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Number of 5-turn bishop's tours on an n X n board summed over all starting positions
1

%I #11 Apr 04 2016 16:14:07

%S 0,0,8,584,5464,26360,91120,252720,603696,1288592,2525400,4620728,

%T 7998984,13219528,21014336,32306400,48256608,70282656,100115880,

%U 139819944,191858360,259112216,344959120,453289232,588596368,755991600

%N Number of 5-turn bishop's tours on an n X n board summed over all starting positions

%C Row 5 of A188777

%H R. H. Hardin, <a href="/A188780/b188780.txt">Table of n, a(n) for n = 1..42</a>

%F Empirical: a(n) = 4*a(n-1) -3*a(n-2) -8*a(n-3) +14*a(n-4) -14*a(n-6) +8*a(n-7) +3*a(n-8) -4*a(n-9) +a(n-10)

%F Contribution from _Vaclav Kotesovec_, Sep 01 2012: (Start)

%F Empirical: G.f.: 8*x^3*(1 + 69*x + 394*x^2 + 790*x^3 + 829*x^4 + 357*x^5 + 84*x^6)/((1-x)^7*(1+x)^3)

%F Empirical: a(n) = 51/4 - 913*n/10 + 69203*n^2/360 - 602*n^3/3 + 1007*n^4/9 - 473*n^5/15 + 631*n^6/180 + (-1)^n*(-51/4 + 25*n/2 - 23*n^2/8)

%F (End)

%e Some solutions for 4X4

%e ..0..5..0..0....0..5..0..0....0..5..0..0....0..0..2..0....0..4..0..0

%e ..1..0..4..0....0..0..1..0....0..0..3..0....0..5..0..3....0..0..3..0

%e ..0..3..0..0....0..2..0..4....0..2..0..4....1..0..4..0....0..2..0..5

%e ..0..0..2..0....0..0..3..0....1..0..0..0....0..0..0..0....0..0..1..0

%K nonn

%O 1,3

%A _R. H. Hardin_ Apr 10 2011