How do I get 0-padded numbers in {} (brace expansion)?

Bash brace expansions could generate the numbers with leading zeros (since bash 4.0 alpha+ ~2009-02-20):

$ echo {001..023}
001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023

So, you can do:

for a in {001..218}; do  echo "bvrprdsve$a; $(ssh -q bvrprdsve$a "echo \$(free -m|grep Mem|/bin/awk '{print \$4}';free -m|grep Swap|/bin/awk '{print \$4}')")"; done >> /tmp/svemem.txt

But, let's look inside the command a little bit:

1. You are calling free twice, using grep and then awk:

   free -m|grep Mem |/bin/awk '{print \$4}';
   free -m|grep Swap|/bin/awk '{print \$4}'
   

   All could be reduced to this one call to free and awk:

   free -m|/bin/awk '/Mem|Swap/{print \$4}'
   

2. Furthermore, the internal command could be reduced to this value:

   cmd="echo \$(free -m|/bin/awk '/Mem|Swap/{print \$4}')"
   

Then, the whole script will look like this:

b=bvrprdsve;
f=/tmp/svemem.txt;
cmd="echo \$(free -m|/bin/awk '/Mem|Swap/{print \$4}')";
for a in {001..218}; do echo "$b$a; $(ssh -q "$b$a" "$cmd")"; done >> "$f";

printf '%04d' "$a" will output a zero-filled four digit string if $a is an integer.

Example:

a=14
printf -v b '%04d' "$a"

printf '%s\n' "$b"

will output

0014

So just loop as before and create your zero-filled variable and use that:

for a in {1..9}; do
  printf -v za '%04d' "$a"
  # rest of code, using $za
done

The assignment to za above could also be done with

printf -v za '%04d' "$a"   # print to variable za

To be more readable to the C programmer

for (( a = 1; a < 10; ++a )); do
  printf -v za '%04d' "$a"
  # rest of code, using $za
done

I haven't looked at the rest of the code.

Note: In ksh one may set an attribute on a variable so that its expansion is always zero-filled:

typeset -Z4 za
for (( a = 1; a < 10; ++a )); do
  za="$a"
  printf "%s\n" "$za"
done

0001
0002
0003
0004
0005
0006
0007
0008
0009

You could use seq -w 1 218 or seq -f "%03g" 1 218 to generate your numbers.