cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A350189 Triangle T(n,k) read by rows: the number of symmetric binary n X n matrices with k ones and no all-1 2 X 2 submatrix.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 1, 3, 6, 10, 9, 9, 4, 1, 4, 12, 28, 46, 72, 80, 80, 60, 16, 1, 5, 20, 60, 140, 296, 500, 780, 1005, 1085, 992, 560, 170, 1, 6, 30, 110, 330, 876, 1956, 4020, 7140, 11480, 16248, 19608, 20560, 16500, 9720, 3276, 360, 1, 7, 42, 182, 665, 2121, 5852, 14792, 33117, 68355, 126994, 214158
Offset: 0

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Author

R. J. Mathar, Mar 09 2022

Keywords

Comments

There are 2^(n^2) binary n X n matrices (entries of {0,1}). There are 2^(n*(n+1)/2) symmetric binary matrices. There are A184948(n,k) symmetric binary n X n matrices with k ones.
This sequence is the triangle T(n,k) of symmetric binary n x n matrices with k ones but no 2 X 2 submatrix with all entries = 1. [So in the display of these matrices there is no rectangle with four 1's at the corners.]
The row lengths minus 1 are 0, 1, 3, 6, 9, 12, 17, 21, 24, 29, ... and indicate the maximum number of 1's than can be packed into a symmetric binary n X n matrix without creating an all-1 quadrangle/submatrix of order 2.

Examples

			The triangle starts
  1;
  1 1;
  1 2 2 2;
  1 3 6 10 9 9 4;
  1 4 12 28 46 72 80 80 60 16;
  1 5 20 60 140 296 500 780 1005 1085 992 560 170;
  ...
To place 4 ones, one can place 2 of them in C(n,2) ways on the diagonal and the other 2 in n*(n-1)/2 ways outside the diagonal, avoiding one matrix that builds an all-1 submatrix, which are C(n,2)*(n*(n-1)/2-1) matrices. One can place all 4 on the diagonal in C(n,4) ways. One can place 2 outside the diagonal (the other 2 mirror symmetrically) in C(n*(n-1)/2,2) ways. Sum of the 3 terms is T(n,4) = C(n,3)*(5*n+3)/2. - _R. J. Mathar_, Mar 10 2022
		

Crossrefs

Cf. A001197 (conjectured row lengths), A352258 (row sums), A352801 (rightmost terms), A350296, A350304, A350237, A352472 (traceless symmetric).

Formula

T(n,0) = 1.
T(n,1) = n.
T(n,2) = A002378(n-1).
T(n,3) = A006331(n-1).
T(n,4) = n*(n-1)*(n-2)*(5*n+3)/12 = A147875(n)*A000217(n-1)/3. - R. J. Mathar, Mar 10 2022
T(n,5) = n*(n-1)*(n-2)*(13*n^2-n-24)/60. T(n,6) = n*(n-1)*(n-2)*(19*n^3-18*n^2-97*n+60)/180. T(n,7) = n*(n-1)*(n-2)*(n-3)*(58*n^3+75*n^2-223*n+180)/1260. - Conjectured by R. J. Mathar, Mar 11 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^u(G) / |Aut(G)| ), where G runs over the connected squarefree graphs with loops, n(G) is the number of nodes in G, u(G) the number of ones in the adjacency matrix of G, and Aut(G) is the automorphism group of G. It follows that F(x,y) = exp(x) * (1 + x*y + x^2*y^2 + (2/3*x^3 + x^2)*y^3 + (5/12*x^4 + 3/2*x^3)*y^4 + (13/60*x^5 + 3/2*x^4 + 3/2*x^3)*y^5 + (19/180*x^6 + 7/6*x^5 + 8/3*x^4 + 2/3*x^3)*y^6 + (29/630*x^7 + 3/4*x^6 + 19/6*x^5 + 10/3*x^4)*y^7 + O(y^8)), implying the above formulas for T(n,k). - Max Alekseyev, Apr 02 2022
Conjecture: the largest k such that T(n,k) is nonzero is k = A072567(n) = A001197(n) - 1. - Max Alekseyev, Apr 03 2022

A345249 Number of labeled simple graphs on n vertices without cycles of length 4.

Original entry on oeis.org

1, 2, 8, 54, 548, 7984, 163440, 4599908, 174204728, 8721120744, 568964631296, 47787888342520, 5112015062311008, 689824902243337856, 116423739687724785152, 24387469030487505651984, 6296486009090647137387200, 1991072810881504185092485408
Offset: 1

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Author

Brendan McKay, Jun 12 2021

Keywords

Crossrefs

EXP transform of A345248. Row sums of A352472.
A006786 counts isomorphism classes.
Cf. A213434, A352258 (loops allowed).

A352801 Number of symmetric n X n 01-matrices with maximum number of 1s and no 2 X 2 all-1 submatrix.

Original entry on oeis.org

1, 1, 2, 4, 16, 170, 360, 840, 82320, 181440, 1512000, 19958400, 79833600, 259459200, 14529715200
Offset: 0

Views

Author

Max Alekseyev, Apr 03 2022

Keywords

Crossrefs

Formula

a(n) is the last element in the n-th row of triangle A350189.
Conjecture: the maximum number of 1s equals A072567(n), and so a(n) = A350189(n, A072567(n)).
Showing 1-3 of 3 results.