TCL 8.5 (71 70 68) (67)

TCL has no real chance of ever winning, but it is a fun way to oil the machine:

proc c x {while \$x>1 {set x [expr $x%2?3*$x+1:$x/2];incr k};set k}

formatted for readability:

proc c x {
    while {$x>1} {
    set x [expr $x%2 ? 3*$x+1 : $x/2]
    incr k
    }
    set k
}

Edits: many suggestions (inspired) by sergiol. I guess the answer is more his than mine, by now :-)

TCL 8.5 (71 70 68) (67)

TCL has no real chance of ever winning, but it is a fun way to oil the machine:

proc c x {while \$x>1 {set x [expr $x%2?3*$x+1:$x/2];incr k};set k}

formatted for readability:

proc c x {
    while {$x>1} {
    set x [expr $x%2 ? 3*$x+1 : $x/2]
    incr k
    }
    set k
}

Edits: many suggestions (inspired) by sergiol. I guess the answer is more theirs than mine, by now :-)

TCL 8.5 (71 70 68) (67)

TCL has no real chance of ever winning, but it is a fun way to oil the machine:

proc c x {while \$x>1 {set x [expr $x%2?3*$x+1:$x/2];incr k};set k}

formatted for readability:

proc c x {
    while {$x>1} {
    set x [expr $x%2 ? 3*$x+1 : $x/2]
    incr k
    }
    set k
}

Edits: many suggestions (inspired) by sergiol. I guess this more their answer than mine, by now :-)

TCL 8.5 (71 70 68) (67)

TCL has no real chance of ever winning, but it is a fun way to oil the machine:

proc c x {while \$x>1 {set x [expr $x%2?3*$x+1:$x/2];incr k};set k}

formatted for readability:

proc c x {
    while {$x>1} {
    set x [expr $x%2 ? 3*$x+1 : $x/2]
    incr k
    }
    set k
}

Edits: many suggestions (inspired) by sergiol. I guess the answer is more his than mine, by now :-)

TCL 8.5 (71) (70)

TCL has no real chance of ever winning, but it is a fun way to oil the machine:

set x $argv;while \$x>1 {set x [expr $x%2?3*$x+1:$x/2];incr k};puts $k

formatted for readability:

set x $argv
while {$x>1} {
   set x [expr $x%2 ? 3*$x+1 : $x/2]
   incr k
}
puts $k

TCL 8.5 (71 70 68) (67)

TCL has no real chance of ever winning, but it is a fun way to oil the machine:

proc c x {while \$x>1 {set x [expr $x%2?3*$x+1:$x/2];incr k};set k}

formatted for readability:

proc c x {
    while {$x>1} {
    set x [expr $x%2 ? 3*$x+1 : $x/2]
    incr k
    }
    set k
}

Edits: many suggestions (inspired) by sergiol. I guess this more their answer than mine, by now :-)