A282011 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.

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, 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, 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

Offset

0,12

Comments

Row n is symmetric if and only if n mod 4 in {0,3} (or if T(n,n) = 1).

Links

Alois P. Heinz, Rows n = 0..200, flattened

Johann Cigler, Some remarks on Rogers-Szegö polynomials and Losanitsch's triangle, arXiv:1711.03340 [math.CO], 2017.

Johann Cigler, Some Pascal-like triangles, 2018.

Formula

T(n,k) = Sum_{j=0..floor((n+1)/4)} C(ceiling(n/2),2*j) * C(floor(n/2),k-2*j).

T(n,k) = A007318(n,k) - A159916(n,k).

Sum_{k=0..n} k * T(n,k) = A057711(n-1) for n>0.

Sum_{k=0..n} (k+1) * T(n,k) = A087447(n) + [n=2].

Example

T(5,0) = 1: {}.
T(5,1) = 2: {2}, {4}.
T(5,2) = 4: {1,3}, {1,5}, {2,4}, {3,5}.
T(5,3) = 6: {1,2,3}, {1,2,5}, {1,3,4}, {1,4,5}, {2,3,5}, {3,4,5}.
T(5,4) = 3: {1,2,3,4}, {1,2,4,5}, {2,3,4,5}.
T(5,5) = 0.
T(7,7) = 1: {1,2,3,4,5,6,7}.
Triangle T(n,k) begins:
  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,   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, 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;

Maple

b:= proc(n, s) option remember; expand(
      `if`(n=0, s, b(n-1, s)+x*b(n-1, irem(s+n, 2))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n, 1)):
seq(T(n), n=0..16);

Mathematica

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 *)

Prog

(PARI) a(n, k)=sum(j=0, floor((n+1)/4), binomial(ceil(n/2), 2*j)*binomial(floor(n/2), k-2*j));
tabl(nn)={for(n=0, nn, for(k=0, n, print1(a(n, k), ", "); ); print(); ); } \\ Indranil Ghosh, Feb 26 2017

Crossrefs

Columns k=0..10 give (offsets may differ): A000012, A004526, A002620, A005993, A005994, A032092, A032093, A018211, A018212, A282077, A282078.

Row sums give A011782.

Main diaginal gives A133872(n+1).

Lower diagonals T(n+j,n) for j=1..10 give: A004525(n+1), A282079, A228705, A282080, A282081, A282082, A282083, A282084, A282085, A282086.

T(2n,n) gives A119358.

Cf. A007318, A057711, A087447, A159916.

Sequence in context: A051950 A341062 A172353 * A245514 A104754 A206827

Adjacent sequences: A282008 A282009 A282010 * A282012 A282013 A282014

Keyword

nonn,tabl

Author

Alois P. Heinz, Feb 04 2017

Status

approved