Spiral table 2021

I don't think this can be solved by hand alone. This reminds me of Ulam's spiral .

Looking at the numbers formed from the right diagonal $1,3,13,31, \ldots$ tells us that it follows the sequence $\{a_n \}_{n\geqslant1} = 4n^2 - 10n + 7$ .

Looking at the numbers formed from the left diagonal $1,5,17,37,65, \ldots$ tells us that it follows the sequence $\{b_n \}_{n\geqslant1} = 4(n-1)^2 + 1$ .

I'll leave the proof of these sequences as an exercise for the readers.

Thus, the $n^\text{th}$ row consists of $N=2n-1$ integers, $4(n-1)^2 + 1, \quad 4(n-1)^2 , \quad4(n-1)^2 - 1, \quad\ldots \quad , \quad4n^2 - 10 n+ 7$

which follows an arithmetic progression , and the sum of each row is $\dfrac N2 \Big[ 2a + (N-1)d \Big] = \dfrac{2n-1}2 \Bigg [ 2\Big[ 4(n-1)^2 + 1 \Big] + (2n-2)(-1) \Bigg ] = (2n-1)(4n^2-9n + 6)$

Hence, $S(n) = \sum_{k=1}^n (2k-1)(4k^2-9k + 6) = \dfrac n6(12n^3 - 20n^2 + 9n + 5)$

by invoking sum of powers . We want to find $\min(n)$ such that $S(n)$ is divisible by 2021.

Running a simple script tells us that the answer is $\boxed{774}$ . I'm interested to see a non-computer assisted solution.

1
2
3
4
5
6
7
8
9

n = 1
while True:
    if n * (12 * (n**3) - 20 * (n**2) + 9 * n + 5)/6 % 2021 == 0:
        print(n)
        break
    else:
        n += 1

# Output: 774