As claimed above by other users, the arrangement is imposible with four sets of four cards. The earliest solution occurs for four sets of cards 1 to 24, for which there are three solutions. These are known as Langford Quads and are described at John Miller's excellent blog on Langford's Problem.
These are Richard Noble's three solutions.
A similar answer to Quintec's (asserting that it is impossible):
Consider that you only have 12 cards that are not a 4, and that you need 12 total cards between the 4s. So the entire arrangement must look like
4 _ _ _ _ 4 _ _ _ _ 4 _ _ _ _ 4
As per Quintec's observation, somewhere in there we must have
_ A _ A _ A _ A _
But you can see that there ...
I claim that no arrangement is possible in the case with cards ace through four.
_A_A_A_A_ (_ is any other card)
must be part of the arrangement. Now consider two cases:
the row starts and ends with aces - this is obviously not
possible, since that arrangement would only hold 7 cards, and we
there are cards to the left and right ...
119? With regard to the main increase amount (ie 7, 12, 16), change of +5 then 0 repeated x 1, change of +4 then 0 repeated x 3, change of 3 then 0 repeated x 5. With the zero repetition being increasing odd numbers and the main increasing amount reducing by 1 each round.