[seqfan] Closed form from recurrence? Isosceles right triangles on a grid

[Ron Hardin]

T(n,k)=Number of isosceles right triangles on a (n+1)X(k+1) grid

Table starts
..4..10..16..22...28...34...40...46...52...58...64...70...76....82....88....94
.10..28..50..74...98..122..146..170..194..218..242..266..290...314...338...362
.16..50..96.150..208..268..328..388..448..508..568..628..688...748...808...868
.22..74.150.244..350..464..582..702..822..942.1062.1182.1302..1422..1542..1662
.28..98.208.350..516..700..896.1100.1308.1518.1728.1938.2148..2358..2568..2778
.34.122.268.464..700..968.1260.1570.1892.2222.2556.2892.3228..3564..3900..4236
.40.146.328.582..896.1260.1664.2100.2560.3038.3528.4026.4528..5032..5536..6040
.46.170.388.702.1100.1570.2100.2680.3300.3952.4628.5322.6028..6742..7460..8180
.52.194.448.822.1308.1892.2560.3300.4100.4950.5840.6762.7708..8672..9648.10632
.58.218.508.942.1518.2222.3038.3952.4950.6020.7150.8330.9550.10802.12078.13372


Empirical: a(n) = k*(k+1)*(k+2)*n + b(k) for n>2*k-2
k=1: a(n) = 6*n - 2
k=2: a(n) = 24*n - 22 for n>2
k=3: a(n) = 60*n - 92 for n>4
k=4: a(n) = 120*n - 258 for n>6
k=5: a(n) = 210*n - 582 for n>8
k=6: a(n) = 336*n - 1140 for n>10
k=7: a(n) = 504*n - 2024 for n>12
k=8: a(n) = 720*n - 3340 for n>14
k=9: a(n) = 990*n - 5210 for n>16
k=10: a(n) = 1320*n - 7770 for n>18
k=11: a(n) = 1716*n - 11172 for n>20
k=12: a(n) = 2184*n - 15582 for n>22
k=13: a(n) = 2730*n - 21182 for n>24
k=14: a(n) = 3360*n - 28168 for n>26

This is k*(k+1)*(k+2)*n - b(k) for some b(k), and if b(k) is known then T(n,k) 
is known at least outside the k,2*k-2 boundary.

Computing the value of T(2*k-1,k)=T(k,2*k-1) to get at b(k)

Number of isosceles right triangles on a 2nX(n+1) grid
a(n)=4 50 208 582 1308 2556 4528 7460 11620 17310 24864 34650 47068 62552 81568 
104616 132228 164970 203440 248270

Empirical: a(n)=4*a(n-1)-5*a(n-2)+5*a(n-4)-4*a(n-5)+a(n-6)

The question is what's a closed form for a(n).


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