# An amazing pattern!

1. Take any number $$m$$ (let's say 3).

2. Write all the numbers starting from $$1$$ in one line.

3. Eliminate numbers at the index $$k$$ ($$k \equiv 0 \pmod{m}$$)

4. Now write for each index, partial sum of the numbers till that index.

5. Eliminate numbers at the index $$k$$ ($$k \equiv m-1 \pmod{m}$$).

6. Now write for each index, partial sum of the numbers till that index.

$$\vdots$$

$$\quad$$2m-1. Eliminate numbers at the index $$k$$ ($$k \equiv 2 \pmod{m}$$).

$$\quad$$2m. Now write for each index, partial sum of the numbers till that index.

For example, for $$m=3$$,

$\begin{array}{ c c c c c c c c c c c c c} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\ 1 & 2 & & 4 & 5 & & 7 & 8 & & 10 & 11 & & 13\\ 1 & 3 & & 7 & 12 & & 19 & 27 & & 37 & 48 & & 61\\ 1 & & & 7 & & & 19 & & & 37 & & & 61\\ 1 & & & 8 & & & 27 & & & 64 & & & 125\\ \end{array}$

$$\displaystyle \text{We get a sequence of} \ n^m \text{!}$$

Prove that this happens for any number of numbers and for any natural number $$m$$.

PS - This is not a challenge. This is seeking help.

Note by Kartik Sharma
10 months ago

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This is seriously magic. Another math-e-magical thing with which you can fascinate a primary school student and can "irritate" a mathematician.

(Another such thing is the Goldbach's Theorem).

- 10 months ago

(1) I have fixed your LaTeX. The correct code here is "\begin{array}".

(2) This result is known as Moessner's Theorem.

- 10 months ago

Oh I see. Thanks! I will look into the references. BTW, can you give a simplified proof?

- 10 months ago

I'm afraid I don't know the proof. I think it's a hard result.

- 10 months ago

Yeah. I am not able to find an understandable proof yet. And its generalizations to the way Conway has done in his book, is magical indeed.

- 10 months ago

- 10 months ago

@Mark Hennings @Pi Han Goh @Ishan Singh Please also fix the latex!

- 10 months ago