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%I #22 Sep 26 2019 03:43:02
%S 1,2,2,1,3,0,2,2,2,0,2,1,4,0,2,1,2,0,2,2,2,2,2,0,4,0,0,2,1,4,1,2,0,6,
%T 0,2,2,1,0,2,0,2,3,2,0,6,0,2,4,4,0,2,0,2,2,0,0,2,1,4,0,6,0,2,4,5,0,6,
%U 0,4,2,0,0,4,1,0,0,2,0,2,2,4,0,2,6,5,0,6,0,8
%N Triangle read by rows: T(n,k) is the number of partitions of weight k among the complements of the partitions of n.
%C The complement of a partition p[1] >= p[2] >=...>= p[k] is p[1]-p[2], p[1]-p[3], ..., p[1]-p[k]. Its Ferrers board emerges naturally from the Ferrers board of the given partition. The weight of a partition of n is n.
%C Sum of entries in row n is A000041(n) (the partition numbers).
%C Apparently, number of entries in row n is A033638(n-1) = 1 + floor((n-1)^2/4).
%C T(n,0) = A000005(n) = number of divisors of n.
%C T(n,1) = A070824(n+1).
%C Sum(k*T(n,k),k>0) = A188814(n).
%H Alois P. Heinz, <a href="/A268192/b268192.txt">Rows n = 1..70, flattened</a>
%F The weight of the complement of a partition p is (number of parts of p)*(largest part of p) - weight of p.
%F For a given q, the Maple program yields the generating polynomial of row q.
%e Row 4 is 3,0,2; indeed, the complements of [4], [3,1], [2,2], [2,1,1], [1,1,1,1] are: empty, [2], empty, [1,1], empty; their weights are 0, 2, 0, 2, 0, respectively.
%e From _Gus Wiseman_, Sep 24 2019: (Start)
%e Triangle begins:
%e 1
%e 2
%e 2 1
%e 3 0 2
%e 2 2 0 2 1
%e 4 0 2 1 2 0 2
%e 2 2 2 2 0 4 0 0 2 1
%e 4 1 2 0 6 0 2 2 1 0 2 0 2
%e 3 2 0 6 0 2 4 4 0 2 0 2 2 0 0 2 1
%e 4 0 6 0 2 4 5 0 6 0 4 2 0 0 4 1 0 0 2 0 2
%e 2 4 0 2 6 5 0 6 0 8 4 0 0 6 2 0 2 2 0 2 0 2 0 0 2 1
%e Row n = 8 counts the following partitions:
%e 8 332 53 62 71 521 4211 611 5111
%e 44 22211 422 2111111 32111 311111 41111
%e 2222 431
%e 11111111 3221
%e 3311
%e 221111
%e (End)
%p q := 10: with(combinat): a := proc (i, j) options operator, arrow: partition(i)[j] end proc: P[q] := 0: for j to numbpart(q) do P[q] := sort(P[q]+t^(nops(a(q, j))*max(a(q, j))-q)) end do: P[q] := P[q];
%p # second Maple program:
%p b:= proc(n, i, l) option remember; expand(`if`(n=0 or i=1,
%p x^(`if`(l=0, 0, n*(l-i))), b(n, i-1, l)+`if`(i>n, 0,
%p x^(`if`(l=0, 0, l-i))*b(n-i, i, `if`(l=0, i, l)))))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2, 0)):
%p seq(T(n), n=1..15); # _Alois P. Heinz_, Feb 12 2016
%t b[n_, i_, l_] := b[n, i, l] = Expand[If[n == 0 || i == 1, x^(If[l == 0, 0, n*(l - i)]), b[n, i - 1, l] + If[i > n, 0, x^(If[l == 0, 0, l - i])*b[n - i, i, If[l == 0, i, l]]]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, n, 0]]; Table[T[n], {n, 1, 15}] // Flatten (* _Jean-François Alcover_, Dec 22 2016, after _Alois P. Heinz_ *)
%t Table[Length[Select[IntegerPartitions[n],Max[#]*Length[#]-n==k&]],{n,1,11},{k,0,Floor[(n-1)/2]*Ceiling[(n-1)/2]}] (* _Gus Wiseman_, Sep 24 2019 *)
%Y Cf. A000005, A000041, A002620, A033638, A070824, A188814, A326844, A326849.
%K nonn,tabf
%O 1,2
%A _Emeric Deutsch_, Feb 12 2016
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