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Irregular triangle read by rows giving the frequency span signature of n.
3

%I #6 May 19 2019 06:20:47

%S 1,1,2,1,1,3,2,1,3,1,3,3,1,1,1,5,3,1,3,1,1,2,2,1,4,1,1,4,4,1,5,2,1,2,

%T 2,1,3,1,1,1,6,2,1,1,2,3,1,1,3,1,5,2,1,1,4,1,2,1,5,1,2,2,1,3,1,1,2,1,

%U 1,1,2,5,1,3,1,1,2,2,1,6,1,2,1,4,1,1,1

%N Irregular triangle read by rows giving the frequency span signature of n.

%C We define the frequency span of an integer partition to be the partition itself if it has no or only one block, and otherwise it is the multiset union of the partition and the frequency span of its multiplicities. For example, the frequency span of (3,2,2,1) is {1,2,2,3} U {1,1,2} U {1,2} U {1,1} U {2} = {1,1,1,1,1,1,2,2,2,2,2,3}. The frequency span of a positive integer is the frequency span of its prime indices (row n of A296150). Row n of this triangle gives an unsorted list of the multiplicities in the frequency span of n. For example, the frequency span of 30 is {1,1,1,1,2,3,3}, so row 30 is (4,1,2).

%e Triangle begins:

%e 1

%e 1

%e 2 1

%e 1

%e 3 2

%e 1

%e 3 1

%e 3

%e 3 1 1

%e 1

%e 5 3

%e 1

%e 3 1 1

%e 2 2 1

%e 4 1

%e 1

%e 4 4

%e 1

%e 5 2 1

%e 2 2 1

%e 3 1 1

%e 1

%e 6 2 1

%e 1 2

%e 3 1 1

%e 3 1

%e 5 2 1

%e 1

%e 4 1 2

%t primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];

%t freqspan[ptn_]:=If[Length[ptn]<=1,ptn,Sort[Join[ptn,freqspan[Sort[Length/@Split[ptn]]]]]];

%t Table[Length/@Split[freqspan[primeMS[n]]],{n,30}]

%Y Row sums are A325249.

%Y Row lengths are A325759.

%Y Run-lengths of A325757.

%Y Row n is the unsorted prime signature of A325760(n).

%Y Cf. A001221, A001222, A056239, A071625, A112798, A181819, A182857, A290822, A323014,A325277.

%K nonn,tabf

%O 1,3

%A _Gus Wiseman_, May 19 2019