1)+k/2+2 j=Xk/2+1 GUIDED BY THE PRIMES - AN EXPLORATION OF VERY TRIANGULAR NUMBERS 13 Therefore, 1 k r 2j 2σ(j) +1 j=k/2+1 j=1 2 X X r = r + (k − (σ(r)+ k/2+2)+1)+ ( k/2+ σ(j) − (σ(j − 1)+ k/2+2)+1) j=2 X + (σ(1) + k/2 − (k/2+1)+1 r = r + k/2 − σ(r) − 1+ (σ(j) − σ(j − 1)) − (r − 1)+ σ(1) j=2 X = k/2 − σ(r) + (σ(r) − σ(1)) + σ(1) = k/2. We now turn our attention to the second summand in (4.1). Since r ⌊log2 ℓ⌋ 2σ(j) +1 ≤ 2j +1 ≤ 2⌊log2 ℓ⌋+1, j=1 j=0 X X we see that r r r 2σ(j) 2σ(j) +1 ≤ 2σ(j)+⌊log2 ℓ⌋+1 < 22(⌊log2 ℓ⌋+1). j=1 j=1 j=1 X X X Consequently, 1 r r 1 ≤ 2σ(j) 2σ(j) +1 j=1 j=1 2 X X ≤ 2⌊log2 ℓ⌋ +1 ≤ 2(log2 k − log2 4) + 1 = 2 log2 k − 3. s(s+1) N Recall that k ∈ T . If we write k = 2 for some s ∈ , then the smallest s(s+1) triangular number greater than k is 2 + (s + 1). On the other hand, s(s + 1) 2 log k − 3 = 2 log − 3 = 2(log s + log (s + 1)) − 5 2 2 2 2 2 ⋆ < 4 log2(s + 1) − 5 5. Future work We invite the reader to carry out investigations similar to ours, either to make our understanding of very triangular numbers more complete, or to extend these results to other sets of integers (or both!) One may start by proving Conjecture 2.4. Using techniques similar to those presented in this paper we were able to show 1 1 that if |(n)2| = 6, then |(tn)2|≥ 4. We suspect however that this approach will not 1 suffice in settling the conjecture for all |(n)2|≥ 7. Another beautiful result yet to be proved is Conjecture 3.7. As mentioned in the paper, numerical evidence is inconclusive regarding the density measure that we might use (along with Szemer´edi’s result). Nonetheless, our sense (based on nu- merical results) is that very triangular numbers are abundant, and hence should have positive upper density in the set of triangular numbers. Also related to arith- metic progressions is the question whether or not one can partition the set VT with finitely many arithmetic progressions, and finally, whether it is possible to find a closed formula for the nth very triangular number. Finding such would answer many of the posed questions in one fell swoop. Finally, a possible generalization of the present work concerns subsets of the natural N N N 1 N ,( )numbers f : ֒→ with the property that s ∈ S ⊂ f( ) implies that |(t)2| ∈ f and how the properties of f effect whether or not S exhibits properties analogous to those of VT ⊂ T . References [1] Joseph Louis Fran¸cois Bertrand, M´emoire sur le nombre de valeurs que peut prendre une fonction quand on y permute les lettres qu’elle renferme. (1845) J. l’Ecole´ Roy. Polytech. 17: 123-140. [2] Christian Elsholtz, Unconditional Prime-Representing Functions, Following Mills. (2020) Amer. Math. Monthly, 127 (7): 639–642. [3] Ben Green and Terence Tao, The primes contain arbitrarily long arithmetic progressions. (2008) Annals of Math., 167 (2): 481-547. doi:10.4007/annals.2008.167.481 [4] Endre Szemer´edi, On sets of integers containing no four elements in arithmetic progression. (1969) Acta Math. Acad. Sci. Hung. 20 (1–2): 89–104. doi:10.1007/BF01894569 [5] Endre Szemer´edi, On sets of integers containing no k elements in arithmetic progression, (1975) Acta Arithmetica 27: 199–245. doi:10.4064/aa-27-1-199-245 [6] James Tanton, Very Triangular and Very Very Triangular Numbers, Essay 28 in “How Round is a Cube? : and other curious mathematical ponderings”, MSRI Mathematical Circles Library, 23. AMS, 2019.