Inspired by Tad

The pattern of the spiral (right 1, up 1, left 2, down 2, right 3, up 3, left 4, down 4, right 5, up 5, left 6, up 6, ...) is easy to solve. The $N$ th point has coordinates $P(N)$ , where $P\big(2n(2n+1)+j\big) \; = \; \left\{ \begin{array}{lll} (-n+j,-n) & \hspace{1cm}& 0 \le j \le 2n+1 \\ (n+1,j-3n-1) & & 2n+2 \le j \le 4n+2 \\ (5n+3-j,n+1) & & 4n+3 \le j \le 6n+4 \\ (-n-1,7n+5-j) & & 6n+5 \le j \le 8n+5 \end{array} \right.$ for all $n \ge 0$ . Thus, for $n \ge 0$ , $f\big(2n(2n+1) + j\big) \; = \; \left\{ \begin{array}{lll} j-2n & \hspace{1cm} & 0 \le j \le 4n+2 \\ 6n+4 - j & & 4n+3 \le j \le 8n+5 \end{array} \right.$ We calculate $\begin{array}{cccc}\hline N & n & j & f \\ \hline 6664001 & 1290 & 5021 & 2441 \\ 8875172 & 1489 & 3710 & 732 \\ 4820669 & 1097 & 4839 & 1747 \\ \hline \end{array}$ making the answer $\boxed{4920}$ .

Solution implemented in python 3

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"Inspired by Tad"

def add(a,b):
    return (a[0]+b[0],a[1]+b[1])

def solve():
    enws = [(1,0),(0,1),(-1,0),(0,-1)]
    target = [6664001,8875172,4820669]
    res = 0
    current = (0,0)
    n = max(target)
    pas = 1
    spas = 0
    sens = 0
    k = 1
    toggle = False
    while k <= n:
        if spas < pas:
            current = add(current,enws[sens])
            if k in target:
                res += sum(current)
            spas +=1
            k += 1
        else:
            spas=0
            if toggle:
                pas += 1
            toggle = not toggle
            sens =  (sens + 1)%4
    return res

print(solve())

Output

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4920

As a programming problem:

$\delta =\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ -1 & 0 \\ 0 & -1 \\ \end{array} \right)$

$\text{interesting}=\{6664001,8875172,4820669\}$

$\max =\max (\text{interesting})$

$\text{position}=\{0,0\}; \\ \text{direction}=0; \\ \text{distance}=1; \\ \text{step}=\text{distance}; \\ z=-1; \\ \text{first}=\text{True}; \\ \text{While}[z\leq \max , \\ \text{++}z; \\ \text{If}[\text{MemberQ}[\text{interesting},z],\text{Print}[\{z,\text{position},\text{Plus}\text{@@}\text{position}\}]]; \\ \text{position}=\delta [[\text{direction}+1]]+\text{position}; \\ \text{--}\text{step}; \\ \text{If}[\text{step}=0, \\ \text{direction}=((\text{direction}+1) \bmod 4); \\ \text{If}[\neg \text{first},\text{++}\text{distance}]; \\ \text{first}=\neg \text{first}; \\ \text{step}=\text{distance};]]$

We can find a pattern in the top-left corners of the diagram. Their numbers count $0, 4, 16, 36, 64, 100...$ , which are the squares of even numbers $0^2, 2^2, 4^2, 6^2, 10^2...$ . Now, if we find the biggest even square $S$ less than $Z$ , then the coordinates of the point $S$ will be $(-\frac{1}{2}\sqrt S, \frac{1}{2}\sqrt S)$ . Starting at this point we move around the rectangle for $(Z-S)$ moves and get to the point $Z$ .

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import math

# A supplementary function, goes around and around from any point and square size
def ineff_f(Z, size=0, px=0, py=0):
    # 'size' is a side of the square already filled
    # (px, py) are coordinates of the pointer
    # Move the pointer Z times
    for n in range(0, Z):
        # Move around the square
        if py == -size and px < size + 1:
            px += 1
        elif px == -size and py != size + 1:
            py -= 1
        elif px == size + 1 and py < size + 1:
            py += 1
        else:
            px -= 1
        if px == -size - 1 and py == size + 1:
            # If a new square is done, increase the size
            size += 1
    # Return pointer coordinates
    return px, py


# The function which substracts the square from Z and uses ineff_f() for any moves left
def eff_f(Z):
    # Find the biggest even square less than Z
    root = math.floor(math.sqrt(Z))
    root = int(root % 2 == 0 and root or root - 1)
    # Run ineff_f() with initial parameters
    return ineff_f(Z - root**2, root // 2, - root // 2, root // 2)


# Finding the answer to the question
result = sum(eff_f(6664001)) + sum(eff_f(8875172)) + sum(eff_f(4820669))
print(result)

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4920
[Finished in 0.2s]