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Triangle read by rows: T(n, k) = floor((n-1)/2)*floor(n/2)*floor((k-1)/2)*floor(k/2).
1

%I #29 Feb 16 2025 08:33:53

%S 0,0,0,0,0,1,0,0,2,4,0,0,4,8,16,0,0,6,12,24,36,0,0,9,18,36,54,81,0,0,

%T 12,24,48,72,108,144,0,0,16,32,64,96,144,192,256,0,0,20,40,80,120,180,

%U 240,320,400,0,0,25,50,100,150,225,300,400,500,625

%N Triangle read by rows: T(n, k) = floor((n-1)/2)*floor(n/2)*floor((k-1)/2)*floor(k/2).

%C T(n, k) is conjectured by Zarankiewicz's conjecture to be the crossing number of the complete bipartite graph K_{k,n}.

%H Robert Israel, <a href="/A298368/b298368.txt">Table of n, a(n) for n = 1..10011</a> (rows 1 to 141, flattened)

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/GraphCrossingNumber.html">Graph Crossing Number</a>.

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/ZarankiewiczsConjecture.html">Zarankiewicz's Conjecture</a>.

%F T(n,n) = A030179(n).

%F From _Robert Israel_, Jan 17 2018: (Start)

%F T(n,k) = A002620(n-1)*A002620(k-1).

%F G.f. as triangle: x^3*y^3*(1+2*x*y+6*x^2*y^2-4*x^3*y-8*x^3*y^2+2*x^4*y+2*x^3*y^3-4*x^4*y^2-2*x^4*y^3+4*x^5*y^2+ x^4*y^4-4*x^5*y^3-2*x^5*y^4+4*x^6*y^3+2*x^7*y^4)/

%F ((1-x*y)^5*(1+x*y)^3*(1-x)^3*(1+x)). (End)

%e First rows are given by:

%e 0;

%e 0, 0;

%e 0, 0, 1;

%e 0, 0, 2, 4;

%e 0, 0, 4, 8, 16;

%e 0, 0, 6, 12, 24, 36;

%e 0, 0, 9, 18, 36, 54, 81;

%e 0, 0, 12, 24, 48, 72, 108, 144;

%e 0, 0, 16, 32, 64, 96, 144, 192, 256;

%e 0, 0, 20, 40, 80, 120, 180, 240, 320, 400;

%p seq(seq(floor((k-1)/2)*floor(k/2)*floor((n-1)/2)*floor(n/2),k=1..n),n=1..12); # _Robert Israel_, Jan 17 2018

%t Table[Floor[(m - 1)/2] Floor[m/2] Floor[(n - 1)/2] Floor[n/2], {n, 11}, {m, n}] // Flatten

%t Table[Times @@ Floor[{m, m - 1, n, n - 1}/2], {n, 11}, {m, n}] // Flatten

%Y Cf. A030179, A191928.

%K nonn,easy,tabl,nice,changed

%O 1,9

%A _Eric W. Weisstein_, Jan 17 2018