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%I #42 Aug 02 2018 18:55:22
%S 1,1,0,1,1,0,1,1,1,1,1,2,2,2,1,1,2,4,6,3,0,1,3,6,10,9,3,0,1,3,9,19,19,
%T 9,3,1,1,4,12,28,38,28,12,4,1,1,4,16,44,66,60,40,20,5,0,1,5,20,60,110,
%U 126,100,60,25,5,0,1,5,25,85,170,226,226,170,85,25,5,1,1,6,30,110,255,396,452,396,255,110,30,6,1
%N Number T(n,k) of k-element subsets of [n] having an even sum; triangle T(n,k), n>=0, 0<=k<=n, read by rows.
%C Row n is symmetric if and only if n mod 4 in {0,3} (or if T(n,n) = 1).
%H Alois P. Heinz, <a href="/A282011/b282011.txt">Rows n = 0..200, flattened</a>
%H Johann Cigler, <a href="https://arxiv.org/abs/1711.03340">Some remarks on Rogers-Szegö polynomials and Losanitsch's triangle</a>, arXiv:1711.03340 [math.CO], 2017.
%H Johann Cigler, <a href="https://homepage.univie.ac.at/johann.cigler/preprints/losanitsch3.pdf">Some Pascal-like triangles</a>, 2018.
%F T(n,k) = Sum_{j=0..floor((n+1)/4)} C(ceiling(n/2),2*j) * C(floor(n/2),k-2*j).
%F T(n,k) = A007318(n,k) - A159916(n,k).
%F Sum_{k=0..n} k * T(n,k) = A057711(n-1) for n>0.
%F Sum_{k=0..n} (k+1) * T(n,k) = A087447(n) + [n=2].
%e T(5,0) = 1: {}.
%e T(5,1) = 2: {2}, {4}.
%e T(5,2) = 4: {1,3}, {1,5}, {2,4}, {3,5}.
%e T(5,3) = 6: {1,2,3}, {1,2,5}, {1,3,4}, {1,4,5}, {2,3,5}, {3,4,5}.
%e T(5,4) = 3: {1,2,3,4}, {1,2,4,5}, {2,3,4,5}.
%e T(5,5) = 0.
%e T(7,7) = 1: {1,2,3,4,5,6,7}.
%e Triangle T(n,k) begins:
%e 1;
%e 1, 0;
%e 1, 1, 0;
%e 1, 1, 1, 1;
%e 1, 2, 2, 2, 1;
%e 1, 2, 4, 6, 3, 0;
%e 1, 3, 6, 10, 9, 3, 0;
%e 1, 3, 9, 19, 19, 9, 3, 1;
%e 1, 4, 12, 28, 38, 28, 12, 4, 1;
%e 1, 4, 16, 44, 66, 60, 40, 20, 5, 0;
%e 1, 5, 20, 60, 110, 126, 100, 60, 25, 5, 0;
%e 1, 5, 25, 85, 170, 226, 226, 170, 85, 25, 5, 1;
%e 1, 6, 30, 110, 255, 396, 452, 396, 255, 110, 30, 6, 1;
%p b:= proc(n, s) option remember; expand(
%p `if`(n=0, s, b(n-1, s)+x*b(n-1, irem(s+n, 2))))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n, 1)):
%p seq(T(n), n=0..16);
%t Flatten[Table[Sum[Binomial[Ceiling[n/2],2j]Binomial[Floor[n/2],k-2j],{j,0,Floor[(n+1)/4]}],{n,0,10},{k,0,n}]] (* _Indranil Ghosh_, Feb 26 2017 *)
%o (PARI) a(n,k)=sum(j=0,floor((n+1)/4),binomial(ceil(n/2),2*j)*binomial(floor(n/2),k-2*j));
%o tabl(nn)={for(n=0,nn,for(k=0,n,print1(a(n,k),", "););print(););} \\ _Indranil Ghosh_, Feb 26 2017
%Y Columns k=0..10 give (offsets may differ): A000012, A004526, A002620, A005993, A005994, A032092, A032093, A018211, A018212, A282077, A282078.
%Y Row sums give A011782.
%Y Main diaginal gives A133872(n+1).
%Y Lower diagonals T(n+j,n) for j=1..10 give: A004525(n+1), A282079, A228705, A282080, A282081, A282082, A282083, A282084, A282085, A282086.
%Y T(2n,n) gives A119358.
%Y Cf. A007318, A057711, A087447, A159916.
%K nonn,tabl
%O 0,12
%A _Alois P. Heinz_, Feb 04 2017
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