"for" and "while" loop problem in "sh"

Hi,
I have a problem with "for" and "while" loop in "sh".
I have:

#!/bin/sh

  for i in $(seq 1 500000); do
    echo $i
  done

and it's working in sh on my ubuntu, but when I try to run this on unix(I have access to my university's unix) it crash:

syntax error at line 2: `$' unexpected

... second problem with while

#!/bin/sh

 i=1;
  while [ $i -le 1000000 ]; do 
  echo $i
  i=$(($i+1)); 
done

It's working on ubuntu but not on unix

syntax error at line 5: `i=$' unexpected

Could anyone help me ?
Thank you for any suggestion

You are probably running a non-standard shell at school. Schools excel at picking oddball teaching environments.

On the school box show the output of:

grep  yourusername /etc/passwd

OR it could be a bourne shell. Modern shells would reject those commands as you found out. You can try substituting
` ` (those are backticks ) for the $( ) construct.

grep yourusername /etc/passwd:

w29185:x:4496:203:MyName:/export/home/w29185:/bin/bash

I tried:

#!/bin/sh

  for i in `seq 1 500000`; do
    echo $i
  done

... and

seq: not found

In the first question you need to write the following script

 
 #!/bin/sh

  for i in `seq 1 10`;
  do
    echo $i
  done

In the second code is working correctly in my system.

seq is an external program which need to be installed on your system.

i=$(($i+1))

is a POSIX syntax, if your shell don't support it you can try:

i=`expr $i + 1`

or:

let i=i+1

i=`expr $i + 1`

ok thx is't working

And what aboute for in sh if my shell doesn't support seq and I need ex 1000 repetition

Something like this?

#!/bin/sh

i=1
while [ $i -le 1000 ]; do 
  echo $i
  let i=i+1
done

let not found ;/

Instead of let you can use this expr.

#!/bin/sh

i=1
while [ $i -le 1000 ]; 
do
      echo $i
      i=`expr $i + 1`
done