Python 2, 193 ... 179 178 bytes

d=10
u=d**2000
v=u*u
def s(n,a=d,i=9):
 while a-i:i,a=a,(a+n/a)/2
 return a
p,q,r,t=s(2*v),s(3*v),1,0
while p:t+=(p+q)%d*r;p/=d;q/=d;r*=d
for i in range(1000):print t/u;t=v/(t%u)

Try it online!

Calculating sqrt(2) and sqrt(3) to such a precision with a short code is a tough job in Python and other languages.

2000 digits is needed to ensure the expansion is correct, and lines 4-6 is the integer square root.

193 > 180 : The digit-wise modulo sum is now carried by a loop instead of array manipulation

180 > 179 : Replaced the 6 occurrences of 10 using d with the cost of defining with 5 bytes, cutting 1 byte in total

179 > 178 : Just realized that a!=i can be replaced by a-i

Python 2, 193 ... 179 178 bytes

d=10
u=d**2000
v=u*u
def s(n,a=d,i=9):
 while a-i:i,a=a,(a+n/a)/2
 return a
p,q,r,t=s(2*v),s(3*v),1,0
while p:t+=(p+q)%d*r;p/=d;q/=d;r*=d
for i in range(1000):print t/u;t=v/(t%u)

Try it online!

Calculating sqrt(2) and sqrt(3) to such a precision with a short code is a tough job in Python and other languages.

2000 digits is needed to ensure the expansion is correct (1020 suffices, but I'm not going to modify it because no improvement), and lines 4-6 is the integer square root.

193 > 180 : The digit-wise modulo sum is now carried by a loop instead of array manipulation

180 > 179 : Replaced the 6 occurrences of 10 using d with the cost of defining with 5 bytes, cutting 1 byte in total

179 > 178 : Just realized that a!=i can be replaced by a-i

Python 2, 193 180 179 bytes

d=10
u=d**2000
v=u*u
def s(n,a=d,i=9):
 while a!=i:i,a=a,(a+n/a)/2
 return a
p,q,r,t=s(2*v),s(3*v),1,0
while p:t+=(p+q)%d*r;p/=d;q/=d;r*=d
for i in range(1000):print t/u;t=v/(t%u)

Try it online!

Calculating sqrt(2) and sqrt(3) to such a precision with a short code is a tough job in Python and other languages.

2000 digits is needed to ensure the expansion is correct, and lines 4-6 is the integer square root.

193 > 180 : The digit-wise modulo sum is now carried by a loop instead of array manipulation

180 > 179 : Replaced the 6 occurrences of 10 using d with the cost of defining with 5 bytes, cutting 1 byte in total

Python 2, 193 ... 179 178 bytes

d=10
u=d**2000
v=u*u
def s(n,a=d,i=9):
 while a-i:i,a=a,(a+n/a)/2
 return a
p,q,r,t=s(2*v),s(3*v),1,0
while p:t+=(p+q)%d*r;p/=d;q/=d;r*=d
for i in range(1000):print t/u;t=v/(t%u)

Try it online!

Calculating sqrt(2) and sqrt(3) to such a precision with a short code is a tough job in Python and other languages.

2000 digits is needed to ensure the expansion is correct, and lines 4-6 is the integer square root.

193 > 180 : The digit-wise modulo sum is now carried by a loop instead of array manipulation

180 > 179 : Replaced the 6 occurrences of 10 using d with the cost of defining with 5 bytes, cutting 1 byte in total

179 > 178 : Just realized that a!=i can be replaced by a-i

Python 2, 193 180 bytes

u=10**2000
v=u*u
def s(n,a=10,i=9):
 while a!=i:i,a=a,(a+n/a)/2
 return a
p,q,r,t=s(2*v),s(3*v),1,0
while p:t+=(p+q)%10*r;p/=10;q/=10;r*=10
for i in range(1000):print t/u;t=v/(t%u)

Try it online!

Calculating sqrt(2) and sqrt(3) to such a precision with a short code 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.

Python 2, 193 180 179 bytes

d=10
u=d**2000
v=u*u
def s(n,a=d,i=9):
 while a!=i:i,a=a,(a+n/a)/2
 return a
p,q,r,t=s(2*v),s(3*v),1,0
while p:t+=(p+q)%d*r;p/=d;q/=d;r*=d
for i in range(1000):print t/u;t=v/(t%u)

Try it online!

Calculating sqrt(2) and sqrt(3) to such a precision with a short code is a tough job in Python and other languages.

2000 digits is needed to ensure the expansion is correct, and lines 4-6 is the integer square root.

193 > 180 : The digit-wise modulo sum is now carried by a loop instead of array manipulation

180 > 179 : Replaced the 6 occurrences of 10 using d with the cost of defining with 5 bytes, cutting 1 byte in total