%I #5 Oct 12 2022 09:00:48
%S 1,0,1,0,0,2,0,0,1,3,0,0,2,2,4,0,0,3,5,3,5,0,0,4,8,10,4,6,0,0,5,11,18,
%T 18,5,7,0,0,6,14,28,36,30,6,8,0,0,7,17,41,63,65,47,7,9,0,0,8,20,58,
%U 104,126,108,70,8,10,0,0,9,23,80,164,230,230,168,100,9,11
%N Triangle read by rows where T(n,k) is the number of integer compositions of n with half-alternating sum k, where k ranges from -n to n in steps of 2.
%C We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ...
%e Triangle begins:
%e 1
%e 0 1
%e 0 0 2
%e 0 0 1 3
%e 0 0 2 2 4
%e 0 0 3 5 3 5
%e 0 0 4 8 10 4 6
%e 0 0 5 11 18 18 5 7
%e 0 0 6 14 28 36 30 6 8
%e 0 0 7 17 41 63 65 47 7 9
%e 0 0 8 20 58 104 126 108 70 8 10
%e Row n = 6 counts the following compositions:
%e (114) (123) (132) (141) (6)
%e (1113) (213) (222) (231) (15)
%e (1122) (1212) (312) (321) (24)
%e (1131) (1221) (1311) (411) (33)
%e (2112) (2211) (42)
%e (2121) (3111) (51)
%e (11121) (11112)
%e (11211) (12111)
%e (21111)
%e (111111)
%t halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
%t Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],halfats[#]==k&]],{n,0,10},{k,-n,n,2}]
%Y Row sums are A011782.
%Y For original alternating sum we have A097805, unordered A344651.
%Y Column k = n-4 appears to be A177787.
%Y The case of partitions is A357637, skew A357638.
%Y The central column k=0 is A357641 (aerated).
%Y The skew-alternating version is A357646.
%Y The reverse version for partitions is A357704, skew A357705.
%Y A351005 = alternately equal and unequal partitions, compositions A357643.
%Y A351006 = alternately unequal and equal partitions, compositions A357644.
%Y A357621 gives half-alternating sum of standard compositions, skew A357623.
%Y A357629 gives half-alternating sum of prime indices, skew A357630.
%Y A357633 gives half-alternating sum of Heinz partition, skew A357634.
%Y Cf. A029862, A035363, A035544, A357136, A357631, A357639.
%K nonn,tabl
%O 0,6
%A _Gus Wiseman_, Oct 12 2022