I'm interested in the limit as you take the square of the number given by the first $N$ digits of $\pi$ (ignoring the decimal point) & then add to it the first $M$ digits of $\pi$ (ignoring the decimal point again), where $M$ is the number of digits in the square of the first $N$ digits. Sorry if this sounds a bit confusing, but here are the examples:

            N                  N^2           M                 N_M              N^2 + N_M




4------------3141--------------9865881---------------7----------------3141592 -------------------13007473

5-----------31415----------986902225---------------9-------------314159265 -----------------1301061490

6---------314159--------98695877281---------------11----------31415926535 --------------130111803816

7--------3141592---9869600294464---------------13--------3141592653589 ----------13011192948053

8------31415926-986960406437476--------------15------314159265358979 --------1301119671796455

What I'm interested in is, does $N^2 + N_M$ converge, and if so, to what value, and if not, how quickly divergent is it?

My best guess is that it converges, or is weakly divergent, but I'm extremely poorly-versed in limits, and know next to nothing about $\pi$'s properties or indeed formulas for it. But am curious nevertheless as to the result. Would appreciate any help. I am very far from my speciality here, which is integer arithmetic, algebra and diophantine equations. It's likely someone has investigated this before, but I don't really have a spare two hours or so to locate the answer using Google search, so would be grateful for any pointers.


closed as off-topic by Andrés E. Caicedo, Namaste, Rohan, Shailesh, JonMark Perry Dec 12 '17 at 7:13

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  • "This question is not about mathematics, within the scope defined in the help center." – Andrés E. Caicedo, Namaste, Rohan, Shailesh, JonMark Perry
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  • 4
    $\begingroup$ Your sequence is approximately $$a_n=10^n \pi^2+10^{2n+1}\pi$$Both numbers tend to infinity, so it doesn't converge $\endgroup$ – John Doe Dec 11 '17 at 12:34
  • $\begingroup$ As integers, they do not converge. However, seen as just a sequence of single digit numbers that increases in length, I see no reason that that each entry shouldn't eventually stabilize. I don't think the sequence means anything, though. $\endgroup$ – Arthur Dec 11 '17 at 12:37
  • 2
    $\begingroup$ @Arthur, note that $\pi^2+\pi\approx13.0111970547$. $\endgroup$ – Barry Cipra Dec 11 '17 at 12:57
  • $\begingroup$ @JohnDoe, I think you've got the wrong exponents on your powers of $10$. In particular, $10^{2n+1}$ quickly dominates $10^n$. $\endgroup$ – Barry Cipra Dec 11 '17 at 12:59
  • $\begingroup$ @JohnDoe The first term is approximately $(10^n\pi)^2$, so your exponent on the $10$ is off. $\endgroup$ – Arthur Dec 11 '17 at 13:01

Turning some comments into an answer (with a shift in notation for the number of digits involved), when you make an integer out of the first $n$ digits of $\pi$ (counting the $3$, which the comments below the OP do not), you have $\lfloor10^{n-1}\pi\rfloor$. Squaring this gives an integer with $2n-1$ digits. The integer formed by the first $2n-1$ digits of $\pi$ is $\lfloor10^{(2n-1)-1}\pi\rfloor$. So the OP's sequence (starting with $n=1$ is


The $n$th term in the sequence has $2n$ digits. Roughly speaking, the first $n$ digits agree with the first $n$ digits of $\pi^2+\pi\approx13.011197054679\ldots$, while the final $n$ digits do not agree with the next $n$ digits of $\pi^2+\pi$. In particular, if you subtract the OP's sequence from $\lfloor10^{2n-2}(\pi^2+\pi)\rfloor$, you get the sequence


This sequence is not (yet) in the OEIS. Aside from the $n$th term having (roughly?) $n$ digits, there's no apparent pattern in the numbers. I daresay there's no reason to expect one.

  • $\begingroup$ Thank you Barry; I appreciate the insight. This only came up as a preliminary to work when I was looking at polygons with an inscribed circle of area 1 unit and therefore diameter of 2/sqrt(pi). In this earlier case I made the radius equal to half of 1 unit in order to make the circumference equal to pi, but this didn't prove very fruitful later on in the calculation, so I switched to a circle of area 1 instead. When messing about with the calculations this came up, so I thought it might be worth posting. But as you say, I doubt there's anything much to it; it's just a curiosity really. $\endgroup$ – Joebloggs Dec 11 '17 at 16:07

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