I'm trying to define a new [silly] calendar, because there aren't enough of them yet.

My calendar, so far, is specified:

epoch (orthodox) is the moment the leading edges of the dinosaur-killing asteroid touched the surface of the earth.

As this is somewhat difficult to define, for practical purposes we'll use:

epoch: same as the unix epoch, i.e., 1970-01-01T00:00:00Z.

From that point, start counting seconds in successive prime-numbered sequence; these are termed 'dinosecs', such that epoch + 2s = dinosec 1, epoch + 2s + 3s = dinosec 2, epoch + 2s + 3s + 5s = dinosec 3, dinosec 3 + 7s = dinosec 4, etc etc.

It is, however, forbidden to use seconds in the actual specification of time - all time must be referred to in dinosecs and fractional dinosecs.

Here's my question: What is a good way to go about defining fractional dinosecs?

One obvious way is to simply take the number of seconds since the last dinosec and divide it by the number of seconds between that dinosec and the next one. I dislike that method because it makes dinosecs nondifferentiable at whole-numbers - i.e. there's a sudden change in how quickly the fractional part is incrementing.

Is there something that could be done with $x/\ln(x)$, for example? Or is there an more rigorous way to do it?

Any ideas would be appreciated.

  • 1
    $\begingroup$ Let me try to rephrase this as mathematically as possible. You want a function which 1) restricts to your prime-sum function at the integers, 2) is infinitely differentiable, 3) is strictly increasing, and presumably 4) is not too difficult to compute. Does that sound right? There are going to be lots of ways to satisfy 1)-3), but I'm not sure there's a single "natural" way, or one that does very well at 4)... $\endgroup$
    – Micah
    Commented Feb 1, 2013 at 19:26
  • $\begingroup$ I first suggested $$ \tau(t)=\sum_{n=1}^\infty\frac{\sin\pi(t-n)}{\pi(t-n)}\sum_{k=1}^np_k\;, $$ but now I realized that doesn't converge because the second factor grows more quickly than the first one decays. $\endgroup$
    – joriki
    Commented Feb 1, 2013 at 19:26
  • $\begingroup$ Micah, yes, that sounds right. And now that you put it that way, I'm guessing this is impossible, because such a function would generate the primes for you. So, I guess I need to rethink this. Any ideas? $\endgroup$
    – Daniel
    Commented Feb 1, 2013 at 20:00

2 Answers 2


You could use a cubic spline, which would give you the requisite smoothness. But calculations might get hairy...


approximating the prime counting function

(part of the inexplicable secrets of creation)

  • 2
    $\begingroup$ What is the plot about? $\endgroup$
    – Mikasa
    Commented Jul 7, 2013 at 9:01

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