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A120399
Consider n x n chessboard. This sequence gives number of chess knight paths from left bottom corner of the board to the right top corner with minimal possible path length (shortest paths).
1
1, 0, 2, 2, 8, 4, 6, 108, 40, 20, 858, 252, 70, 5596, 1344, 252, 32814, 6600, 924, 179696, 30888, 3432, 937794, 140140, 12870, 4721964, 622336, 48620, 23127208, 2720952, 184756, 110809672, 11757200, 705432, 521527428, 50338904, 2704156, 2418585240, 213955200
OFFSET
1,3
COMMENTS
Generating function for this sequence obeys the following second order algebraic equation: x^7*(4*x^3 - 1)^5*g(x)^2 - 2*(4*x^3 - 1)^5*(2*x^10 - 2*x^9 - 2*x^7 - 3*x^6 + 4*x^4 + 36*x^3 - 19)*g(x) + (4096*x^28 - 8192*x^27 + 4096*x^26 - 13312*x^25 + 6144*x^24 + 32768*x^23 + 38400*x^22 + 189696*x^21 + 48128*x^20 - 10112*x^19 - 300288*x^18 - 76800*x^17 - 46320*x^16 + 189888*x^15 + 27824*x^14 + 52492*x^13 - 65080*x^12 + 1876*x^11 - 24840*x^10 + 13181*x^9 - 2404*x^8 + 6074*x^7 - 1512*x^6 + 296*x^5 - 756*x^4 + 76*x^3 + 38*x)=0.
FORMULA
Let K(n) equal shortest path length. Then for n>3 K(n)=2[(n+1)/3], where [x]=floor(x). a(n)=2*(K-2)*binomial(K-1,K/2-2), if n mod 3 = 0 a(n)=binomial(K,K/2), if n mod 3 = 1 a(n)=(K-2)*(K-3)*binomial(K-2,K/2-1)+2*((K-2)*binomial(K-1,K/2-2)- 2*binomial(K-2,K/2-3))+2*(binomial(K-2,2)*binomial(K-2,K/2-4)- 2*binomial(K-3,K/2-5)), if n mod 3 = 2
In the above expressions binomial(x,y) = x!/(y!(x - y)!). g(x) = ( - (2*x^9 + 3*x^6 - 36*x^3 + 19) + 2*(7*x^9 - 14*x^6 + 7*x^3 - 1)/(1 - 4*x^3)^(1/2) + 2*x^3*(10*x^9 - 36*x^6 + 21*x^3 - 3)/(1 - 4*x^3)^(3/2) + (16*x^15 + 860*x^12 - 1710*x^9 + 1039*x^6 - 250*x^3 + 21)/(1 - 4*x^3)^(5/2))/x^7 + x/(1 - 4*x^3)^(1/2) - 2 + 4/x^3 + 2*x^3 - 2*(2*x^3 - 1)*(9*x^3 - 2)/x^3/(1 - 4*x^3)^(3/2);
EXAMPLE
a(2)=0 because final path point is out of reach,
for n=3 K(3)=4, a(3)=2 (a1-b3-c1-a2-c3) or (a1-c2-a3-b1-c3)
a(4)=2 (a1-b3-d4) or (a1-c2-d4)
CROSSREFS
Sequence in context: A369771 A239677 A331333 * A144816 A134812 A144847
KEYWORD
nonn
AUTHOR
Sergey Perepechko, Jul 02 2006
EXTENSIONS
a(18)-a(39) from Alois P. Heinz, Jan 28 2015
STATUS
approved