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%I #15 Apr 15 2015 05:27:08
%S 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,
%T 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,
%U 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
%N Number of partitions of n having T(n,k) odd parts in excess on even places over odd places.
%C Row lengths are 2*floor((3 + sqrt(1+8*n))/4), k runs from -floor((3 + sqrt(1+8*n))/4) up to floor((-1 + sqrt(1+8*n))/4); row sums are A000041.
%C P. D. Hanna remarks that "zig-zag" diagonals/antidiagonals produce A077028 (Rascal triangle).
%e In the table below, replace each integer i with A000720(i) to get the current sequence:
%e -3 -2 -1 0 1 2 (= k)(n= )
%e 0 1 0
%e 1 0 1
%e 0 2 2
%e 0 2 0 1 3
%e 0 0 3 0 4
%e 0 3 0 2 5
%e 1 0 4 0 6
%e 0 4 0 3 7
%e 2 0 5 0 8
%e 0 5 0 4 9
%e 0 3 0 6 0 1 10
%e 0 0 6 0 5 0 11
%e 0 4 0 7 0 2 12
%e 0 0 7 0 6 0 13
%e 0 5 0 8 0 3 14
%e 1 0 8 0 7 0 15
%e ...
%e The table then starts as:
%e 0 0,1
%e 1 1,0
%e 2 0,2
%e 3 0,2,0,1
%e 4 0,0,5,0
%e 5 0,5,0,2
%e 6 1,0,10,0
%e 7 0,10,0,5
%e 8 2,0,20,0
%e 9 0,20,0,10
%e 10 0,5,0,36,0,1
%e ...
%e 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.
%t 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 *)
%t 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]}]
%Y Cf. A000720, A077028.
%K nonn,tabf
%O 0,6
%A _Wouter Meeussen_, Apr 20 2013
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