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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. 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,5
COMMENTS
Symmetry and traceless mean that the number of 1's is always even; the corresponding zeros for odd numbers are not shown here.
LINKS
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
EXAMPLE
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;
CROSSREFS
Cf. A350189 (allows nonzero trace), A345249 (row sums), A006855 (row lengths minus 1), A191966 (rightmost values).
Sequence in context: A296186 A232967 A144163 * A080858 A144228 A083029
KEYWORD
nonn,tabf
AUTHOR
R. J. Mathar, Mar 17 2022
STATUS
approved

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Last modified April 25 01:35 EDT 2024. Contains 371964 sequences. (Running on oeis4.)