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1, 1, 2, 3, 2, 3, 3, 2, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2

a(60)-a(80) from Jinyuan Wang, Jun 26 2020

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The case of non-strong case of compositions is A332296.

allocated for Gus WisemanNumber of narrowly totally strongly normal integer partitions of n.

1, 1, 2, 3, 2, 3, 3, 2, 2, 2, 3, 3, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 3, 2, 2, 2

0,3

A partition is narrowly totally strongly normal if either it is empty, a singleton (narrow), or it covers an initial interval of positive integers (normal) and has weakly decreasing run-lengths (strong) that are themselves a narrowly totally strongly normal partition.

The a(1) = 1, a(2) = 2, a(3) = 3, and a(55) = 4 partitions:

(1) (2) (3) (55)

(1,1) (2,1) (10,9,8,7,6,5,4,3,2,1)

(1,1,1) (5,5,5,5,5,4,4,4,4,3,3,3,2,2,1)

(1)^55

For example, starting with the partition (3,3,2,2,1) and repeatedly taking run-lengths gives (3,3,2,2,1) -> (2,2,1) -> (2,1) -> (1,1) -> (2). The first four are normal and have weakly decreasing run-lengths, and the last is a singleton, so (3,3,2,2,1) is counted under a(11).

tinQ[q_]:=Or[q=={}, Length[q]==1, And[Union[q]==Range[Max[q]], GreaterEqual@@Length/@Split[q], tinQ[Length/@Split[q]]]];

Table[Length[Select[IntegerPartitions[n], tinQ]], {n, 0, 30}]

Normal partitions are A000009.

The non-totally normal version is A316496.

The widely alternating version is A332292.

The case of non-strong compositions is A332296.

The case of compositions is A332336.

The wide version is a(n) - 1 for n > 1.

Cf. A001462, A025487, A100883, A181819, A182850, A317081, A317245, A317256, A317491, A332275, A332277, A332278, A332291, A332337.

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Gus Wiseman, Feb 15 2020

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allocated for Gus Wiseman

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