Return last number in sub-sequences in a list of integers

You can use Split:

Last /@ Split[z, #2 == # + 1&]

{113, 121, 476, 529, 538, 544}

First /@ Split[z, #2 == # + 1&]

{113, 117, 475, 529, 538, 542}

Use a "background" element of the desired difference in SparseArray on the differences, and "AdjacencyLists" will pick out what you desire:

Last in sequence:

positions = Append[SparseArray[Differences@z, Automatic, 1], 0]["AdjacencyLists"]
(*  {1, 6, 8, 9, 10, 13}  *)

z ~Part~ positions // Normal
(*  {113, 121, 476, 529, 538, 544}  *)

More efficient version:

Append[
 z ~Part~
   SparseArray[Rest@z - Most@z, Automatic, 1]["AdjacencyLists"],
 Last@z]

First in sequence:

Prepend[
 Reverse@z ~Part~
  Reverse[SparseArray[Rest@z - Most@z, Automatic, 1]]["AdjacencyLists"],
 First@z]
(*  {113, 117, 475, 529, 538, 542}  *)

My answer is similar to Mr.Wizard's answer to Find subsequences of consecutive integers inside a list, which asks for the essentially same thing as the OP but with the output in a different form.

Pick in combination with vectorized arithmetic is as fast as SparseArray:

myLastInSequenceV3[z_] := Join[Pick[Most[z], Unitize[Rest[z]-Most[z]-1], 1], {Last[z]}]

Alternative version. Shorter, but slower:

myLastInSequence[z_] := Pick[z, Join[Differences[z], {0}], Except[1, _Integer]]

Test:

LastinSequence[z] == myLastInSequence[z] == myLastInSequenceV3[z]

True

Timing for SparseArray-based solution:

data = NestList[(# + RandomInteger[9]) &, 1, 100000];

lastInSequenceSparseUpd[z_] := Append[z ~Part~SparseArray[Rest@z - Most@z, Automatic, 1]["AdjacencyLists"],Last@z]
First@RepeatedTiming[lastInSequenceSparseUpd[data];]

0.0018

This method:

First@RepeatedTiming[myLastInSequenceV3[data];]

0.0019