login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A196810 Number of ways to place 2 nonattacking nightriders on an n X n cylindrical board 3
0, 4, 18, 80, 200, 420, 756, 1472, 2358, 3860, 5500, 8304, 11232, 15484, 21090, 27392, 34816, 44604, 55404, 69840, 84294, 102124, 122452, 147264, 173800, 203476, 237762, 276752, 318304, 368340, 418500, 478208, 541398, 611524, 689780, 774576, 863136, 963148 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

A nightrider is a fairy chess piece that can move (proportionate to how a knight moves) in any direction.

REFERENCES

S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-queens problem I. General theory, January 26, 2013; http://www.math.binghamton.edu/zaslav/Tpapers/qq1.pdf. - From N. J. A. Sloane, Feb 16 2013

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-queens problem, arXiv:1303.1879 [math.CO]

V. Kotesovec, Non-attacking chess pieces

FORMULA

G.f.: -(2*x^2*(2 + 17*x + 96*x^2 + 384*x^3 + 1203*x^4 + 3100*x^5 + 6917*x^6 + 13670*x^7 + 24466*x^8 + 39974*x^9 + 60206*x^10 + 83709*x^11 + 107667*x^12 + 128088*x^13 + 141070*x^14 + 143882*x^15 + 136037*x^16 + 119239*x^17 + 96892*x^18 + 72808*x^19 + 50428*x^20 + 31926*x^21 + 18321*x^22 + 9388*x^23 + 4223*x^24 + 1622*x^25 + 514*x^26 + 127*x^27 + 22*x^28 + 2*x^29))/((-1+x)^5*(1+x)^3*(1+x^2)^3*(1+x+x^2)^3*(1+x+x^2+x^3+x^4)^3).

Recurrence: a(n) = a(n-32) + 4*a(n-31) + 10*a(n-30) + 17*a(n-29) + 20*a(n-28) + 11*a(n-27) - 15*a(n-26) - 54*a(n-25) - 90*a(n-24) - 99*a(n-23) - 63*a(n-22) + 18*a(n-21) + 116*a(n-20) + 188*a(n-19) + 194*a(n-18) + 123*a(n-17) - 123*a(n-15) - 194*a(n-14) - 188*a(n-13) - 116*a(n-12) - 18*a(n-11) + 63*a(n-10) + 99*a(n-9) + 90*a(n-8) + 54*a(n-7) + 15*a(n-6) - 11*a(n-5) - 20*a(n-4) - 17*a(n-3) - 10*a(n-2) - 4*a(n-1).

Explicit formula: a(n) = -n/4+(572*n^2)/225-(3*n^3)/2+n^4/2+(-1)^n*(n/4+n^2/2)+1/2*n^2*cos((n*Pi)/2)+16/25*n^2*cos((4*n*Pi)/5)+4/9*n^2*cos((4*n*Pi)/3)+16/25*n^2*cos((8*n*Pi)/5).

Chaiken et al. give a 4th degree quasi-polynomial formula. - N. J. A. Sloane, Feb 16 2013

Note that cited formula is for normal chessboard (not cylindrical), see sequence A172141. - Vaclav Kotesovec, Dec 09 2013

MATHEMATICA

Table[(143*n^2)/30-(79*n^3)/15+n^4/2+16/5*n^2*Floor[n/5]+n^2*Floor[n/4]+4/3*n^2*Floor[n/3]+(n+2*n^2)*Floor[n/2]+8/5*n^2*Floor[(1+n)/5]+n^2*Floor[(1+n)/4]+2/3*n^2*Floor[(1+n)/3]+8/5*n^2*Floor[(2+n)/5]+8/5*n^2*Floor[(3+n)/5], {n, 1, 100}]

CROSSREFS

Cf. A172141, A196812.

Sequence in context: A240342 A208309 A112619 * A177755 A037965 A045902

Adjacent sequences:  A196807 A196808 A196809 * A196811 A196812 A196813

KEYWORD

nonn,easy

AUTHOR

Vaclav Kotesovec, Oct 06 2011

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified September 21 00:17 EDT 2019. Contains 327252 sequences. (Running on oeis4.)