While idly thinking while heading back from the office, and then more later while thinking after dinner with my academic little brother Alex Walker and my future academic little sister-in-law Sara Schulz, we began to think about $2017$, the number.

## General Patterns

- 2017 is a prime number. 2017 is the 306th prime. The 2017th prime is 17539.
- As 2011 is also prime, we call 2017 a sexy prime.
- 2017 can be written as a sum of two squares, $$ 2017 = 9^2 +44^2,$$ and this is the only way to write it as a sum of two squares.
- Similarly, 2017 appears as the hypotenuse of a primitive Pythagorean triangle, $$ 2017^2 = 792^2 + 1855^2,$$ and this is the only such right triangle.
- 2017 is uniquely identified as the first odd prime that leaves a remainder of $2$ when divided by $5$, $13$, and $31$. That is, $$ 2017 \equiv 2 \pmod {5, 13, 31}.$$
- In different bases, $$ \begin{align} (2017)_{10} &= (2681)_9 = (3741)_8 = (5611)_7 = (13201)_6 \notag \\ &= (31032)_5 = (133201)_4 = (2202201)_3 = (11111100001)_2 \notag \end{align}$$ The base $2$ and base $3$ expressions are sort of nice, including repetition.

## Counting to 20

$$\begin{array}{ll} 1 = 2\cdot 0 + 1^7 & 11 = 2 + 0! + 1 + 7 \\ 2 = 2 + 0 \cdot 1 \cdot 7 & 12 = 20 - 1 - 7 = -2 + (0! + 1)\cdot 7 \\ 3 = (20 + 1)/7 = 20 - 17 & 13 = 20 - 1 \cdot 7 \\ 4 = -2 + 0 - 1 + 7 & 14 = 20 - (-1 + 7) \\ 5 = -2 + 0\cdot 1 + 7 & 15 = -2 + 0 + 17 \\ 6 = -2 + 0 + 1 + 7 & 16 = -(2^0) + 17 \\ 7 = 2^0 - 1 + 7 & 17 = 2\cdot 0 + 17 \\ 8 = 2 + 0 - 1 + 7 & 18 = 2^0 + 17 \\ 9 = 2 + 0\cdot 1 + 7 & 19 = 2\cdot 0! + 17 \\ 10 = 2 + 0 + 1 + 7 & 20 = 2 + 0! + 17. \end{array}$$

In each expression, the digits $2, 0, 1, 7$ appear, in order, with basic mathematical symbols. I wonder what the first number is that can't be nicely expressed (subjectively, of course)?

## Iterative Maps on 2017

Now let's look at less-common manipulations with numbers.

- The digit sum of $2017$ is $10$, which has digit sum $1$.
- Take $2017$ and its reverse, $7102$. The difference between these two numbers is $5085$. Repeating gives $720$. Continuing, we get $$ 2017 \mapsto 5085 \mapsto 720 \mapsto 693 \mapsto 297 \mapsto 495 \mapsto 99 \mapsto 0.$$ So it takes seven iterations to hit $0$, where the iteration stabilizes.
- Take $2017$ and its reverse, $7102$. Add them. We get $9119$, a palindromic number. Continuing, we get $$ \begin{align} 2017 &\mapsto 9119 \mapsto 18238 \mapsto 101519 \notag \\ &\mapsto 1016620 \mapsto 1282721 \mapsto 2555542 \mapsto 5011094 \mapsto 9912199. \notag \end{align}$$ It takes one map to get to the first palindrome, and then seven more maps to get to the next palindrome. Another five maps would yield the next palindrome.
- Rearrange the digits of $2017$ into decreasing order, $7210$, and subtract the digits in increasing order, $0127$. This gives $7083$. Repeating once gives $8352$. Repeating again gives $6174$, at which point the iteration stabilizes. This is called Kaprekar's Constant.
- Consider Collatz: If $n$ is even, replace $n$ by $n/2$. Otherwise, replace $n$ by $3\cdot n + 1$. On $2017$, this gives $$\begin{align} 2017 &\mapsto 6052 \mapsto 3026 \mapsto 1513 \mapsto 4540 \mapsto \notag \\ &\mapsto 2270 \mapsto 1135 \mapsto 3406 \mapsto 1703 \mapsto 5110 \mapsto \notag \\ &\mapsto 2555 \mapsto 7666 \mapsto 3833 \mapsto 11500 \mapsto 5750 \mapsto \notag \\ &\mapsto 2875 \mapsto 8626 \mapsto 4313 \mapsto 12940 \mapsto 6470 \mapsto \notag \\ &\mapsto 3235 \mapsto 9706 \mapsto 4853 \mapsto 14560 \mapsto 7280 \mapsto \notag \\ &\mapsto 3640 \mapsto 1820 \mapsto 910 \mapsto 455 \mapsto 1366 \mapsto \notag \\ &\mapsto 683 \mapsto 2050 \mapsto 1025 \mapsto 3076 \mapsto 1538 \mapsto \notag \\ &\mapsto 769 \mapsto 2308 \mapsto 1154 \mapsto 577 \mapsto 1732 \mapsto \notag \\ &\mapsto 866 \mapsto 433 \mapsto 1300 \mapsto 650 \mapsto 325 \mapsto \notag \\ &\mapsto 976 \mapsto 488 \mapsto 244 \mapsto 122 \mapsto 61 \mapsto \notag \\ &\mapsto 184 \mapsto 92 \mapsto 46 \mapsto 23 \mapsto 70 \mapsto \notag \\ &\mapsto 35 \mapsto 106 \mapsto 53 \mapsto 160 \mapsto 80 \mapsto \notag \\ &\mapsto 40 \mapsto 20 \mapsto 10 \mapsto 5 \mapsto 16 \mapsto \notag \\ &\mapsto 8 \mapsto 4 \mapsto 2 \mapsto 1 \notag \end{align}$$ It takes $69$ steps to reach the seemingly inevitable $1$. This is much shorter than the $113$ steps necessary for $2016$ or the $113$ (yes, same number) steps necessary for $2018$.
- Consider the digits $2,1,7$ (in that order). To generate the next number, take the units digit of the product of the previous $3$. This yields $$2,1,7,4,8,4,8,6,2,6,2,4,8,4,\ldots$$ This immediately jumps into a periodic pattern of length $8$, but $217$ is not part of the period. So this is preperiodic.
- Consider the digits $2,0,1,7$. To generate the next number, take the units digit of the sum of the previous $4$. This yields $$ 2,0,1,7,0,8,6,1,5,0,2,8,\ldots, 2,0,1,7.$$ After 1560 steps, this produces $2,0,1,7$ again, yielding a cycle. Interestingly, the loop starting with $2018$ and $2019$ also repeat after $1560$ steps.
- Take the digits $2,0,1,7$, square them, and add the result. This gives $2^2 + 0^2 + 1^2 + 7^2 = 54$. Repeating, this gives $$ \begin{align} 2017 &\mapsto 54 \mapsto 41 \mapsto 17 \mapsto 50 \mapsto 25 \mapsto 29 \notag \\ &\mapsto 85 \mapsto 89 \mapsto 145 \mapsto 42 \mapsto 20 \mapsto 4 \notag \\ &\mapsto 16 \mapsto 37 \mapsto 58 \mapsto 89\notag\end{align}$$ and then it reaches a cycle.
- Take the digits $2,0,1,7$, cube them, and add the result. This gives $352$. Repeating, we get $160$, and then $217$, and then $352$. This is a very tight loop.

## A Few Matrices

- One can make $2017$ from determinants of basic matrices in a few ways. For instance, $$ \begin{align} \left \lvert \begin{pmatrix} 1&2&3 \\ 4&6&7 \\ 5&8&9 \end{pmatrix}\right \rvert &= 2, \qquad \left \lvert \begin{pmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{pmatrix}\right \rvert &= 0\notag \\ \left \lvert \begin{pmatrix} 1&2&3 \\ 4&7&6 \\ 5&9&8 \end{pmatrix}\right \rvert &= 1 , \qquad \left \lvert \begin{pmatrix} 1&2&3 \\ 4&5&7 \\ 6&8&9 \end{pmatrix}\right \rvert &= 7\notag \end{align}$$ The matrix with determinant $0$ has the numbers $1$ through $9$ in the most obvious configuration. The other matrices are very close in configuration.
- Alternately, $$ \begin{align} \left \lvert \begin{pmatrix} 1&2&3 \\ 5&6&9 \\ 4&8&7 \end{pmatrix}\right \rvert &= 20 \notag \\ \left \lvert \begin{pmatrix} 1&2&3 \\ 6&8&9 \\ 5&7&4 \end{pmatrix}\right \rvert &= 17 \notag \end{align}$$ So one can form $20$ and $27$ separately from determinants.
- One cannot make $2017$ from a determinant using the digits $1$ through $9$ (without repetition).
- If one uses the digits from the first $9$ primes, it is interesting that one can choose configurations with determinants equal to $2016$ or $2018$, but there is no such configuration with determinant equal to $2017$.

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Comments (1)2017-01-22 enginistSounds like a cricket match between Hardy and Ramanujan.