# [seqfan] Triangular rooks in triangular array and polynomials

Ron Hardin rhhardin at att.net
Tue Aug 16 20:47:58 CEST 2011

```Number of ways to arrange k points in an nXnXn triangular array with fewer than
z points in any row or diagonal (empirical)

z=2 k=1 : a(n) = (1/2)*n^2 + (1/2)*n
z=3 k=1 : a(n) = (1/2)*n^2 + (1/2)*n
z=4 k=1 : a(n) = (1/2)*n^2 + (1/2)*n

z=2 k=2 : a(n) = (1/8)*n^4 - (1/4)*n^3 - (1/8)*n^2 + (1/4)*n
z=3 k=2 : a(n) = (1/8)*n^4 + (1/4)*n^3 - (1/8)*n^2 - (1/4)*n
z=4 k=2 : a(n) = (1/8)*n^4 + (1/4)*n^3 - (1/8)*n^2 - (1/4)*n

z=3 k=3 : a(n) = (1/48)*n^6 + (1/16)*n^5 - (3/16)*n^4 + (1/48)*n^3 + (1/6)*n^2 -
(1/12)*n
z=4 k=3 : a(n) = (1/48)*n^6 + (1/16)*n^5 - (1/16)*n^4 - (11/48)*n^3 + (1/24)*n^2
+ (1/6)*n

z=3 k=4 : a(n) = (1/384)*n^8 + (1/96)*n^7 - (5/64)*n^6 + (13/240)*n^5 +
(27/128)*n^4 - (23/96)*n^3 - (13/96)*n^2 + (7/40)*n
z=4 k=4 : a(n) = (1/384)*n^8 + (1/96)*n^7 - (1/64)*n^6 - (13/120)*n^5 +
(19/128)*n^4 + (7/96)*n^3 - (13/96)*n^2 + (1/40)*n

z=3 k=5 : a(n) = (1/3840)*n^10 + (1/768)*n^9 - (7/384)*n^8 + (37/1920)*n^7 +
(737/3840)*n^6 - (2347/3840)*n^5 + (101/192)*n^4 + (93/320)*n^3 - (7/10)*n^2 +
(3/10)*n
z=4 k=5 : a(n) = (1/3840)*n^10 + (1/768)*n^9 - (1/384)*n^8 - (59/1920)*n^7 +
(281/3840)*n^6 + (149/3840)*n^5 - (5/24)*n^4 + (29/320)*n^3 + (11/80)*n^2 -
(1/10)*n

z=4 k=6 : a(n) = (1/46080)*n^12 + (1/7680)*n^11 - (1/3072)*n^10 -
(137/23040)*n^9 + (871/46080)*n^8 + (3107/161280)*n^7 - (5573/46080)*n^6 +
(1157/23040)*n^5 + (2627/11520)*n^4 - (1121/5760)*n^3 - (181/1440)*n^2 +
(11/84)*n

z=4 k=7 : a(n) = (1/645120)*n^14 + (1/92160)*n^13 - (1/30720)*n^12 -
(79/92160)*n^11 + (101/30720)*n^10 + (757/129024)*n^9 - (3049/92160)*n^8 -
(34099/645120)*n^7 + (6613/15360)*n^6 - (16859/23040)*n^5 + (1043/3840)*n^4 +
(2759/5040)*n^3 - (753/1120)*n^2 + (13/56)*n

The results are polynomial (order n^(2k)) only up to some k limit, with k limit
increasing with z, which is the interesting (or strange) result.

The upper k limit for z=4 is not known (haven't got enough terms to tell if
k=8.. are polynomial).

rhhardin at mindspring.com
rhhardin at att.net (either)

```