First spiral, then diagonal

Question

Given a positive input number n, construct a spiral of numbers from 1 to n^2, with 1 in the top-left, spiraling inward clockwise. Take the sum of the diagonals (if n is odd, the middle number n^2 is counted twice) and output that number.

Example for n = 1:

1

(1) + (1) = 2

Example for n = 2:

1 2
4 3

(1+3) + (4+2) = 4 + 6 = 10

Example for n = 4:

 1  2  3 4
12 13 14 5
11 16 15 6
10  9  8 7

(1+13+15+7) + (10+16+14+4) = 36 + 44 = 80

Example of n = 5:

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

(1+17+25+21+9) + (13+23+25+19+5) = 73 + 85 = 158

Further rules and clarifications

• This is OEIS A059924 and there are some closed-form solutions on that page.
• The input and output can be assumed to fit in your language's native integer type.
• The input and output can be given in any convenient format.
• You can choose to either 0-index or 1-index, as I am here in my examples, for your submission. Please state which you're doing.
• Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
• If possible, please include a link to an online testing environment so other people can try out your code!
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Answer 1

R, 43 34 bytes

function(n)(8*n^3-3*n^2+4*n+3)%/%6

Try it online!

The OEIS page lists the following formula for a(n):

(16*n^3 - 6*n^2 + 8*n + 3 - 3*(-1)^n)/12

However, I skipped right over that to get to the PROG section where the following PARI code is found:

floor((16*n^3 - 6*n^2 + 8*n + 3 - 3*(-1^n))/12))

Naturally, +3-3*(-1^n) is the same as +6 so we can simplify the linked formula, first by reducing it to

(16*n^3-6*n^2+8*n+6)/12 -> (8*n^3-3*n^2+4*n+3)/6

and using %/%, integer division, rather than / to eliminate the need for floor.

Answer 2

Python 2, 30 bytes

Saved some bytes by porting Giuseppe's approach.

lambda n:((8*n-3)*n*n+4*n+3)/6

Try it online!

Python 2,  36  34 bytes

Saved some more bytes thanks to @LeakyNun.

lambda n:((8*n-3)*n*n+4*n+n%2*3)/6

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Answer 3

Mathematica, 19 bytes

((8#-3)#*#+4#+3)/6&

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Saved some bytes by porting Giuseppe's approach.

Mathematica, 58 bytes

I always enjoy questions with given answers thanx to oeis (for the nice question and answer)

LinearRecurrence[{3,-2,-2,3,-1},{0,2,10,34,80},2#][[#+1]]&

Answer 4

Mathematica, 19 bytes

((8#-3)#*#+4#+3)/6&

You can run it with the following syntax:

((8#-3)#*#+4#+3)/6&[5]

Where 5 can be replaced with the input.

You can Try it in the Wolfram Sandbox (Copy-Paste + Evaluate Cells)

Answer 5

Jelly, 11 10 bytes

1 byte thanks to Jonathan Allan.

⁽ø\DN2¦ḅ:6

Try it online!

Answer 6

SOGL V0.12, 25 15 14 bytes

8*3-**4.*+3+6÷

Try it Here!

Translation of Mr.Xcoder's Python answer which is using Giuseppe's approach. SOGL's not winning anything here :p

Answer 7

MATL, 21 bytes

tt2^Qw6Y3YL-wXytP+*ss

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Answer 8

Cubix, 33 bytes

I:u8**.../\*3t3n*3t+u@O,6+3+t3*4p

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cube version:

      I : u
      8 * *
      . . .
/ \ * 3 t 3 n * 3 t + u
@ O , 6 + 3 + t 3 * 4 p
. . . . . . . . . . . .
      . . .
      . . .
      . . .

Implements the same algorithm as my R answer. I suspect this can be golfed down.

Answer 9

05AB1E,  13  12 bytes

Uses the same base-conversion technique as Leaky Nun's Jelly submission

^{Maybe there is a shorter way to create the list of coefficients?} 

-1 byte thanks to Datboi (use spaces ans wrap to beat compression(!))

8 3(4 3)¹β6÷

Try it online!

How?

8 3(4 3)¹β6÷               stack: []
8            - literal            ['8']
  3          - literal            ['8','3']
   (         - negate             ['8',-3]
    4        - literal            ['8',-3,'4']
      3      - literal            ['8',-3,'4','3']
       )     - wrap               [['8',-3,'4','3']]
        ¹    - 1st input (e.g. 4) [['8',-3,'4','3'], 4]
         β   - base conversion    [483]
          6  - literal six        [483,6]
           ÷ - integer division   [80]
             - print TOS           80

---

My 13s...

•VŠ•S3(1ǝ¹β6÷

•2ùë•₂в3-¹β6÷

•мå•12в3-¹β6÷

All using compressions to find the list of coefficients.

Answer 10

Java 8, 24 bytes

n->((8*n-3)*n*n+4*n+3)/6

Port of @Giuseppe's R answer.
Note that /6 floors by default when calculating with integers in Java.

Try it here.

Answer 11

Excel, 35 30 bytes

Saved 5 bytes using Giuseppe's approach.

=INT((8*A1^3-3*A1^2+4*A1+3)/6)

---

First attempt:

=(8*A1^3-3*A1^2+4*A1+3*MOD(A1,2))/6

Evolved from a direct implementation of formula from OEIS (37 bytes):

=(16*A1^3-6*A1^2+8*A1+3-3*(-1)^A1)/12

+3-3*(-1)^A1 logic can be changed to 6*MOD(A1,2).

=(16*A1^3-6*A1^2+8*A1+6*MOD(A1,2))/12

Does not save bytes, but allows removal of a common factor for 2 bytes.

Answer 12

Pyth, 17 bytes

/+3+*^Q2-*8Q3*4Q6

Try it here.

Answer 13

Pyke, 14 bytes

8*3-**Q4*+3+6f

Try it here!

Answer 14

05AB1E, 14 bytes

8*3-**4I*+3+6÷

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Answer 15

MATL, 12 bytes

[DcKI]6/iZQk

Try it at MATL Online!

Explanation

Same approach as Giuseppe's answer.

[DcKI]   % Push array [8, -3, 4, 3]
6/       % Divide each entry by 6
i        % Push input
ZQ       % Evaluate polynomial
k        % Round down. Implicitly display

Answer 16

Gaia, 14 bytes

8×3⁻××@4×+3+6/

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