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%I #35 Dec 19 2023 18:51:54
%S 1,1,0,1,2,0,2,1,0,1,6,0,6,3,2,3,2,1,0,1,24,0,24,12,8,18,8,10,3,6,3,2,
%T 1,0,1,120,0,120,60,40,90,64,50,39,42,23,28,13,10,8,6,3,2,1,0,1,720,0,
%U 720,360,240,540,384,420,234,372,198,208,168,124,98,75,60,35,34,13,16,8,6,3
%N Triangle read by rows: T(n,k) is the number of permutations of [n] for which the sum of the positions of the left-to-right maxima is k (1 <= k <= n(n+1)/2).
%C Row n contains n*(n+1)/2 = A000217(n) entries.
%C Sum of entries in row n = n! = A000142(n).
%H Alois P. Heinz, <a href="/A143946/b143946.txt">Rows n = 1..50, flattened</a>
%H I. Kortchemski, <a href="http://arxiv.org/abs/0804.0446">Asymptotic behavior of permutation records</a>, arXiv: 0804.0446 [math.CO], 2008-2009.
%F T(n,1) = T(n,3) = (n-1)! for n>=2.
%F Sum_{k=1..n*(n+1)/2} k * T(n,k) = n! * n = A001563(n).
%F Generating polynomial of row n is t(t^2+1)(t^3+2)...(t^n+n-1).
%F Sum_{k=1..n*(n+1)/2} (n*(n+1)/2-k) * T(n,k) = A001804(n). - _Alois P. Heinz_, Dec 19 2023
%e T(4,6)=3 because we have 1243, 1342 and 2341 with left-to-right maxima at positions 1,2,3.
%e Triangle starts:
%e 1;
%e 1, 0, 1;
%e 2, 0, 2, 1, 0, 1;
%e 6, 0, 6, 3, 2, 3, 2, 1, 0, 1;
%e 24, 0, 24, 12, 8, 18, 8, 10, 3, 6, 3, 2, 1, 0, 1;
%e ...
%p P:=proc(n) options operator, arrow: sort(expand(product(t^j+j-1,j=1..n))) end proc: for n to 7 do seq(coeff(P(n),t,i),i=1..(1/2)*n*(n+1)) end do; # yields sequence in triangular form
%p # second Maple program:
%p b:= proc(n) option remember; `if`(n=0, 1,
%p expand(b(n-1)*(x^n+n-1)))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n)):
%p seq(T(n), n=1..7); # _Alois P. Heinz_, Aug 05 2020
%t row[n_] := CoefficientList[Product[t^k + k - 1, {k, 1, n}], t] // Rest;
%t Array[row, 7] // Flatten (* _Jean-François Alcover_, Nov 28 2017 *)
%Y Cf. A000142, A000217, A001563, A001804, A143947.
%Y T(n,n) gives A368246.
%K nonn,tabf
%O 1,5
%A _Emeric Deutsch_, Sep 21 2008
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