A350946 Heinz numbers of integer partitions with as many even parts as odd parts and as many even conjugate parts as odd conjugate parts.
1, 6, 65, 84, 210, 216, 319, 490, 525, 532, 731, 1254, 1403, 1924, 2184, 2340, 2449, 2470, 3024, 3135, 3325, 3774, 4028, 4141, 4522, 5311, 5460, 7030, 7314, 7315, 7560, 7776, 7942, 8201, 8236, 9048, 9435, 9464, 10659, 10921, 11484, 11914, 12012, 12025, 12740
Offset: 1
Keywords
Examples
The terms together with their prime indices begin: 1: () 6: (2,1) 65: (6,3) 84: (4,2,1,1) 210: (4,3,2,1) 216: (2,2,2,1,1,1) 319: (10,5) 490: (4,4,3,1) 525: (4,3,3,2) 532: (8,4,1,1) 731: (14,7) 1254: (8,5,2,1) 1403: (18,9) 1924: (12,6,1,1) 2184: (6,4,2,1,1,1) 2340: (6,3,2,2,1,1) 2449: (22,11) 2470: (8,6,3,1) For example, the prime indices of 532 are (8,4,1,1), even/odd counts 2/2, and the prime indices of the conjugate 3024 are (4,2,2,2,1,1,1,1), with even/odd counts 4/4; so 532 belongs to the sequence.
Crossrefs
For the first condition alone:
- ordered version (compositions) A098123
- ranked by A325698
There are four statistics:
There are four other possible pairings of statistics:
There are two other possible double-pairings of statistics:
A122111 represents partition conjugation using Heinz numbers.
A195017 = # of even parts - # of odd parts.
A316524 = alternating sum of prime indices.
Programs
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Mathematica
primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]; conj[y_]:=If[Length[y]==0,y,Table[Length[Select[y,#>=k&]],{k,1,Max[y]}]]; Select[Range[1000],#==1||Mean[Mod[primeMS[#],2]]== Mean[Mod[conj[primeMS[#]],2]]==1/2&]
Comments