# [Python 2], 193 bytes

<!-- language-all: lang-python -->

    u=10**2000
    v=u*u
    def s(n,a=10,i=9):
     while a!=i:i,a=a,(a+n/a)/2
     return[int(j)for j in `a`[:-1]]
    t=int("".join(`(a+b)%10`for a,b in zip(s(2*v),s(3*v))))
    for i in range(1000):print t/u;t=v/(t%u)

[Try it online!][TIO-je042j1z]

[Python 2]: https://docs.python.org/2/
[TIO-je042j1z]: https://tio.run/##Fc5BDoMgEAXQPaegJiaDpQXpqjacxJiAqdYxDRoEm/byFmYzi/f/ZNZvmBanjiPqWlaVklKSXccqkucw0g0ctwk46jtrCP1M@B6oPWlsMIHlYM9OWCYUoX4I0bsWXYCZjYunM0VHjTVtc6m7jgSdqSiu84IOTGr2rKylyVHL@xz@4QobqGpnfINbWmlIdszqrXsNUKcPWbP6dIwGER9B7wJCGdlx/AE "Python 2 – Try It Online"

# [Python 3], 199 bytes

<!-- language-all: lang-python -->

    u=10**2000
    v=u*u
    def s(n,a=10,i=9):
     while a!=i:i,a=a,(a+n//a)>>1
     return[int(j)for j in str(a)]
    t=int("".join(str((a+b)%10)for a,b in zip(s(2*v),s(3*v))))
    for i in range(1000):print(t//u);t=v//(t%u)

[Try it online!][TIO-je03s2w4]

[Python 3]: https://docs.python.org/3/
[TIO-je03s2w4]: https://tio.run/##Hc7BDoIwDAbg@55ikpC0c3FDTmLGixgPI4KUmEHGhtGXx81emvT723T5hHF29b5HU2khzlprtpkoInv0A1/BSZtAkrlgw/h7pFfP7cFQQwmsBHt0Slls24px34fo3Y1cgAmH2fOJk@Nr8GDxzoLJUBSnaSYHeZqWOywr/c9a2eX0lxZY4Sw2lCvUqaVi2Smrt@7ZQ5WexGbx@V5QKuI1mE0pCGXEff8B "Python 3 – Try It Online"

Calculating sqrt(2) and sqrt(3) to such a precision is a tough job in Python and other languages.
2000 digits is needed to ensure the expansion is correct, and the middle 3 lines is the integer square root.