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%I #18 Oct 08 2022 00:01:41
%S 1,2,2,1,2,3,2,2,3,2,2,6,1,2,2,4,3,4,2,2,4,6,2,6,2,2,4,4,2,3,9,4,2,2,
%T 4,4,2,6,6,3,12,1,2,2,4,4,2,4,6,3,6,6,12,5,2,2,4,4,2,4,6,6,4,6,3,18,2,
%U 4,10,2,2,4,4,2,4,6,4,4,6,3,6,12,2,6
%N Array: row n gives the number of occurrences of each possible antidiagonal partition of n, arranged in reverse-Mathematica order.
%C Suppose that p is a partition of n, let F(p) be its Ferrers matrix, as defined at A237981, and let mXm be the size of F(p). The numbers of 1's in each of the 2m-1 antidiagonals of F(p) form a partition of n. Any partition which is associated with a partition of n in this manner is introduced here as an antidiagonal partition of n. A000041(n) = sum of the numbers in row n; A000009(n) = number of terms in row n, since the antidiagonal partitions of n are the conjugates of the strict partitions of n.
%H Clark Kimberling, <a href="/A238325/b238325.txt">Table of n, a(n) for n = 1..1000</a>
%H Clark Kimberling and Peter J. C. Moses, <a href="http://faculty.evansville.edu/ck6/GalleryThree/Introduction3.html">Ferrers Matrices and Related Partitions of Integers</a>
%e The Mathematica ordering of the 6 antidiagonal partitions of 8 follows: 3221, 32111, 22211, 221111, 2111111, 11111111. Frequencies of these among the 22 partitions of 8 are given in reverse Mathematica ordering as follows: 11111111 occurs 2 times, 2111111 occurs 2 times, 221111 occurs 4 times, 22211 occurs 6 times, 32111 occurs 2 times, and 3221 occurs 6 times, so that row 8 of the array is 2 2 4 6 2 6.
%e ...
%e First 12 rows:
%e 1;
%e 2;
%e 2, 1;
%e 2, 3;
%e 2, 2, 3;
%e 2, 2, 6, 1;
%e 2, 2, 4, 3, 4;
%e 2, 2, 4, 6, 2, 6;
%e 2, 2, 4, 4, 2, 3, 9, 4;
%e 2, 2, 4, 4, 2, 6, 6, 3, 12, 1;
%e 2, 2, 4, 4, 2, 4, 6, 3, 6, 6, 12, 5;
%e 2, 2, 4, 4, 2, 4, 6, 6, 4, 6, 3, 18, 2, 4, 10;
%t z = 20; ferrersMatrix[list_] := PadRight[Map[Table[1, {#}] &, #], {#, #} &[Max[#, Length[#]]]] &[list]; antiDiagPartNE[list_] := Module[{m = ferrersMatrix[list]}, Map[Diagonal[Reverse[m], #] &, Range[-#, #] &[Length[m] - 1]]]; a1[n_] := Last[Transpose[Tally[Map[DeleteCases[Reverse[Sort[Map[Count[#, 1] &, antiDiagPartNE[#]]]], 0] &, IntegerPartitions[n]]]]];
%t t = Table[a1[n], {n, 1, z}]; TableForm[Table[a1[n], {n, 1, z}]] (* A238325, array *)
%t u = Flatten[t] (* A238325, sequence *)
%t (* _Peter J. C. Moses_, 18 February 2014 *)
%Y Cf. A238326.
%K nonn,tabf,easy
%O 1,2
%A _Clark Kimberling_ and _Peter J. C. Moses_, Feb 24 2014
%E Example corrected by _Peter J. Taylor_, Apr 10 2022
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