Be Still My Pulsating Sequence

Can you infer the simple rule behind a number sequence that spikes up and down like the beating of a heart?

Olena Shmahalo/Quanta Magazine

21

Our Insights problem this month is based on a sequence of integers which comes from Neil Sloane, a mathematician who is arguably the world’s greatest authority on such sequences. Contributing writer Erica Klarreich interviewed Sloane for Quanta Magazine in August, in “The Connoisseur of Number Sequences”:

Neil Sloane is considered by some to be one of the most influential mathematicians of our time.

That’s not because of any particular theorem the 75-year-old Welsh native has proved, though over the course of a more than 40-year research career at Bell Labs (later AT&T Labs) he won numerous awards for papers in the fields of combinatorics, coding theory, optics and statistics. Rather, it’s because of the creation for which he’s most famous: the Online Encyclopedia of Integer Sequences (OEIS), often simply called “Sloane” by its users.

We asked Sloane to help us construct a puzzle based on a sequence that is not in the OEIS. Here is what we came up with:

Question 1:

Consider the sequence of numbers shown below:

13    26    2      4     6     3      9     12     8     10     5     15     18     14      7      21      24      16 …

Unlike most mathematical functions, this sequence, graphed in the picture above, spikes up and down like the pulsations of the heart. Can you figure out the simple rule behind it? (Update: The solution is now available here.)

I love puzzles like this one because they are truly “insight” problems in the psychological sense. Most of the math we do in school involves taking a general rule — a formula or equation — and plugging in the particulars of a problem to get the solution. This process requires deductive reasoning, in which we go from the general to the particular. In the above question, though, you are given a particular set of numbers and you have to come up with a general rule that fits them. Here you are going in reverse — from the particular to the general. This is inductive reasoning. You can remember how it goes with the aid of the mnemonic PIG (particular -> inductive -> general).

A monthly puzzle celebrating the sudden insights and unexpected twists of scientific problem solving. Your guide is Pradeep Mutalik, a medical research scientist at the Yale Center for Medical Informatics and a lifelong puzzle enthusiast.

Problems that use inductive reasoning require a different approach from the deductive-reasoning problems we normally encounter. You have to take the particulars you are given and spend some time playing around with them, looking for patterns and trying out various candidate rules, many of which will be unsuccessful. Then suddenly, the solution may just come to you in a flash of insight — an “aha” or “eureka” moment.

If that magic moment hasn’t blessed you yet and you feel that you aren’t making any progress, it may help to put the problem aside and do something else for a while. A recent book titled “The Eureka Factor” by John Kounios and Mark Beeman, two of the leading researchers on this phenomenon, describes the multistage interplay between the conscious and unconscious mind that gives rise to the insight phenomenon. First you immerse your conscious mind in the problem. Often you may reach an impasse. At this point, it is best to engage in a different activity as a diversion and let your unconscious mind chew on the problem — a stage called incubation. Then, if you’re patient and lucky, this underground activity can burst to the surface like a volcanic eruption and you’ll have your insight and the answer. Try it and see.

If you’re still not having any luck, we have a hint for you, which you can read by highlighting the blank space below (or changing the font color). Read it and try again, and may the insight strike you this time!

Look at each pair of adjacent numbers in the sequence. Do they have something in common?

Once you figure out the rule behind the sequence, you are ready for the next question. (If you still can’t figure it out, wait for a day, and then we will release readers’ comments.)

Question 2:

Do you think that every positive integer greater than 1 will appear in this sequence, or will some integers be skipped? Can you prove your answer?

Here’s a question that may require more work:

Question 3:

How far down the sequence do you have to go to encounter a pair of adjacent numbers that share more than one prime factor?

And finally, a tricky question:

Question 4:

Why on earth does this sequence begin with the number 13? (Outlandish reasons are encouraged!)

To those who are chomping at the bit for more, we will discuss this sequence in more detail when we publish the solution in a couple of weeks. We’ll also present an unsolved problem about it from Neil Sloane, for those intrepid readers who dream of being immortalized in the mathematical treasure trove that is the OEIS.

Happy puzzling, and may the insight strike you!

Editor’s notes: The reader who submits the most interesting, creative or insightful solution (as judged by the columnist) in the comments section will receive a Quanta Magazine T-shirt. (Update: The solution is now available here.)

If you’d like to suggest a favorite puzzle for a future Insights column, submit it as a comment below, clearly marked “NEW PUZZLE SUGGESTION” (it will not appear online, so solutions to the puzzle questions above should be submitted separately).

Note that we will hold comments for the first day to allow for independent contributions.

• Q1: a(n + 1) is the least term coprime to a(n) that doesn't already appear in the sequence.

Q2: Every integer > 1 will appear in the sequence. (Proved in the paper The EKG Sequence by J. C. Lagarias, E. M. Rains, N. J. A. Sloane.)

Q3: a(577) = 620; a(578) = 610. These share prime factors 2 and 5.

Q4: Because the sequence starting with 12 is already is OEIS! (A169855)

• Tomasz Miller says:

The rule seems to be the following. If x_n is the n-th term of the sequence, then x_(n+1) is the least integer sharing a prime factor with x_n that hasn't appeared in the sequence so far (i.e. x_(n+1) must be different than x_1, x_2, …, x_n).

The answer to Question 2 seems to be positive (all integers greater than 1 eventually show up), whereas for Question 3 my conjecture is: it will never happen. Tried hard to prove them both rigorously but failed…

And why 13? Maybe to give a hint about the rule, because 13, being followed by 26, immediately draws attention. Besides, it's already 11-13-15 when I'm typing this (CET). Coincidence? I don't think so.

• Deepak Kamlesh says:

First of all, let's note that a sequence is not uniquely determined by just providing the first few terms of the sequence since there are infinitely many distinct sequences which agree with the given sequence on these starting terms but differ later on.

With the above in mind, we give one plausible algorithm that generates the given sequence starting from the third term, that is 2 onwards.

Note: We put the following additional constraint in our algorithm that no number is to be repeated twice. So, whenever our algorithm generates a number that has already been generated, we add two to it and carry on with the algorithm. With this in mind, we give the algorithm.

Algorithm –
0. Take the starting term of the sequence as the current number and go to step 1. (Note: We are starting with 2)
1. Check if the current number is twice an odd prime.
2a. If the answer to 1 is no, then add two to the current number and take that as the current number. Go to step 1.
2b. If the answer to 1 is yes, then the next number is current number divided by two. With this new number as the current number, the next number is thrice the current number. With this new number as the current number, the next number is three plus the current number. With this new number as the current number the next number is the current number divided by 3. Now again go to step 1.

It can be easily checked that the above algorithm along with the additional constrained mentioned in the note generates the given sequence starting from the term 2 onwards.

As such, it is possible that the first two terms, that is, 13, and 26 are perhaps included to lead us astray and have no significance on how the rest of the sequence is defined. On the other extreme it is also entirely possible that I have missed to see the relevance of the starting terms in the sequence and thus have no idea what I'm talking about.

As to the second question, the answer is no, atleast with respect to our algorithm, that is, not every integer greater than 1 will appear in the sequence. For example, 17 will never appear in the sequence. This is because using our algorithm, after some steps we reach see that the sequence is 22, 11, 33, 36, 24 etc and since we have crossed 34, the sequence will never come down to a value smaller than the half of 34 which is 17.

As for the third question, note from the last paragraph, that the sequence has 36 and 24 as adjacent terms of the sequence, and this pair has more than 1 common prime factor. On writing down the sequence using our algorithm, one can see that this is indeed the first such pair with the required property and their positions can be counted from the sequence.

Finally, for the fourth question, I repeat the same reason as before, that perhaps the first term, that is 13, rather the first two terms have been included to mislead the readers.

• Daniel McLaury says:

QUESTION 1

Of course the sequence could be absolutely anything, but here's one sequence which agrees with the given terms and sounds fairly satisfying from a mathematical perspective:

The first term is chosen to be 13 by fiat. After this, each term of the sequence is the smallest positive integer which has not appeared before in the sequence and which is not relatively prime to the previous term.

QUESTION 2

Will this sequence hit every positive integer greater than one? Yes.

Why? Well, suppose not. In particular, say n>1 is never a term of the sequence. Then for any prime divisor p of n, the sequence can contain only finitely many terms which are multiples of p. In particular, it cannot even contain n such terms, because each such term would have to be followed by a unique number smaller than n.

But then this actually holds for every prime: if the sequence contains only finitely many multiples of p, and q is some other prime, then every sufficiently large multiple of p q isn't a term of the sequence. Taking such a number for n in the previous paragraph, we see that only finitely many multiples of q are terms of the sequence.

Let m be the smallest even number which is not a term of the sequence, and let s be a prime number which 2s >= m. We can conclude that s is not a term of the sequence, because otherwise the next term of the sequence would be 2s. (We know that 2s didn't appear before in the sequence because we've already assumed it doesn't appear in the sequence at all.)

But in fact this means that no multiple of s appears in the sequence. If one did, then we could take b*s to be the smallest multiple of s appearing in the sequence. Now the only way for this to happen would be if the previous term of the sequence was divisible by b and moreover that every number smaller than b*s which was not relatively prime to b had already appeared in the sequence. (Caveat: we've implicitly assumed that s is not 13, the first term of the sequence. But of course we can do this since we know that 13 appears in the sequence). Consequently, s itself would be the smallest number sharing a common factor with b*s which was not a previous term of the sequence.

Therefore, the sequence is comprised by finitely many multiples of finitely many primes, and is therefore finite. But this is impossible, because the way the sequence was defined means it's infinite — given a number, we can always find arbitrarily large numbers which share a prime factor with it. So it follows that every integer greater than 1 appears as a term of the sequence.

QUESTION 3

Terms 575-580 of the sequence are

…, 608, 589, 620, 610, 614, 307, …

Notice that 620 and 610 share a factor of ten. There doesn't seem to be a clean way of explaining this without simply computing the first several terms, although heuristically it helps that 620 = 2^2 * 5 * 31 has two small prime factors and one large one.

QUESTION 4

Why does the sequence start with 13? Well, it cannot start with 1, because every positive integer is relatively prime to 1. And we can't start from 2, because that sequence is in the OEIS (A064413). In fact, it appears that the corresponding sequences starting from 3, 4, 5, … 12 are all in the OEIS, so this is the first sequence where you can't just get the answer from a search.

Of course any number thirteen or larger could have been chosen by this rationale, so perhaps it was also chosen on the strength of this article being posted on 11/12 (using the American system for writing dates), something I likely only noticed because today is my mother's birthday.

• Marcos Costa Santos Carreira says:

Question 1:
Start with a list initialized as {13}
For the next number: If the number of prime factors of the last number is 1, look for the next multiple not yet on the list; if not, look for the smallest prime factor of the last number not on the list; if all the prime factors are already on the list, look for the smallest multiple of those prime factors that is not already on the list.
Plotting the list one can see a sudden drop when hitting a prime, followed by a jump and then back to a trend.
Question 2:
It should cover all the numbers, as all the primes will appear and then its multiples. The range defined by x=Complement[Range[2,Max[list]]; {Min[x],Max[x]} can be approximated by {0.525*Length[list],1.55*Length[list]}. There seems to be no number left behind.
Question 3:
{620, 610} are found in positions {577,578} on the sequence, and have both 2 and 5 as common prime factors.
Question 4:
13 was chosen because the sequence looks like the Josephus problem, and we are looking for unlucky numbers.

• maurizio codogno says:

I eventually found the rule – but I must admit that I needed to read Question 3 to have a clue.

Each number following the first one is the smallest number not already present in the sequence and which has one factor in common with the previous one.

• maurizio codogno says:

I forgot: 13 is the first number because Sloane loves it 🙂

• Paul says:

Next numbers are 20 22 11

• Joe says:

I'll take a crack at this, although I doubt that I solved it fully as you will see in a second.
Rule: The next number in the sequence shares a factor with the previous number. (The factor shared could be either number too.) As for how the specific numbers are picked, I'm a bit baffled. Eeny, meeny, miny, moe? Roll a dice? Have a taste test to see which number tastes best in cookie form?
Question 2: According to my rule, every number should be picked. The specific nuisance would be prime numbers, as you can only get them by multiplying by number itself or one, neither of which seems to be options. To get prime numbers you would have to get that number multiplied by two (i.e. 34 for the prime number 17), which then can be followed by the prime itself.
Question 3: Well, you could argue that all numbers are also multiplied by one an infinite amount of times, which means that the first two would share more than one prime factor, but lets stick to the definition. In reality, it looks like 12 and 8 are the lucky winners. 12's prime factorization is 2*2*3, and 8's prime factorization is 2*2*2, so they share two prime factors (even though those prime factors are the same number)
Question 4: Are we looking for Illuminati-proving outlandish? Or just funny outlandish? I'll go with the latter as we all know the Illuminati exist anyways. Clearly, Mr. Sloane took a recent vacation to Italy and got a little to excited at the casino. He was in there for so long that he finally hit the lucky number 13 (well, it's lucky in Italy), only to find out that the jackpot is, you guessed it, a measly 26 euros. This whole sequence originated in Mr. Sloane's mind as he walked out with his 26 euros, where he thought, for the briefest time, that he would be not only famous but rich as well.
(I hope readers could detect my sarcasm, but I just want to repeat that I DO NOT believe the Illuminati is confirmed to exist.)

• Pace says:

Answer to question 1: If we call the sequence a_1, a_2, …, then the pattern appears to be "a_n is the smallest integer that has not yet appeared in the list and shares a prime factor with a_{n-1}."

Answer to question 2: Every positive integer greater than 1 will occur in the sequence (no matter what the starting value is).

First, assume by way of contradiction that the prime divisors of each a_n are among some finite set p_1, p_2, …, p_k. Fix some prime q not on that list. As there are only finitely many numbers less than M=p_1*p_2*…*p_k*q, we can find some index n so that both a_n and a_{n+1} are bigger than M. But a_n shares a common factor with M, and since M<a_{n+1}, this contradicts the definition of the sequence.

Thus, we now know that there are infinitely many (and in particular, arbitrarily large) prime divisors in the sequence. Next, we prove that for each prime p, there are infinitely many a_n's divisible by p. Assuming otherwise, then there is some power p^k which does not divide any entry in the sequence. But there is also some prime q>p^k which does divide some a_n. We may as well assume that n>1, choosing a different q as needed. As a_n*(p^k/q) is an integer that shares a common factor with a_{n-1}, and it hasn't shown up previously in the sequence since it is divisible by p^k, and a_n*(p^k/q)<a_n, this contradicts the definition of the sequence.

Finally, let M>1 be an arbitrary integer. Since the smallest prime factor p of M divides infinitely many of the a_n, and there are only finitely many integers less than M, we see that M must eventually show up among the a_n.

Answer to question 3: The first two consecutive numbers in the sequence which share at least two *distinct* prime factors are 620 and 610 (which occur at the 577th and 578th positions). Of course the numbers 12 and 8 (occurring at the 8th and 9th positions respectively) are the first to share two *non*-distinct prime factors.

Answer to question 4: Apparently 13 is the smallest number for which we can use the simple rule described above, and the sequence does not (yet) appear in the OEIS!

• Bjorn Hope says:

1. X(n+1) = min(A*B) where A is a prime factor of X(n) and X(n+1) is not in {X(1),…,X(n)}.
2. Yes. Let's assume k is the smallest positive integer greater than 1 that hasn't appeared yet. Then X(n+1) = min(A*B) <= A*k. If X(n+1)=k*A, then X(n+2)=k. If X(n+1)=A*B<k*A, then A*B<k, which is a contradiction of the assumption.
3. Pretty far if you do it by hand.
4. Because the answer to this puzzle will be revealed today, the 13th.

• Gary Allan says:

Here are the rules or method for creating the sequence, as I see it.

Choose an arbitrary number; you chose 13

The first, n = 1, entry is 13 x 1

For each subsequent entry, n + 1 in the sequence, consider the unique prime factors of the nth entry excluding 1, and compare them to all previous entries up to the (n – 1)th which share that prime factor. For each find the largest such entry and add the prime factor to that value. From this set of new possible entries, choose the smallest. That becomes (n + 1)th entry, provided it is not in the sequence already; if it is then add the prime factor associated with the possible entry again and recheck set for smallest. Repeat

Thus
13 = 13 x 1 (Prime factor is 13, since 13 is already in the sequence 13 + 13 or 2 x 13 is next)
26 = 13 x 2 (Prime factors are 13, 2, next possibles are 3 x 13 = 39, or 2 x 1 = 2, which is least)
2 = 2 x 1 (Prime factor is 2, since 2 is already in the sequence 2 + 2 or 2 x 2 is next)
4 = 2 x 2 (Prime factor is 2, since 2 & 4 are already in the sequence 2 x 3 is next)
6 = 2 x 3 (Prime factors are 3, 2, next possibles are 2 x 4 = 8, or 3 x 1 = 3 which is least)
3 = 3 x 1 (Prime factor is 3, since 3 is already in the sequence 3 + 3 or 2 x 3 is next, but 6 is already in the sequence, so go to 3 x 3)
9 = 3 x 3 (Prime factor is 3, since 9 is already in the sequence, 3 x 4 is next)
12 = 3 x 4 (Prime factors are 3, 2, next possibles are 2 x 4 = 8, which is least, or 3 x 5 = 15)
8 = 2 x 4 (Prime factor is 2, since 8 is in the sequence 2 x 5 is next)
10 = 2 x 5 (Prime factors are 2, 5, next possibles are 2 x 6 = 12, or 5 x 1 = 5 which is least)
5 = 5 x 1 (Prime factor is 5, since 5, 10 are in the sequence 3 x 5 is next)
15 = 5 x 3 (Prime factors are 3, 5, next possibles are 3 x 6 = 18, or 5 x 4 = 20)
18 = 3 x 6 (Prime factors are 3, 2, next possibles are 3 x 7 = 21, or 2 x 7 = 14)
14 = 2 x 7 (Prime factors are 7, 2, next possibles are 7 x 1 = 7, or 2 x 8 = 16)
7 = 7 x 1 (Prime factors are 7, next possibles are 7 x 3 = 21)
21 = 7 x 3 (Prime factors are 7, 3, next possibles are 7 x 4 = 28, or 3 x 8 = 24)
24 = 3 x 8 (Prime factors are 3, 2, next possibles are 3 x 9 = 27, or 2 x 8 = 16)
16 = 2 x 8 (Prime factor is 2, next entry is 2 x 10 = 20)
20 = 2 x 10 (Factors are 2, 5, next possible entries, 22 and 25)
22 = 2 x 11 (2, 11, next possibles, 26, 11)
11 = 11 x 1 (11, next possible 33)
33 = 11 x 3 (11, 3, next possibles 44, 27)
27 = 3 x 9 (3, next possible 30)
30 = 3 x 10 (3, 2, 5, next possibles 36, 28, 25)
25 = 5 x 5 (5, next possible 35)
35 = 5 x 7 (5, 7, next possibles 40, 28)
28 = 7 x 4 (7, 2, next possibles 42, 32)
32 = 2 x 16 (2, next possible 34)
34 = 2 x 17 (2, 17, next possibles 36, 17)
17 = 17 x 1 (17, n p 51)
51 = 17 x 3 (17, 3, n p 68, 36)
36 = 3 x 12 (3, 2, n p 39, 38)
38 = 2 x 19 (2, 19, n p 40, 19)
19 = 19 x 1 (19, n p 57)
57 = 3 x 19 (3, 19, n p 76, 39)
39 = 3 x 13 (3, 13, n p 42, 52)
42 = 3 x 14 (3, 2, 7, n p 45, 40, 49)
40 = 2 x 20 (2, 5, n p 44, 45)
44 = 2 x 22 (2, 11, n p 46, 55)
46 = 2 x 23 (2, 23, n p 50, 23)
23 = 23 x 1 (23 n p 69)
69 = 23 x 3 (23, 3, n p 92, 45)
45 = 3 x 15 (3, 5, n p 48, 50)
48 = 3 x 16 (3, 2, n p 54, 50)
50 = 2 x 25 (2, 5, n p 52, 55)
52 = 2 x 26 (2, 13, n p 54, 65)
54 = 2 x 27 (2, 3, n p 56, 60)
56 = 2 x 28 (2, 7, n p 58, 63)
58 = 2 x 29 (2, 29, n p 60, 29)
29 = 29 x 1 (29, n p 87)
87 = 29 x 3 (29, 3, n p 116, 60)
60 = 3 x 20 (3, 2, 5, n p 63, 62, 55)
55 = 5 x 11 (5, 11, n p 65, 66)
65 = 5 x 13 (5, 13, n p 70, 78)
70 = 5 x 14 (5, 2, 7, n p 75, 62, 63)
62 = 2 x 31 (2, 31, n p 64, 31)
31 = 31 x 1 (31, n p 93)
93 = 31 x 3 (31, 3, n p 124, 63)
63 = 3 x 21 (3, 7, n p 66, 77)
66 = 3 x 22 (3, 2, 11, n p 72, 68, 77)

2) Yes, I think all positive integers will appear. By the way the sequence is defined all multiples of any prime number which is a factor of any entry must show up in the sequence sooner or later, whether even or odd. Since there will always be further multiples of 2 in the sequence sooner or later, all multiples of 2 by all integers must show up in the sequence which means all odd primes must sooner or later make an appearance. Since all positive integers are either 2, or an odd prime, or the product of primes, by the way the sequence is constructed all positive integers must show up sooner or later.

3)I am doubtful that any two consecutive entries in the sequence share more than one prime factor, but I cannot prove it.

4)It is Friday the 13th.

• Ravi Kumar Meduri says:

Question 1

Simple sequence I could not find. But the next number in my opinion is 20. Here are the set of rules, I used to get that:

1) Any two adjacent numbers should have at least one common factor above 1 i.e. no two adjacent numbers can be co-prime
2) No number is repeated
3) Keeping one factor as common, increase/ decrease by 1, the other factor or as minimum as possible to get the next number. Try to get the lowest next number to determine which factor is common and which other factor is changed
4) Keeping one factor as a common, go back to 1 or the lowest other factor if the number has not already come in the sequence
5) When a number is not a product of two prime factors, express the number in the all possible factors for determining the next number

Question 2

Some higher primes are the suspects. For example in the above sequence, the number 11 can come even though it is a prime. If after 20 (2*10) , you get 22 (2*11) and then 11 (11*1) . Similarly, 17 can come after 33 (11*3), 27 (3*9), 30 (3*10/ 5*6), 25 (5*5), 35 (5*7), 28 (4*7/ 2*14), 32 (2*16), 34 (2*17) and then 17 (1*17). Not sure about the higher primes but.

Question 3

It will be inevitable since teh numbers with only one prime factor would have already occurred.

Question 4

Some possible reasons for starting with 13 is that it is considered an unlucky number and therefore better to get over with it first. Other reason is that it would have considerable time for it to occur being a prime.

• eJ says:

1. Each term is the smallest positive integer not yet published that shares a factor with the term before it.
2. See 4.
3. Terms #577,578 are 620 and 610 respectively, with gcd 10.
4. To make it trickier to discover OEIS sequence A064413 and http://neilsloane.com/doc/g4g7.pdf !

• Next few terms of the sequence after 16 are: ….,22,11,33,36,34,17,51,54,38,19,57,60,46,23,69,72,32,…

Ans1. We have powers of 2 as checkpoints, whenever we encounter them we restart our algorithm. We start with "seeds" 13, 26, 2; and then we place powers of 2 with spacing equal to previous power of 2. Thus we get:
13,26,2,4,_,_,_,_,8,_,_,_,_,_,_,_,_,16,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,32,…
Then we start inserting primes (note that 13 and 2 have already been used), with their double preceding them and their triple succeeding them. Then, we add 3 to triple of prime number to get its successor. Thus we get: (3 being our prime number, and 8 being the checkpoint)
13,26,2,4,6,3,9,12,8,_,_,_,_,_,_,_,_,16,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,32,…
Now since we reached the checkpoint we restart our process (jumping over it):
13,26,2,4,6,3,9,12,8,10,5,15,18,14,7,21,24,16,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,32,…
Since, each step adds 4 numbers to the sequence and we have gap of multiple of 4 between two powers of 2, so the above algorithm will work flawlessly.

Ans2. No, every positive integer greater than one won't appear in this sequence. For example, 39 will never appear in this sequence, since it's 13*3, and 13 has been used in the very beginning, thus no chance of it's occurrence. Also, 42 will never occur (since it's predecessor doesn't occur). Further, if p and q be two consecutive primes with |p-q|>2, then the numbers from (p+2)*3 to (q-1)*3 will not occur in this sequence (in addition to 39 and 42). Also, all those composite numbers which are "not divisible by 3, are not powers of 2 and are divisible by 2^k with k=0 or k>1", will not occur in this sequence (for example: 20,25,28,…).

Ans3. We will never encounter a pair of adjacent numbers that share more than one prime factor.

Ans4. This sequence starts with 13, so that it becomes easy to answer Question 2.

• Frosty says:

Lowest unused integer using any of the prime factors from the previous number. Yes, every positive integer: All multiples of 2 will be used, and every prime times 2 will be used, so every prime will be used. 30, 24 (share 2 and 3 as prime factors). Start with 13: confuse the solver? Other starts are already in Sloane? Friday the 13th? Implies an alphanumeric solution? Prettier heart graph?

• Daniel McLaury says:

Was there any way of doing the third problem aside from brute force? That's what I was hoping to learn from the other comments.

• mclaren says:

No, successive numbers have nothing in common. It's obvious that there is no pattern in the sequence.

• Kevin says:

Q1: Numbers correspond to buses I take to work in Glasgow thanks to its SimpliCITY scheme to optimise travel (or maximis revenue)

Q2: no, clearly there is a finite set of Scottish and Polish bus drivers, as well as buses

Q3: I am stuck on this one, sadly.

Q4: the number 13 stops next to my house. Sure I could walk to the number 26 stop, but that's almost 200m away …

• Klavs Hansen says:

With a finite number of entries there is no way one can identify a unique rule. I think there is even a theorem about this. Incidentially, that's why all these Mensa test are idiotic (in the original sense of the word). If you really want a solution, just make a sum of delta functions and add some arbitrary term as the next one. It does not have to be a number or even something you can write on a screen. I declare that the next term in the series is a bowl of gazpazzo soup, which I will eat now. Good night.

• Antonio Sanz says:

Hello everyone, I´m Antonio Sanz, from Spain. First, thanks to Mr Mutalik and Mr Sloane for this amazing challenge. Here we go with the answers:

Regarding Question 1, I think that this is the sequence of the smallest intergers that share a common factor to the previous item, starting with thirteen.

Concerning to Question 2, the fact that the third position of the sequence was the number two results from this specific feature. I explain myself:
(a) The series of multiples of any prime number includes the compound numbers containing whole series of prime numbers repeated from 1 to infinite times.
(b) Because any number is prime or a composite of prime numbers, infinite series of prime numbers and their multiples include all number greater than 1.
Given any number (p1), the next number in the sequence (p2) is the smallest not yet selected included in the series of multiples of the prime numbers that compose p1 or one of these prime numbers. Considering (a), p2 shares a factor with another number (p3) of the same series or any other series -therefore, with any integer greater than 1, by (b) – which, in turn, is the smallest that has not yet been selected, and so on until p-infinity. Thus, all integers greater than 1 are in the sequence.

In respect to Question 3, they seem to be 620 (position 577) and 610 (position 578), as they share two and five. I think that for this Question, we can find a more simple method to answer it, but my neurons are exhausted.

Finally, regarding Question 4, why not starting with thirteen since the rule applies equally regardless of the first number? If we consider one as prime number, the sequence of natural numbers greater than zero also follows this rule, because all natural numbers greater than zero share interger 1 with the preceding.

Thank you very much.
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