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Number of widely recursively normal integer partitions of n.
8

%I #6 Feb 18 2020 04:47:52

%S 1,1,2,3,4,6,6,10,12,17,21,30,34,48,54,74,86,113,132,169,200,246,293,

%T 360,422,512,599,726,840,1009,1181,1401,1631,1940,2240,2636,3069,3567,

%U 4141,4846,5556,6470,7505,8627,9936,11523,13176,15151,17430,19935,22846

%N Number of widely recursively normal integer partitions of n.

%C A sequence is widely recursively normal if either it is all 1's (wide) or its run-lengths cover an initial interval of positive integers (normal) and are themselves a widely recursively normal sequence.

%e The a(1) = 1 through a(8) = 12 partitions:

%e (1) (2) (3) (4) (5) (6) (7) (8)

%e (11) (21) (31) (32) (42) (43) (53)

%e (111) (211) (41) (51) (52) (62)

%e (1111) (221) (321) (61) (71)

%e (311) (411) (322) (332)

%e (11111) (111111) (331) (422)

%e (421) (431)

%e (511) (521)

%e (3211) (611)

%e (1111111) (3221)

%e (4211)

%e (11111111)

%e For example, starting with y = (4,3,2,2,1) and repeatedly taking run-lengths gives (4,3,2,2,1) -> (1,1,2,1) -> (2,1,1) -> (1,2) -> (1,1), all of which have normal run-lengths, so y is widely recursively normal. On the other hand, starting with y and repeatedly taking multiplicities gives (4,3,2,2,1) -> (2,1,1,1) -> (3,1), so y is not fully normal (A317491).

%e Starting with y = (5,4,3,3,2,2,2,1,1) and repeatedly taking run-lengths gives (5,4,3,3,2,2,2,1,1) -> (1,1,2,3,2) -> (2,1,1,1) -> (1,3), so y is not widely recursively normal. On the other hand, starting with y and repeatedly taking multiplicities gives (5,4,3,3,2,2,2,1,1) -> (3,2,2,1,1) -> (2,2,1) -> (2,1) -> (1,1), so y is fully normal (A317491).

%t recnQ[ptn_]:=Or[ptn=={},Union[ptn]=={1},And[Union[Length/@Split[ptn]]==Range[Max[Length/@Split[ptn]]],recnQ[Length/@Split[ptn]]]];

%t Table[Length[Select[IntegerPartitions[n],recnQ]],{n,0,30}]

%Y The narrow version is A000012.

%Y Partitions with normal multiplicities are A317081.

%Y The Heinz numbers of these partitions are a proper superset of A317492.

%Y Accepting any constant sequence instead of just 1's gives A332272.

%Y The total (instead of recursive) version is A332277.

%Y The case of reversed partitions is this same sequence.

%Y The alternating (instead of recursive) version is this same sequence.

%Y Dominated by A332576.

%Y Cf. A000009, A001462, A181819, A182850, A317245, A317491, A329746, A329747, A332276.

%K nonn

%O 0,3

%A _Gus Wiseman_, Feb 16 2020