/* 

  Sum of a range ... (Code golf) in Picat.
  
  https://codegolf.stackexchange.com/questions/262647/sum-of-a-range-of-a-sum-of-a-range-of-a-sum-of-a-range-of-a-sum-of-a-range-of-a?page=2&tab=scoredesc#tab-top
  """
  Sum of a range of a sum of a range of a sum of a range of a sum of a range of a sum of

  Inspired by the fact that a few related challenges to this could be answered by Vyxal in 
  0 Bytes using a special flag combination.

  Given only one input integer n, calculate f(n,n)
  where
     f(x,y)={xf((∑xk=1k), y−1)if y=0otherwise
  If you want an explanation in plain English, here it is, quoted from OEIS:

  Let T(n) be the n-th triangular number n∗(n+1)/2; 
  then a(n)  = n-th iteration [of] T(T(T(...(n))))
    .

  Note that a(n) is the function.

  This is also A099129(n)
  , but with the case for n=0
  . This is code-golf, so as long as you make your answer short, it doesn't matter whether 
    it times out on TIO (my computer can't calculate n=6
    within five minutes!). Yes, standard loopholes apply.

  Test cases:
  0 -> 0
  1 -> 1
  2 -> 6
  3 -> 231
  4 -> 1186570
  5 -> 347357071281165
  6 -> 2076895351339769460477611370186681
  7 -> 143892868802856286225154411591351342616163027795335641150249224655238508171
  """

  This program was created by Hakan Kjellerstrand, hakank@gmail.com
  See also my Picat page: http://www.hakank.org/picat/

*/

main => go.

go ?=>
  garbage_collect(200_000_000),
  foreach(N in 1..10)
    println(N=f(N,N))
  end,
  
  nl.
go => true.


f(N) = f(N,N).

% a little slower with table. It's just Y iterations since Y is decremented each time.
% table
% f(X,0) = X.
% f(X,Y) = f([K : K in 1..X].sum,Y-1). % very slow
% f(X,Y) = f(sum(1..X),Y-1). % still very slow
% f(X,Y) = f(X*(X+1)//2,Y-1). % much faster

% golfed: 35 chars
f(X,0)=X. f(X,Y)=f(X*(X+1)//2,Y-1).