Progressing Two's

Python 2, 52 bytes

n=p=0
exec"n+=1;r=n-n**.5//1;print p+r;p=r;"*input()

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54 bytes

lambda N:[n-~n-n**.5//1-(n+1)**.5//1for n in range(N)]

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It's a formula!

$$f(n) = 2n+1 - \lfloor \sqrt n\rfloor - \lfloor \sqrt {n+1} \rfloor$$

This can also be split up as

$$f(n) = \sum_{k \in \{n,n+1\}}\left({k-\lfloor \sqrt k\rfloor}\right)$$

Note that \$k-\lfloor \sqrt k\rfloor\$ is the number of non-squares from \$1\$ to \$k\$ inclusive.

APL (Dyalog Extended), ¹⁴ 12 bytes

0,2+/⍳-⌊∘√∘⍳

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Uses xnor's formula of

$$ f(n) = \sum_{k \in \{n,n+1\}}\left({k-\lfloor \sqrt k\rfloor}\right) $$

How it works

0,2+/⍳-⌊∘√∘⍳
     ⍳-       ⍝ 1..n minus...
       ⌊∘√∘⍳  ⍝ floor(sqrt(1..n))
  2+/         ⍝ Add two consecutive pairs
              ⍝ giving first n items of the sequence except leading 0
0,            ⍝ Prepend the leading 0

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APL (Dyalog Extended), 14 bytes

⊢↑2(∧+/,2××/)⍳

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Based on the observation that the sequence is the union of all odd numbers and the numbers in the form of \$2n(n+1), n \ge 0\$. Uses ⎕IO←0.

How it works

⊢↑2(∧+/,2××/)⍳  ⍝ Input: positive integer n
             ⍳  ⍝ Generate 0..n-1
  2(      ×/)   ⍝ Pairwise product (0×1, 1×2, ..., (n-2)×(n-1))
        2×      ⍝ Double it
     +/,        ⍝ Concat with pairwise sum (0+1, 1+2, ..., (n-2)+(n-1))
    ∧           ⍝ Ascending sort the 2(n-1) numbers in total
⊢↑              ⍝ Take the first n numbers
                ⍝ For n=1, "overtake" from zero elements, giving single 0

Haskell, 42 bytes

(`take`q 4)
q k=0:[1,3..k]++map(k+)(q$k+4)

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Uses a version of Bubbler's observation that the sequence alternates runs of consecutive odd numbers with an even number directly in between.

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Haskell, 43 bytes

(`take`scanl(+)0(q[2]))
q r=1:r++1:q(2:2:r)

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Generates an infinite list of 1's and 2's, take the cumulative sums, and truncates to the input length.