Awk: print an undefined number of columns

I have a file that contains several fields separated by a tab. I am trying to print all the columns except the first, but I want to print them in just one column with AWK. File format

col 1   col 2   ... col n

There are at least 2 columns in one row.

Example

2012029754      901749095
2012028240      901744459       258789
2012024782      901735922
2012026032      901738573       257784
2012027260      901742004
2003062290      901738925       257813  257822
2012026806      901741040
2012024252      901733947       257493
2012024365      901733700
2012030848      901751693       260720  260956  264843  264844

So, I want to say awk to print column 2 in column n for n greater than 2 without printing blank rows when there is no information in column n of this row, all in one column, as shown below.

901749095
901744459
258789
901735922
901738573
257784
901742004
901738925
257813
257822
901741040
901733947
257493
901733700
901751693
260720
260956
264843
264844

This is the first time I'm using awk, so bear with me. I wrote this from the command line, which works:

awk '{i=2; 
while ($i ~ /[0-9]+/)
{ 
    printf "%s\n", $i
    i++
}
}' bth.data

This is more of an endorsement for the search than the question of whether this is the right way to do something similar in AWK or is there a better / shorter way to do this.

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