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A239302 Triangular array: T(n,k) = number of partitions x(1) > x(2) > ... > x(k) of n+2 such that x(1) = x(2) + k, for n >= 1.

%I A239302 #9 Mar 17 2014 13:24:16
%S A239302 1,0,1,1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,2,1,1,1,1,0,1,2,2,1,1,1,1,
%T A239302 0,1,2,2,2,1,1,1,1,0,1,3,2,2,2,1,1,1,1,0,1,3,3,2,2,2,1,1,1,1,0,1,4,3,
%U A239302 3,2,2,2,1,1,1,1,0,1,5,4,3,3,2,2,2,1
%N A239302 Triangular array:  T(n,k) = number of partitions x(1) > x(2) > ... > x(k) of n+2 such that x(1) = x(2) + k, for n >= 1.
%C A239302 The columns are identical, and the limit of the reversal of rows is A087897(n+3).
%C A239302 Sum of row n equals A111133(n+3).
%H A239302 Clark Kimberling, <a href="/A239302/b239302.txt">Table of n, a(n) for n = 1..1000</a>
%e A239302 First 17 rows:
%e A239302 1
%e A239302 0 1
%e A239302 1 0 1
%e A239302 1 1 0 1
%e A239302 1 1 1 0 1
%e A239302 1 1 1 1 0 1
%e A239302 2 1 1 1 1 0 1
%e A239302 2 2 1 1 1 1 0 1
%e A239302 2 2 2 1 1 1 1 0 1
%e A239302 3 2 2 2 1 1 1 1 0 1
%e A239302 3 3 2 2 2 1 1 1 1 0 1
%e A239302 4 3 3 2 2 2 1 1 1 1 0 1
%e A239302 5 4 3 3 2 2 2 1 1 1 1 0 1
%e A239302 5 5 4 3 3 2 2 2 1 1 1 1 0 1
%e A239302 6 5 5 4 3 3 2 2 2 1 1 1 1 0 1
%e A239302 8 6 5 5 4 3 3 2 2 2 1 1 1 1 0 1
%e A239302 8 8 6 5 5 4 3 3 2 2 2 1 1 1 1 0 1
%e A239302 To account for row 7, start with the strict partitions (A000009) of 9 that have more than one part:  81, 72, 63, 621, 54, 531, 432.  Next, form (part 1) - (part 2) for each of those partitions, getting 7, 5, 3, 4, 1, 2, 1; finally, note that the numbers of occurrences of 1,2,3,4,5,6,7, respectively, are 2,1,1,1,1,0,1.
%t A239302 z = 25; d[n_] := d[n] = Rest[Select[IntegerPartitions[n], DeleteDuplicates[#] == # &]]; t[n_] := t[n] = Table[d[n][[k, 1]] - d[n][[k, 2]], {k, 1, -1 + PartitionsQ[n]}]; u = Table[Count[t[n], j], {n, 3, z}, {j, 1, n - 2}]; TableForm[u] (* A239302 as an array *)
%t A239302 v = Flatten[u]  (* A239302 as a sequence *)
%Y A239302 Cf. A000009, A111133, A087897.
%K A239302 nonn,tabl,easy
%O A239302 1,22
%A A239302 _Clark Kimberling_, Mar 14 2014