A325268 Triangle read by rows where T(n,k) is the number of integer partitions of n with omicron k.
1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 3, 0, 1, 0, 1, 5, 0, 0, 1, 0, 1, 7, 2, 0, 0, 1, 0, 1, 12, 1, 0, 0, 0, 1, 0, 1, 17, 2, 1, 0, 0, 0, 1, 0, 1, 24, 4, 0, 0, 0, 0, 0, 1, 0, 1, 33, 5, 1, 1, 0, 0, 0, 0, 1, 0, 1, 44, 9, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 57, 14, 3, 0, 1
Offset: 0
Examples
Triangle begins: 1 0 1 0 1 1 0 1 1 1 0 1 3 0 1 0 1 5 0 0 1 0 1 7 2 0 0 1 0 1 12 1 0 0 0 1 0 1 17 2 1 0 0 0 1 0 1 24 4 0 0 0 0 0 1 0 1 33 5 1 1 0 0 0 0 1 0 1 44 9 1 0 0 0 0 0 0 1 0 1 57 14 3 0 1 0 0 0 0 0 1 0 1 76 20 3 0 0 0 0 0 0 0 0 1 Row n = 8 counts the following partitions. (8) (44) (431) (2222) (11111111) (53) (521) (62) (71) (332) (422) (611) (3221) (3311) (4211) (5111) (22211) (32111) (41111) (221111) (311111) (2111111)
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)
Crossrefs
Programs
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Mathematica
Table[Length[Select[IntegerPartitions[n],Switch[#,{},0,{},1,,NestWhile[Sort[Length/@Split[#]]&,#,Length[#]>1&]//First]==k&]],{n,0,10},{k,0,n}]
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PARI
omicron(p)={if(!#p, 0, my(r=1); while(#p > 1, my(L=List(), k=0); r=#p; for(i=1, #p, if(i==#p||p[i]<>p[i+1], listput(L,i-k); k=i)); listsort(L); p=L); r)} row(n)={my(v=vector(1+n)); forpart(p=n, v[1 + omicron(Vec(p))]++); v} { for(n=0, 10, print(row(n))) } \\ Andrew Howroyd, Jan 18 2023
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