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.

A194494 Number of ways to arrange 4 nonattacking queens on the lower triangle of an n X n board.

Original entry on oeis.org

0, 0, 0, 0, 0, 8, 118, 802, 3708, 13280, 39734, 104000, 244948, 530632, 1072776, 2048056, 3723314, 6492342, 10915254, 17777372, 28147380, 43465356, 65624634, 97098802, 141050688, 201509798, 283514112, 393348562, 538725268, 729098516
Offset: 1

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Author

R. H. Hardin Aug 26 2011

Keywords

Comments

Column 4 of A194498

Examples

			Some solutions for 6X6
..0............0............0............0............0............0
..1.0..........0.0..........0.0..........0.1..........0.0..........0.1
..0.0.0........1.0.0........0.0.1........0.0.0........0.1.0........0.0.0
..0.1.0.0......0.0.1.0......1.0.0.0......1.0.0.0......0.0.0.1......0.0.1.0
..0.0.0.0.1....0.0.0.0.1....0.0.0.1.0....0.0.1.0.0....1.0.0.0.0....1.0.0.0.0
..0.0.1.0.0.0..0.1.0.0.0.0..0.1.0.0.0.0..0.0.0.0.1.0..0.0.1.0.0.0..0.0.0.1.0.0
		

Formula

Empirical: a(n) = 4*a(n-2) +3*a(n-3) -5*a(n-4) -11*a(n-5) -3*a(n-6) +11*a(n-7) +14*a(n-8) +6*a(n-9) -7*a(n-10) -16*a(n-11) -14*a(n-12) +14*a(n-14) +16*a(n-15) +7*a(n-16) -6*a(n-17) -14*a(n-18) -11*a(n-19) +3*a(n-20) +11*a(n-21) +5*a(n-22) -3*a(n-23) -4*a(n-24) +a(n-26), [R. H. Hardin, Aug 26 2011]
G.f.: -2*x^6*(287*x^18 + 1545*x^17 + 4929*x^16 + 11689*x^15 + 22673*x^14 + 36995*x^13 + 51875*x^12 + 63203*x^11 + 67465*x^10 + 63168*x^9 + 51807*x^8 + 36900*x^7 + 22544*x^6 + 11587*x^5 + 4879*x^4 + 1606*x^3 + 385*x^2 + 59*x + 4)/((x-1)^9*(x+1)^5*(x^2+1)*(x^2+x+1)^3*(x^4+x^3+x^2+x+1))
Explicit formula: n^8/384 - n^7/16 + 23*n^6/36 - 1301*n^5/360 + 14125*n^4/1152 - 8013*n^3/320 + 83147*n^2/2880 - 2089*n/160 + (n^4/32 - 17*n^3/24 + 193*n^2/32 - 2209*n/96 + 439/16)*floor(n/2) + (n^2/3 - 13*n/3 + 136/9)*floor(n/3) - 28/9*floor((n+1)/3) + 23/4*floor(n/4) - 3*floor((n+1)/4) + 4/5*floor(n/5) - 2/5*floor((n+2)/5) - 2/5*floor((n+3)/5), [Vaclav Kotesovec, Apr 08 2012]