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Number of integer partitions of n with adjusted frequency depth ceiling(sqrt(n)).
2

%I #4 Apr 23 2019 09:20:47

%S 1,1,1,1,2,2,4,4,6,8,17,26,25,44,53,63,83,128,168,212,273,344,429,525,

%T 662,796,684,910,1211,1595,2060,2663,3406,4315,5426,6784,8417,0,0,0,0,

%U 0,1,5,14,36,76,143,269,446,0,0,0,0,0,0,0,0,0,0,0

%N Number of integer partitions of n with adjusted frequency depth ceiling(sqrt(n)).

%C The adjusted frequency depth of an integer partition is 0 if the partition is empty, and otherwise it is one plus the number of times one must take the multiset of multiplicities to reach a singleton. For example, the partition (32211) has adjusted frequency depth 5 because we have: (32211) -> (221) -> (21) -> (11) -> (2).

%e The a(2) = 1 through a(11) = 26 partitions:

%e 11 111 22 32 42 43 53 54 433 443

%e 1111 41 51 52 62 63 442 533

%e 321 61 71 72 622 551

%e 2211 421 431 81 811 722

%e 521 432 3331 911

%e 3311 531 4222 3332

%e 621 7111 5222

%e 222111 61111 8111

%e 222211 32222

%e 322111 33311

%e 331111 44111

%e 511111 71111

%e 2221111 222221

%e 4111111 322211

%e 22111111 332111

%e 31111111 422111

%e 211111111 611111

%e 2222111

%e 3221111

%e 3311111

%e 5111111

%e 22211111

%e 41111111

%e 221111111

%e 311111111

%e 2111111111

%t fdadj[ptn_List]:=If[ptn=={},0,Length[NestWhileList[Sort[Length/@Split[#1]]&,ptn,Length[#1]>1&]]];

%t Table[Length[Select[IntegerPartitions[n],fdadj[#]==Ceiling[Sqrt[n]]&]],{n,0,30}]

%Y Cf. A117571, A181819, A225485, A323014, A323023, A325245, A325246, A325252, A325258, A325271, A325278, A325280, A325282.

%K nonn

%O 0,5

%A _Gus Wiseman_, Apr 22 2019