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We can roll up the natural numbers in a rectangular spiral:

 17--16--15--14--13
  |               |
 18   5---4---3  12
  |   |       |   |
 19   6   1---2  11
  |   |           |
 20   7---8---9--10
  |
 21--22--23--24--25

But now that we have them on a rectangular grid we can unwind the spiral in a different order, e.g. going clockwise, starting north:

 17  16--15--14--13
  |   |           |
 18   5   4---3  12
  |   |   |   |   |
 19   6   1   2  11
  |   |       |   |
 20   7---8---9  10
  |               |
 21--22--23--24--25

The resulting sequence is clearly a permutation of the natural numbers:

1, 4, 3, 2, 9, 8, 7, 6, 5, 16, 15, 14, 13, 12, 11, 10, 25, 24, 23, 22, 21, 20, 19, 18, 17, ...

Your task is to compute this sequence. (OEIS A020703, but spoiler warning: it contains another interesting definition and several formulae that you might want to figure out yourself.)

Fun fact: all 8 possible unwinding orders have their own OEIS entry.

The Challenge

Given a positive integer n, return the nth element of the above sequence.

You may write a program or function, taking input via STDIN (or closest alternative), command-line argument or function argument and outputting the result via STDOUT (or closest alternative), function return value or function (out) parameter.

Standard rules apply.

Test Cases

1       1
2       4
3       3
4       2
5       9
6       8
7       7
8       6
9       5
100     82
111     111
633     669
1000    986
5000    4942
9802    10000
10000   9802

For a complete list up to and including n = 11131 see the b-file on OEIS.

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up vote 6 down vote accepted

Jelly, 11 10 bytes

’ƽð²+ḷ‘Ḥ_

Another Jelly answer on my phone.

’ƽð²+ḷ‘Ḥ_   A monadic hook:
’ƽ          Helper link. Input: n
’             n-1
 ƽ            Atop integer square root. Call this m.
   ð         Start a new dyadic link. Inputs: m, n
    ²+ḷ‘Ḥ_    Main link:
    ²+ḷ       Square m, add it to itself,
       ‘      and add one.
        Ḥ     Double the result
         _    and subtract n.

Try it here.

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Any tips on getting started with Jelly? I can't tell how the forks/hooks get parsed at all. – Lynn Feb 2 at 20:26
    
Learn APL or J first. Chains are actually easier than trains because the functions all have fixed arity. – lirtosiast Feb 2 at 20:30
    
I see. Yeah, I have J experience. I suppose I will try to read jelly.py and figure out which chains are supported. – Lynn Feb 2 at 20:30
2  
How the hell did you type that on your phone!? That's more impressive than the code itself is! – Dr Green Eggs and Iron Man Feb 3 at 3:55

Japt, 20 19 16 bytes

V=U¬c)²-V *2-U+2

Test it online!

Based on the observation that

F(N) = ceil(N^.5) * (ceil(N^.5)-1) - N + 2

Or, rather, that

F(N) = the first square greater than or equal to N, minus its square root, minus N, plus 2.

I don't know if this explanation is on the OEIS page, as I haven't looked at it yet.

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Julia, 28 bytes

n->2((m=isqrt(n-1))^2+m+1)-n

This is a lambda function that accepts an integer and returns an integer. To call it, assign it to a variable.

We define m to be the largest integer such that m2n-1, i.e. the integer square root of n-1 (isqrt). We can then simplify the OEIS expression 2 (m + 1) m - n + 2 down to simply 2 (m2 + m + 1) - n.

Try it online

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CJam, 14 bytes

qi_(mQ7Ybb2*\-

Using Alex's approach: 2*(m^2+m+1)-n where m = isqrt(n-1).

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ES7, 31 28 26 bytes

n=>(m=--n**.5|0)*++m*2-~-n

I had independently discovered Alex's formula but I can't prove it because I wasn't near a computer at the time.

Edit: Saved 3 bytes partly thanks to @ETHproductions. Saved a further 2 bytes.

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n=>((m=--n**.5|0)+m*m)*2-n+1 would work, I think. – ETHproductions Feb 3 at 0:28
    
@ETHproductions Thanks, I was wondering to myself how to get that --n in there... – Neil Feb 3 at 0:53
    
@ETHproductions Heh, I managed to shave 2 bytes from your answer. – Neil Feb 3 at 1:03

Pyth, 21 bytes

K2-h+^.E@QKK^t.E@QKKQ

Try it online!

Nothing fancy going on. Same method as in the JAPT answer.

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MATL, 13 16 bytes

qX^Y[tQ*Q2*G-

Based on Lynn's CJam answer.

Try it online!

q       % input n. Subtract 1
X^      % square root
Y[      % floor
tQ      % duplicate and add 1
*       % multiply
Q       % add 1
2*      % multiply by 2
G-      % subtract n

This uses a different approach than other answers (16 bytes):

6Y3iQG2\+YLt!G=)

It explicitly generates the two spiral matrices (actually, vertically flipped versions of them, but that doesn't affect the output). The first one is

17    16    15    14    13
18     5     4     3    12
19     6     1     2    11
20     7     8     9    10
21    22    23    24    25

and the second one traces the modified path:

25    10    11    12    13
24     9     2     3    14
23     8     1     4    15
22     7     6     5    16
21    20    19    18    17

To find the n-th number of the sequence it suffices to find n in the second matrix and pick the corresponding number in the first. The matrices need to be big enough so that n appears, and should have odd size so that the origin (number 1) is in the same position in both.

Try it online too!

6Y3      % 'spiral' string
i        % input n
QG2\+    % round up to an odd number large enough
YL       % generate spiral matrix of that size: first matrix
t!       % duplicate and transpose: second matrix
G=       % logical index that locates n in the second matrix
)        % use that index into first matrix
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Brachylog, 20 bytes

-1$r$[I*I+I+1=*2-?=.

This uses the same technique as pretty much all other answers.

Explanation

-1                   § Build the expression Input - 1
  $r                 § Square root of Input - 1
    $[I              § Unify I with the floor of this square root
       *I+I+1        § Build the expression I * I + I + 1
             =*2-?   § Evaluate the previous expression (say, M) and build the expression
                     § M * 2 - Input
                  =. § Unify the output with the evaluation of M * 2 - Input

A midly interesting fact about this answer is that it is easier and shorter to use = rather than parentheses.

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