A224937 Number of partitions of n having T(n,k) odd parts in excess on even places over odd places.
0, 1, 1, 0, 0, 2, 0, 2, 0, 1, 0, 0, 5, 0, 0, 5, 0, 2, 1, 0, 10, 0, 0, 10, 0, 5, 2, 0, 20, 0, 0, 20, 0, 10, 0, 5, 0, 36, 0, 1, 0, 0, 36, 0, 20, 0, 0, 10, 0, 65, 0, 2, 0, 0, 65, 0, 36, 0, 0, 20, 0, 110, 0, 5, 1, 0, 110, 0, 65, 0, 0, 36, 0, 185, 0, 10, 2, 0, 185, 0, 110, 0, 0, 65, 0, 300, 0, 20
Offset: 0
Examples
In the table below, replace each integer i with A000720(i) to get the current sequence: -3 -2 -1 0 1 2 (= k)(n= ) 0 1 0 1 0 1 0 2 2 0 2 0 1 3 0 0 3 0 4 0 3 0 2 5 1 0 4 0 6 0 4 0 3 7 2 0 5 0 8 0 5 0 4 9 0 3 0 6 0 1 10 0 0 6 0 5 0 11 0 4 0 7 0 2 12 0 0 7 0 6 0 13 0 5 0 8 0 3 14 1 0 8 0 7 0 15 ... The table then starts as: 0 0,1 1 1,0 2 0,2 3 0,2,0,1 4 0,0,5,0 5 0,5,0,2 6 1,0,10,0 7 0,10,0,5 8 2,0,20,0 9 0,20,0,10 10 0,5,0,36,0,1 ... The partitions of n=5 then give (0,5,0,2) for k=(-2,-1,0,1); this corresponds to 5 partitions with -1 excess odd parts on even over odd positions, and 2 with 1 excess, namely (4,1') and (2,1',1,1') where odd parts on even positions are marked by a quote.
Programs
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Mathematica
Table[ CoefficientList[ x^Floor[(3+Sqrt[1+8*n])/4]* Tr[x^Tr[(-1)^Mod[Flatten[Position[#,_?OddQ]],2]]&/@Partitions[n]],x],{n,0,12}]; (* or *) a712[n_Integer]:= a712[n] =If[n<0, 0, (# . Reverse[#])& [PartitionsP[ Range[0, n] ]]]; Table[If[Mod[n+k,2]==1,0,a712[-1+Max[0,(2+n-k*(2*k+1))/2]]],{n,0,12},{k,-Floor[(3+Sqrt[1+8*n])/4],Floor[(-1+Sqrt[1+8*n])/4]}]
Comments