# Integer Sequence Review Mêlée Hyper-Battle DX 2000 (Bracket 1)

After taking a couple of weeks off from reviewing integer sequences, we’ve decided to shake up the format. Prepare yourself for…

We’re going to review six sequences each week for four weeks, picking a winner from each. Then, we’ll pick one sequence from the ones we’ve already reviewed individually, plus a wildcard. Finally, a single sequence will be crowned the Integest Sequence 2013!

We’re still judging each sequence on four axes: Aesthetics, Completeness, Explicability, and Novelty.

Without further ado, here we go!

#### A002210 Decimal expansion of Khintchine’s constant.

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

Christian: This one’s going to blow your mind. Take a real number. Write it out as a continued fraction. Take the geometric mean of the coefficients. It’s almost always the same constant!

David: We have a winner.

Christian: Don’t be so hasty. While I’m also certain this is going to win, we’ve set up this tournament thing so we’d better go through with it. Explicability is high, …

David: It’s a decimal expansion so it should lose a bit on Aesthetics. Which series sum gives an approximation to it the quickest?

Christian: Borwein and Bailey have worked that out. BUT WE DON’T HAVE TIME TO READ THAT PAPER!

David: Seven!

Christian: That’s not how we score these.

Aesthetics $\frac{2}{5}$ $\frac{4}{5}$ $\frac{5}{5}$ $\frac{5}{5}$ $\frac{16}{20} = \frac{4}{5}$

#### A189055 Primes of the form $(n+1)^{11} – n^{11}$.

313968931, 6612607849, 68618940391, 2257404775627, 26360313735014491, 130898631716248441, 11736367906285382977, 28945284114821573731, 229761141540921525811, 202978059247932180748537, 228398127589553102936371, 476213535986962784582617, 1627839264198988265272849, 3421374091098795513254497, ...

Christian: I recognise this one!

David: That’s because its discoverer is a good friend of yours and a genius. And you calculated most of these primes. It’s the primes which are the difference of eleventh powers of natural numbers.

Christian: So it’s a zero for novelty. Did you know what you were doing when you picked this?

David: Eleven was the lowest power that wasn’t in the encyclopedia.

Christian: So, aesthetically, it’s worse than the same sequence for, say, fifth powers.

It’s OK, we’re trained mathematicians.

David: Ooh! I like the ones which are the difference of fifth powers. They all end in the digit 1. We should review those instead.

Christian: Denied. You picked this one.

Aesthetics $\frac{3}{5}$ $\frac{2}{5}$ $\frac{5}{5}$ $\frac{1}{5}$ $\frac{11}{20}$

#### A023186 Lonely (or isolated) primes: increasing distance to nearest prime.

2, 5, 23, 53, 211, 1847, 2179, 3967, 16033, 24281, 38501, 58831, 203713, 206699, 413353, 1272749, 2198981, 5102953, 10938023, 12623189, 72546283, 142414669, 162821917, 163710121, 325737821, 1131241763, 1791752797, 3173306951, 4841337887, 6021542119, 6807940367, 7174208683, 8835528511, 11179888193, 15318488291, 26329105043, 31587561361, 45241670743, ...

Christian: So as you’re going through the primes, these are the ones where the gap to the next or previous prime is the biggest you’ve yet seen.

David: It’s taken you a very long time to both understand this and explain it to me. Very low Explicability.

Christian: I suppose it’s a tiny bit subtle. I was expecting it to be just the gap to the next prime. What about the other categories?

David: It’s got Erdős’s name on it, so who can argue it’s not pretty?

Christian: Novelty?

David: It’s only got primes in it. That makes it either unnovel or novel, I’m not sure. Can we just score it? This form of article is much longer than I was expecting. Can we just give the trophy to the Khintchine constant already?

Christian: No. We’ve got five weeks of this ahead of us. Oh god…

Aesthetics $\frac{4}{5}$ $\frac{4}{5}$ $\frac{2}{5}$ $\frac{3}{5}$ $\frac{13}{20}$

#### A199949 Decimal expansion of least $x$ satisfying $x^2+\cos(x)=2 \sin(x)$.

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

Christian: Now, I know what you’re going to say –

David: This one’s rubbish.

Christian: Correct. But I wanted to include it because this is what most of the OEIS is – tedious, boring, stupid sequences included just for completeness. This one is part of a huge family of roots of equations. There’s a table in its description.

David: This is a sequence review, Christian, not an encyclopedia review.

Christian: I won’t argue. It doesn’t even need any explanation. I should say that the OEIS gives Mathematica code to produce a plot of the implicit sequence between the solutions to this whole family of equations, and I bet it’s pretty. But, since we don’t have Mathematica, we can’t judge it on that. Let’s give it a low score.

Aesthetics $\frac{0}{5}$ $\frac{5}{5}$ $\frac{1}{5}$ $\frac{1}{5}$ $\frac{7}{20}$

#### A120498 Numbers C from the ABC conjecture.

9, 32, 49, 64, 81, 125, 128, 225, 243, 245, 250, 256, 289, 343, 375, 512, 513, 539, 625, 676, 729, 961, 968, 1025, 1029, 1216, 1331, 1369, 1587, 1681, 2048, 2057, 2187, 2197, 2304, 2312, 2401, 2500, 2673, 3025, 3072, 3125, 3136, 3211, 3481, 3584, 3773, 3888, ...

David: The abc conjecture is, in my opinion, the most important problem in mathematics.

Christian: Sigh…

David: It is! It’s how numbers work.

Christian: Can we explain the whole thing here? Probably not.

David: We can have a link to my blog post explaining it.

Christian: Done.

David: The reason I picked this sequence to represent the abc conjecture is that lots of the numbers are pretty – lots of them are perfect powers, such as 32 and 64, or $a(n)$ and $a(n)-1$ are pretty nice-looking, such as 9 and 8, or 49 and 48.

Christian: Eh?

David: They’ve got lots of divisors.

Christian: Completeness?

David: It’s very easy to compute.

Aesthetics $\frac{4}{5}$ $\frac{5}{5}$ $\frac{1}{5}$ $\frac{3}{5}$ $\frac{13}{20}$

#### A006472 $\displaystyle{ \frac{n! \cdot (n-1)!}{2^{n-1}} }$

1, 1, 3, 18, 180, 2700, 56700, 1587600, 57153600, 2571912000, 141455160000, 9336040560000, 728211163680000, 66267215894880000, 6958057668962400000, 834966920275488000000, 113555501157466368000000, 17373991677092354304000000, ...

David: I don’t like it. Too many zeroes, and why does it start with two 1s?

Christian: Slow your roll! What if I told you that $a(n)$ is the product of the first $n$ triangular numbers?

David: That’s pretty cool. But I bet I can think of my own sequence of triangular numbers which is better. Stay tuned folks.

Christian: I remain tuned. I like it. It grows quickly. Like your mum.

David: I was just going to say that!

Christian: Have we explained it? Why is it the first $n$ triangular numbers? Oh! From the formula. Clearly.

David: Twenty.

Christian: In a particular category, or overall, or have you just gone off these numbers and started saying other ones?

David: I’m losing interest. The first sequence we saw is the winner anyway, can we just get it over with?

Aesthetics $\frac{2}{5}$ $\frac{3}{5}$ $\frac{4}{5}$ $\frac{3}{5}$ $\frac{12}{20} = \frac{3}{5}$

### And the winner is…

A002210, the decimal expansion of Khintchine’s constant!

A002210 advances to the final with an impressive score of $\frac{4}{5}$.

We’ll be back with Bracket 2 next week. In the mean time, please leave suggestions for sequences you think we should review in the comments.

### Ipso Post Facto Navigato

• #### Christian Lawson-Perfect

Mathematician, koala fan, Aperiodical editor. Usually found paddling in the North Sea, or fiddling with computers.
• #### David Cushing

Mathematician / magician / origami enthusiast. Wanted for fraud in at least one branch of Subway.

### 6 Responses to “Integer Sequence Review Mêlée Hyper-Battle DX 2000 (Bracket 1)”

1. John

A002211 is the real winner. I wonder what the geometric mean of its terms is.