Last week A002210, the decimal expansion of Khintchine’s constant, emerged victorious from Bracket 1. Now, get ready for round 2 of…

Here are the rules: we’re judging each sequence on four axes: Aesthetics, Completeness, Explicability, and Novelty. We’re reviewing six sequences each week for four weeks, picking a winner from each. Then, we’ll pick one sequence from the ones we reviewed individually before this thing started, plus a wildcard. Finally, a single sequence will be crowned the Integest Sequence 2013! 

A006037
Weird numbers: abundant (A005101) but not pseudoperfect (A005835).

70, 836, 4030, 5830, 7192, 7912, 9272, 10430, 10570, 10792, 10990, 11410, 11690, 12110, 12530, 12670, 13370, 13510, 13790, 13930, 14770, 15610, 15890, 16030, 16310, 16730, 16870, 17272, 17570, 17990, 18410, 18830, 18970, 19390, 19670, ...

Christian: Lots of mumbo-jumbo. What are abundant and pseudoperfect, apart from my virtues, and imitations of myself, respectively?

David: I… do you want me to take a guess, or do you want me to look it up?

Christian: Take a guess and we’ll see if you’re right.

David: I would say that they were numbers the sum of whose divisors is more than one greater than themselves. Because abundant for numbers the sum is just greater than the original, while for pseudoperfect numbers it’s exactly one more.

Christian: Survey says…

David: Ooh! A number is pseudoperfect if the sum of some of its divisors is equal to the original number. I like pseudoperfect numbers. But unfortunately, this sequence doesn’t contain any. I’m not a fan of this sequence. And abundant numbers are just greedy.

Christian: I’m fairly sure greedy numbers already exist. Anyway, we’ve spent too long on this sequence already. Explicability: 2?

David: It’s not too bad, actually. 3.

Christian: Aesthetics?

David: I know I say it a lot, but the last digit of each number is really annoying me. It’s pretty much a zero all the time.

Christian: I don’t like it much at all. I can go 2.

David: Yeah.

Christian: Completeness and Novelty both 3. Happy?

A064442
Decimal expansion of number with continued fraction expansion 2, 3, 5, 7, 11, 13, 17, 19, … = 2.3130367364335829063839516 …

2, 3, 1, 3, 0, 3, 6, 7, 3, 6, 4, 3, 3, 5, 8, 2, 9, 0, 6, 3, 8, 3, 9, 5, 1, 6, 0, 2, 6, 4, 1, 7, 8, 2, 4, 7, 6, 3, 9, 6, 6, 8, 9, 7, 7, 1, 8, 0, 3, 2, 5, 6, 3, 4, 0, 2, 1, 0, 1, 2, 4, 4, 4, 2, 1, 4, 4, 5, 6, 4, 7, 3, 1, 7, 7, 6, 2, 7, 2, 2, 4, 3, 6, 9, 5, 3, 2, 2, 0, 1, 7, 2, 3, 8, 3, 2, 8, 1, 7, 5, ...

Christian: ANOTHER decimal expansion. Zero for everything.

David: It’s a pretty cool constant though, isn’t it?

Christian: Yes, it’s eerily close to Rényi’s parking constant.

David: What’s that and why aren’t we reviewing it?

Christian: It’s the proportion of space typically left empty on a street when it fill up with parked cars.

David: Does this sequence have any cool facts like that?

Christian: I actually have a paper about this saved somewhere… “Continued fractions constructed from prime numbers” by Marek Wolf.

David: Should we read it or give this a rubbish score?

Christian: I’ll just check… Oh! That paper gives a different number! In this one the two is an integer, while –

David: This is a shambles. Can we not for once have an unshambolic review? Let’s give it a score now and end this madness.

A001220
Wieferich primes: primes p such that $p^2$ divides $2^{p-1} – 1$.

1093, 3511, ...?

Christian: WHAT. IS. THIS.

David: From Fermat’s little theorem, every prime $p$ divides $2^{p-1}-1$. So, it’s a natural question to ask when $p^2$ does. Turns out, not very often.

Christian: Do you know if this is finite?

David: No. In fact, we don’t even know if there’s a point after which every prime belongs to the sequence.

Christian: Does the OEIS say how far has been checked?

David: $6.7 \times 10^{15}$.

(ring ring)

Christian: Hold on, it’s Doron Zeilberger on the phone… he says that’s proof enough that the list is finite.

David: It’s conjectured that the number of Wieferich primes less than $x$ is approximately $\log \log x$. And $\log \log (10^{15})$ is approximately… 3.5. So we aren’t too far out. The point is they don’t appear very often, but they do appear. Possibly.

Christian: SCORES DAVID.

David: LOTS AND LOTS OF SCORES CHRISTIAN.

A117727
Partial sums of A051109.

1, 3, 8, 18, 38, 88, 188, 388, 888, 1888, 3888, 8888, 18888, 38888, 88888, 188888, 388888, 888888, 1888888, 3888888, 8888888, 18888888, 38888888, 88888888, 188888888, 388888888, 888888888, 1888888888, 3888888888, 8888888888, ...

Christian: I. Like. Big. Butts and I can not lie. If I was a bingo caller, I’d describe this sequence as “Bigg market hen party”.

David: What’s this? What’s A051109?

Christian: A051109 is the hyperinflation sequence for banknotes. It’s the denominations you’d  issue if your currency was losing value very very quickly. This sequence gives the smallest amounts of money that can only be made with at least $n$ notes.

David: I hate it.

Christian: Why?

David: I hate both the definition of the original sequence, sequences which are partial sums of other sequences, and that number of 8s is very very nauseating.

Christian: They please me. I refer you to my previous comments re big butts.

David: This is a family show, Christian. Did you know that teachers in Australia are encouraged to show this to their students?

Christian: It slipped my mind. I hereby retract all previous statements on the subject of backsides and the bigness thereof.

David: Scores. Aesthetics: 1. I feel physically sick looking at it.

Christian: Do you suffer from octophobia? I’ll respect your wishes and go for a 2.

David: Completeness: 1.

Christian: Why?

David: Why not?

Christian: Because it’s complete!

David: Complete with ugly things! Explicability: 2.

Christian: I give up. I don’t care enough about this sequence. Straight 2s?

David: It adds up to 8! That is beautiful.

Christian: Normally we’d do some chicanery based on that, but we’ve got more work to do.

A055999
$a(n)=n \times \frac{n+7}{2}$.

0, 4, 9, 15, 22, 30, 39, 49, 60, 72, 85, 99, 114, 130, 147, 165, 184, 204, 225, 247, 270, 294, 319, 345, 372, 400, 429, 459, 490, 522, 555, 589, 624, 660, 697, 735, 774, 814, 855, 897, 940, 984, 1029, 1075, 1122, 1170, 1219, 1269, 1320, 1372, 1425, 1479, ...

Christian: This is rubbish.

David: Can I pick which ones to review next week?

Christian: I’m fairly sure you picked this one.

David: Pretty sure I didn’t. Why would I pick drivel like this? Is this like when Jedward got pretty far through the X Factor and nobody knew how?

Christian: OK, zero aesthetics, zero explicability?

David: What is it?

Christian: A comment on the OEIS says,

If $X$ is an $n$-set and $Y$ a fixed $(n-4)$-subset of $X$ then $a(n-3)$ is equal to the number of $2$-subsets of $X$ intersecting $Y$.

David: Is zero not a bit generous?

…

Christian: Sorry, I fell asleep typing that out. Let’s write some small numbers in the boxes.

A120820
Primes with $n$ consecutive digits ascending beginning with the digit two.

1, 2, 8, 82, 118, 158, 2122, 2242, 2388, ...?

David: I’ve made a huge mistake. I think the sequence we should be reviewing is a sequence created by this one. I’ll review this one anyway.

Christian: Go on then.

David: Write out 234567890123456… and stop whenever you get a prime. Count the number of digits in that prime. That’s a number in the sequence.

Christian: 5 for Explicability then.

David: Is it 5 for Explicability? I struggled to come up with good words.

Christian: But you done speak good in end David. Aesthetics? I don’t like that they’re all even.

David: I don’t either! I prefer the sequence which was the actual primes –

Christian: A089987?

David: No! That’s a truncation of Champernowne’s constant. Slightly different to what we’re after, but it is sexy nonetheless.

Christian: You should submit your one. Anyway, Novelty?

David: One. Because there are seven more sequences!

Christian: Don’t you mean eight? There’s one for each digit numbers can start with.

David: The sequence with first digit 1 isn’t there!

Christian: Let’s not think about why that might be. So a low Novelty score. Completeness?

David: The OEIS doesn’t mention any facts about it, so we’ll have to assume it might be infinitely big, so it might not be complete.

Christian: OK. I think we can do this.

And the winner is…

A001220, the Wieferich primes!

A001220 advances to the final with a decent score of $\frac{3}{4}$.

We’ll be back with Bracket 3 next week. In the mean time, please leave suggestions for sequences you think we should review in the comments.