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The stay number of a partition P is defined as follows. Let U be the ordering of the parts of P in non-increasing nonincreasing order, and let V be the reverse of U. The stay number of P is the number of numbers whose position in V is the same as in U. (1st column) = A238479. When the rows of the array are read in reverse order, it appears that the limiting sequence is A008483.

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Cf. A000041, A008483, A238479.

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allocated for Clark KimberlingTriangular array: each row partitions the partitions of n into n parts; of which the k-th part is the number of partitions having stay number k-1; see Comments.

1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 2, 3, 1, 0, 0, 1, 3, 3, 2, 2, 0, 0, 1, 4, 6, 2, 1, 1, 0, 0, 1, 5, 8, 4, 1, 2, 1, 0, 0, 1, 8, 10, 4, 4, 1, 1, 1, 0, 0, 1, 10, 14, 8, 3, 2, 2, 1, 1, 0, 0, 1, 13, 20, 9, 5, 3, 2, 1, 1, 1, 0, 0, 1, 18, 25, 12, 8, 5, 2

1,12

The stay number of a partition P is defined as follows. Let U be the ordering of the parts of P in non-increasing order, and let V be the reverse of U. The stay number of P is the number of numbers whose position in V is the same as in U. (1st column) = A238479. When the rows of the array are read in reverse order, it appears that the limiting sequence is A008483.

The first 8 rows:

1

0 1

0 1 1

1 1 0 1

1 2 1 0 1

2 3 1 0 0 1

3 3 2 2 0 0 1

4 6 2 1 1 0 0 1

5 8 4 1 2 1 0 0 1

For n = 5, P consists of these partitions:

[5], with reversal [5], thus, 1 stay number

[4,1], with reversal [1,4], thus 0 stay numbers

[3,2], with reversal [2,3], thus 0 stay numbers

[2,2,1], with reversal [1,2,2], thus 1 stay number

[2,1,1,1], with reversal [1,1,1,2], thus 2 stay numbers

[1,1,1,1,1], thus, 5 stay numbers.

As a result, row 5 of the array is 2 3 1 0 0 1

Map[BinCounts[#, {0, Last[#] + 1, 1}] &, Map[Map[Count[#, 0] &, # - Map[Reverse, #] &[IntegerPartitions[#]]] &, Range[0, 35]]]

(* Peter J. C. Moses, May 14 2019 *)

Cf. A000041, A008483.

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nonn,tabl,easy

Clark Kimberling, May 16 2019

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allocated for Clark Kimberling

recycled

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