OFFSET
0,4
COMMENTS
A sequence of integers is widely totally co-strongly normal if either it is constant 1's (wide) or it covers an initial interval of positive integers (normal) with weakly increasing run-lengths (co-strong) which are themselves a widely totally co-strongly normal sequence.
Is this sequence bounded?
EXAMPLE
The a(1) = 1 through a(20) = 2 partitions:
1: (1)
2: (11)
3: (21),(111)
4: (211),(1111)
5: (11111)
6: (321),(111111)
7: (1111111)
8: (11111111)
9: (32211),(111111111)
10: (4321),(322111),(1111111111)
11: (11111111111)
12: (111111111111)
13: (1111111111111)
14: (11111111111111)
15: (54321),(111111111111111)
16: (1111111111111111)
17: (11111111111111111)
18: (111111111111111111)
19: (1111111111111111111)
20: (4332221111),(11111111111111111111)
MATHEMATICA
totnQ[ptn_]:=Or[ptn=={}, Union[ptn]=={1}, And[Union[ptn]==Range[Max[ptn]], LessEqual@@Length/@Split[ptn], totnQ[Length/@Split[ptn]]]];
Table[Length[Select[IntegerPartitions[n], totnQ]], {n, 0, 30}]
CROSSREFS
Not requiring co-strength gives A332277.
The strong version is A332297(n) - 1 for n > 1.
The narrow version is a(n) - 1 for n > 1.
The alternating version is A332289.
The Heinz numbers of these partitions are A332293.
The case of compositions is A332337.
KEYWORD
nonn,more
AUTHOR
Gus Wiseman, Mar 05 2020
EXTENSIONS
a(71)-a(78) from Jinyuan Wang, Jun 26 2020
STATUS
approved