The FindRepetitions(L) function inspects the elements in list L and returns all the elements which previously appeared in the list.

If the optional second argument is specified, FindRepetitions(L, N) removes N copies of each unique element in list L (if possible).

By default, operands are compared by using boolean comparison. If a third argument is specified, then the function f(x, y, opt1, opt2, ...) is called to check if x and y should be considered equal. This function should implement an equivalence relation (i.e. it should be reflexive, commutative, and transitive).  If it is not, then the result may not be valid.

$\mathrm{with}\left(\mathrm{ListTools}\right)\:$

$L≔\left[1\,2\,3\,4\,3\,2\,3\,4\,5\,4\,3\,4\,5\,6\right]$

$L≔\left[1\,2\,3\,4\,3\,2\,3\,4\,5\,4\,3\,4\,5\,6\right]$

$\mathrm{FindRepetitions}\left(L\right)$

$\left[3\,2\,3\,4\,4\,3\,4\,5\right]$

$\mathrm{FindRepetitions}\left(L\,0\right)$

$\left[1\,2\,3\,4\,3\,2\,3\,4\,5\,4\,3\,4\,5\,6\right]$

$\mathrm{FindRepetitions}\left(L\,2\right)$

$\left[3\,4\,3\,4\right]$

$\mathrm{FindRepetitions}\left(L\,'\mathrm{\infty }'\right)$

$L≔\left[0\,\left(x-1\right)\left(x+1\right)\,0.\,x^2-1\,-0.\,-\left(1-x\right)\left(x+1\right)\right]$

$L≔\left[0\,\left(x-1\right)\left(x+1\right)\,0.\,x^2-1\,\mathrm{-0.}\,-\left(1-x\right)\left(x+1\right)\right]$

$\mathrm{FindRepetitions}\left(L\right)$

$\mathrm{FindRepetitions}\left(L\,1\,\mathrm{verify}\,\mathrm{expand}\right)$

$\left[0.\,x^2-1\,\mathrm{-0.}\,-\left(1-x\right)\left(x+1\right)\right]$