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One kind of figurate numbers are, starting with $\color{blue}{n=1}$, $$P_1(n) = n = 1, 2, 3, 4, 5,\dots$$ $$P_2(n) = \tfrac{n(n+1)}{2!} = 1, 3, 6, 10, 15,\dots$$ $$P_3(n) = \tfrac{n(n+1)(n+2)}{3!} = 1, 4, 10, 20, 35,\dots$$ $$P_4(n) = \tfrac{n(n+1)(n+2)(n+3)}{4!} =1, 5, 15, 35, 70,\dots$$ namely the linear, triangular, tetrahedral, pentatope, etc.

Using $(-1)^n$ to generate an alternating series is well-known. But what if we replace the exponent of $(-1)^n$ with higher figurate numbers $P_k(n) $?

Thus we get the sequences, $$\begin{aligned} S_1(n) &=-(-1)^{P_1(n)} = \color{blue}{1,-1},1,-1,\dots\\ S_2(n) &= -(-1)^{P_2(n)}= \color{blue}{1, 1, -1, -1}, 1, 1, -1, -1,\dots\\ S_3(n) &= -(-1)^{P_3(n)}= \color{blue}{1, -1, -1, -1}, 1, -1, -1, -1,\dots\\ S_4(n) &= -(-1)^{P_4(n)}= \color{blue}{1, 1, 1, 1, -1, -1, -1, -1},\dots \end{aligned}$$

and so on. The periods $\omega_k$ for $k=1,2,3,\dots8$ are $\omega_k = 2, 4, 4, 8, 8, 8, 8, 16.$

Q: How can we express the period $\omega_k$ as a function of $k$?

P.S. This was inspired by this post.

Community
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Tito Piezas III
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  • Does the pattern of periods continue -- $n$ copies of $2^n$? – pjs36 Dec 15 '16 at 03:20
  • @pjs36: I didn't check beyond $k=8$. I was getting cross-eyed looking at the signs. :) – Tito Piezas III Dec 15 '16 at 03:22
  • Exactly why I didn't check myself :P Maybe I'll write a program with Sage if I get some time, just to see. – pjs36 Dec 15 '16 at 03:27
  • Update: Yes, it appears that the sequence of periods is the concatenation of $n$ copies of $2^n$; it is true that $\omega_k = 2^{\lceil\log_2(k)\rceil}$ (I think that's the formula I want: it's the smallest power of $2$ that's at least $k$) for $k \le 257$ (so $\omega_{257}$ seems to be $512$ as expected). If I could figure out how to share the Sage notebook page I would. I'll leave it to someone else to show that this is true... unless I really need to put off grading finals :) – pjs36 Dec 15 '16 at 04:05
  • @pjs36: I guess it is this OEIS sequence. – Tito Piezas III Dec 15 '16 at 04:30
  • The period is certainly a divisor of $2^{\lceil\log_2(k)\rceil}$. Perhaps I can return to write the proof if there is time tonight. I have no proof though that it would be exactly $2^{\lceil\log_2(k)\rceil}$. In the mean time, perhaps you can work it out yourself. You need to think about the powers of $2$ dividing $k$ consecutive numbers. – 2'5 9'2 Dec 15 '16 at 04:30
  • You might be interested in this post: https://math.stackexchange.com/questions/2336334/finding-the-periods-of-a-sequence-of-integers-reduced-mod-m – mr_e_man Jan 05 '23 at 01:34

1 Answers1

2

A comment instead of an answer, but too long for the "comment"-format -

I understand things that the problem is the parity of the binomial-coefficients; that coefficients are along columns in the lower triangular Pascal matrix P, so the columns, expressed by $P \pmod 2$ should give the pattern of the signs in your question.
Here is a "graphical" expression for the patterns; the columns are to be read downwards beginning at the diagonal. The "1" give the sign $(-1)^1=-1$, the dots the positions of $(-1) ^0=1$ : 

  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  .  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  1  1  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  .  .  .  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  1  .  .  1  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  .  1  .  1  .  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  1  1  1  1  1  1  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  .  .  .  .  .  .  .  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  1  .  .  .  .  .  .  1  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  .  1  .  .  .  .  .  1  .  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  1  1  1  .  .  .  .  1  1  1  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  .  .  .  1  .  .  .  1  .  .  .  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  1  .  .  1  1  .  .  1  1  .  .  1  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  .  1  .  1  .  1  .  1  .  1  .  1  .  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .  1  1  .  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  .  1  .  .  .  .  .  .  .  .  .  .  .  .  .  1  .  1  .  .  .  .  .  .  .  .  .  .  .  .  .
  1  1  1  1  .  .  .  .  .  .  .  .  .  .  .  .  1  1  1  1  .  .  .  .  .  .  .  .  .  .  .  .
  1  .  .  .  1  .  .  .  .  .  .  .  .  .  .  .  1  .  .  .  1  .  .  .  .  .  .  .  .  .  .  .
  1  1  .  .  1  1  .  .  .  .  .  .  .  .  .  .  1  1  .  .  1  1  .  .  .  .  .  .  .  .  .  .
  1  .  1  .  1  .  1  .  .  .  .  .  .  .  .  .  1  .  1  .  1  .  1  .  .  .  .  .  .  .  .  .
  1  1  1  1  1  1  1  1  .  .  .  .  .  .  .  .  1  1  1  1  1  1  1  1  .  .  .  .  .  .  .  .
  1  .  .  .  .  .  .  .  1  .  .  .  .  .  .  .  1  .  .  .  .  .  .  .  1  .  .  .  .  .  .  .
  1  1  .  .  .  .  .  .  1  1  .  .  .  .  .  .  1  1  .  .  .  .  .  .  1  1  .  .  .  .  .  .
  1  .  1  .  .  .  .  .  1  .  1  .  .  .  .  .  1  .  1  .  .  .  .  .  1  .  1  .  .  .  .  .
  1  1  1  1  .  .  .  .  1  1  1  1  .  .  .  .  1  1  1  1  .  .  .  .  1  1  1  1  .  .  .  .
  1  .  .  .  1  .  .  .  1  .  .  .  1  .  .  .  1  .  .  .  1  .  .  .  1  .  .  .  1  .  .  .
  1  1  .  .  1  1  .  .  1  1  .  .  1  1  .  .  1  1  .  .  1  1  .  .  1  1  .  .  1  1  .  .
  1  .  1  .  1  .  1  .  1  .  1  .  1  .  1  .  1  .  1  .  1  .  1  .  1  .  1  .  1  .  1  .
  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1

The first column follows the sequence -1,-1,-1,-1,... ; th second -1,1,-1,1,... , the third -1,-1,+1,+1,... and so on and confirms the $2^k$ - blockwise length of periodicity which you have already found. Of course, this pattern is well known and has been explored in extreme broadth&width; I'll see whether I can find some well known way of proof online ...

Gottfried Helms
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    For a proof, consider the recursive definition of the binomial coefficients, $\binom{n+1}k=\binom nk+\binom n{k-1}$, modulo $2$. Suppose for induction that the $2^n\times2^n$ block of the matrix has only $1$'s in its bottom row (the row labelled $2^n-1$). Then the recursion shows that the next row (labelled $2^n$) has two $1$'s (at positions $0$ and $2^n$) and $0$'s everywhere else, and those two $1$'s generate two copies of the original $2^n\times2^n$ block. Then these form a $2^{n+1}\times2^{n+1}$ block with only $1$'s in its bottom row, completing the induction. – mr_e_man Jan 04 '23 at 20:43
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    Notice the Sierpiński triangle – lhf Jan 06 '23 at 13:48