A352472 Triangle T(n,k) read by rows: the number of traceless symmetric binary n X n matrices with 2k one's and no all-1 2 X 2 submatrix.

1, 1, 1, 1, 3, 3, 1, 1, 6, 15, 20, 12, 1, 10, 45, 120, 195, 162, 15, 1, 15, 105, 455, 1320, 2508, 2680, 900, 1, 21, 210, 1330, 5880, 18564, 40474, 54750, 35595, 6615, 1, 28, 378, 3276, 20265, 93240, 320040, 795120, 1333080, 1323840, 619920, 90720, 1, 36, 630, 7140, 58527, 364896

Offset: 1

Author

R. J. Mathar, Mar 17 2022

Keywords

• nonn
• tabf

Comments

Symmetry and traceless mean that the number of 1's is always even; the corresponding zeros for odd numbers are not shown here.

Examples

			The triangle starts at 1 X 1 matrices and 0,2,4,... ones as
1: 1;
2: 1  1;
3: 1  3   3    1;
4: 1  6  15   20    12;
5: 1 10  45  120   195    162      15;
6: 1 15 105  455  1320   2508    2680     900;
7: 1 21 210 1330  5880  18564   40474   54750    35595     6615;
8: 1 28 378 3276 20265  93240  320040  795120  1333080  1323840   619920    90720;
9: 1 36 630 7140 58527 364896 1763076 6578640 18514935 37535932 50808870 40684140 15892065 1995840;
		

Links

• Max Alekseyev, Table of n, a(n) for n = 1..205
• Index entries for sequences of squarefree graphs

Crossrefs

Cf. A350189 (allows nonzero trace), A345249 (row sums), A006855 (row lengths minus 1), A191966 (rightmost values).

Formula

T(n,1) = A000217(n-1). - R. J. Mathar, Mar 25 2022

T(n,2) = 3*A000332(n+1). T(n,3) = A093566(n+1). - Conjectured by R. J. Mathar, Mar 25 2022; proved by Max Alekseyev, Apr 02 2022

G.f.: F(x,y) = Sum_{n,k} T(n,k)*(x^n/n!)*y^k = exp( Sum_G x^n(G) * y^e(G) / |Aut(G)| ), where G runs over the connected squarefree graphs (cf. A077269), n(G) and e(G) are the numbers of nodes and edges in G, and Aut(G) is the automorphism group of G. It follows that F(x,y) = exp(x) * (1 + (1/2)*x^2*y + ((1/2)*x^3 + (1/8)*x^4)*y^2 + ((1/6)*x^3 + (2/3)*x^4 + (1/4)*x^5 + (1/48)*x^6)*y^3 + O(y^4)), implying the above formulas for T(n,2) and T(n,3). - Max Alekseyev, Apr 02 2022