6
$\begingroup$

Ten years ago I blogged:

I picked up my old, battered copy of G.H. Hardy's Pure Mathematics. I haven't spent as much time reading this book as I should have; it's full of good stuff. There didn't seem to be anything in there about the Gaussian integers (digression: What's next in the sequence 1, 2, 4, 6, 10, 14, 16, 24, 26?) but while scanning the index I noticed there was an entry for Ramanujan, so I checked it out.

What is next in the sequence 1, 2, 4, 6, 10, 14, 16, 24, 26? I can't remember. Google search finds only this one mention from my blog. OEIS doesn't know; the best I can do is A228898 which I'm sure is not what I had in mind. There is probably something to it, and the answer will not be of the type "If you use Lagrange interpolation to fit the following 9th-degree polynomial… the next value is -163”, or the stops on the 7th avenue IRT train, or anything similarly obnoxious.

It must have something to do with Hardy's book, but I have no idea what.

  • 3
    If $20$ were in that sequence, I'd say the next number after $26$ would be $36$. Are you sure you didn't accidentally forget $20$? (It would fit Gaussian integers, though, not Hardy's book.) – 2017-01-12
  • 2
    That is surely the answer: [“nonnegative integers such that $a(n)+i$ is a Gaussian prime”](http://oeis.org/A005574). Thank you! – 2017-01-12

1 Answers 1

2

Daniel Fischer's comment is almost certainly the right answer:

If 20 were in that sequence, I'd say the next number after 26 would be 36. Are you sure you didn't accidentally forget 20?

because these are positive integers $n$ such that $n+i$ is a gaussian prime. Or, more simply, $n^2+1$ is prime. See OEIS A005574.

Perhaps I computed the values by hand and made a mistake on $20+i$.