Prove that the given sequence contains all natural numbers

This is not a complete solution.

For $a_1 = 1$.

Hint: List out the first 20 terms.

Find a pattern. The pattern doesn't start immediately, which is why I asked for 20 terms.
That should be enough for you to guess what $a_i$ is, then prove it.

Here's my guess of how the general case will proceed:

• Show that for any $a_1$, there exists an $N=4k+1$ such that the first $N$ terms in the sequence are the first $N$ integers.
• Then, the result follows from $a_1 = 1$.

NOT AN ANSWER

Here for everyone to see is the first 100 elements of this sequence, with $a_1=1$ (generated with Python)

[  1.   2.   4.   3.   7.   5.   9.   6.   8.  11.  13.  10.  12.  15.
  17.  14.  16.  19.  21.  18.  20.  23.  25.  22.  24.  27.  29.  26.
  28.  31.  33.  30.  32.  35.  37.  34.  36.  39.  41.  38.  40.  43.
  45.  42.  44.  47.  49.  46.  48.  51.  53.  50.  52.  55.  57.  54.
  56.  59.  61.  58.  60.  63.  65.  62.  64.  67.  69.  66.  68.  71.
  73.  70.  72.  75.  77.  74.  76.  79.  81.  78.  80.  83.  85.  82.
  84.  87.  89.  86.  88.  91.  93.  90.  92.  95.  97.  94.  96.  99.
 101.  98.]

If you want to generate lists of your own, here is the code to do so:

import numpy as np
N=100
a=np.zeros(N)
a[0]=3
for n in range(1,N):
    S=np.sum(a[:n])
    k_found=False
    k=2
    while k_found==False:
        in_a=True
        while in_a==True:
            if k in a:
                k+=1
            else:
                in_a=False
        if np.gcd(int(S),int(k))==1:
            a[n]=k
            k_found=True
        else:
            k+=1
print(a)

Here a[0] is the starting element (I have used $0$ because of the way Python indexes lists) and N is the number of terms you wish to compute.