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A384125 Array read by antidiagonals: T(n,m) is the number of edges in the n X m rook graph K_n X K_m.

%I A384125 #12 May 23 2025 01:09:30
%S A384125 0,1,1,3,4,3,6,9,9,6,10,16,18,16,10,15,25,30,30,25,15,21,36,45,48,45,
%T A384125 36,21,28,49,63,70,70,63,49,28,36,64,84,96,100,96,84,64,36,45,81,108,
%U A384125 126,135,135,126,108,81,45,55,100,135,160,175,180,175,160,135,100,55
%N A384125 Array read by antidiagonals: T(n,m) is the number of edges in the n X m rook graph K_n X K_m.
%H A384125 Andrew Howroyd, <a href="/A384125/b384125.txt">Table of n, a(n) for n = 1..1275</a> (first 50 antidiagonals)
%H A384125 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/EdgeCount.html">Edge Count</a>.
%H A384125 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RookGraph.html">Rook Graph</a>.
%F A384125 T(n,m) = n*binomial(m,2) + m*binomial(n,2).
%F A384125 T(n,m) = binomial(n*m,2) - 2*binomial(n,2)*binomial(m,2).
%F A384125 T(n,m) = T(m,n).
%e A384125 Array begins:
%e A384125 =======================================
%e A384125 n\m |  1  2   3   4   5   6   7   8 ...
%e A384125 ----+----------------------------------
%e A384125   1 |  0  1   3   6  10  15  21  28 ...
%e A384125   2 |  1  4   9  16  25  36  49  64 ...
%e A384125   3 |  3  9  18  30  45  63  84 108 ...
%e A384125   4 |  6 16  30  48  70  96 126 160 ...
%e A384125   5 | 10 25  45  70 100 135 175 220 ...
%e A384125   6 | 15 36  63  96 135 180 231 288 ...
%e A384125   7 | 21 49  84 126 175 231 294 364 ...
%e A384125   8 | 28 64 108 160 220 288 364 448 ...
%e A384125   ...
%t A384125 Table[#*Binomial[m, 2] + m*Binomial[#, 2] &[n - m + 1], {n, 11}, {m, n}] // Flatten (* _Michael De Vlieger_, May 22 2025 *)
%o A384125 (PARI) T(n,m) = n*binomial(m,2) + m*binomial(n,2)
%Y A384125 Main diagonal is A045991.
%Y A384125 Columns 1..6 are A000217(n-1), A000290, A045943, A054000, A269457(n-1), A067707.
%Y A384125 Cf. A003991 (number of vertices), A360855 (triangles), A384120 (all cliques).
%K A384125 nonn,tabl,easy
%O A384125 1,4
%A A384125 _Andrew Howroyd_, May 20 2025