[seqfan] Recurrences of Periodic Order

[Ron Hardin]

Take a free (n+1)X(k+1) binary array y(i,j)

Form a nXk 0..3 array with element x(i,j)=y(i,j)+y(i+1,j)+y(i,j+1)
from the sum of the same element, the element east of it, and the element south of it.

The number T(n,k) of arrays x appears to have recurrences by column of periodic order 4,6,6,4,6,6,...

/tmp/djg
T(n,k)=Number of nXk 0..3 arrays of the sum of the corresponding element, the element to the east and the element to the south in a larger (n+1)X(k+1) 0..1 array

Table starts
....4......14.........48..........164............560..............1912
...14.....128.......1064.........8592..........68672............546752
...48....1064......19124.......319340........5212236..........84210828
..164....8592.....319340.....10624396......345788172.......11156280332
..560...68672....5212236....345788172....22494002188.....1451228983308
.1912..546752...84210828..11156280332..1451228983308...187243393024012
.6528.4346752.1353901580.358453456908.93250181644300.24062262212427788

Empirical for column k:
k=1: a(n)=4*a(n-1)-2*a(n-2)
k=2: a(n)=12*a(n-1)-36*a(n-2)+32*a(n-3)-16*a(n-4)
k=3: a(n)=25*a(n-1)-152*a(n-2)+144*a(n-3)-368*a(n-4)+1888*a(n-5)-1536*a(n-6) for n>9
k=4: a(n)=49*a(n-1)-560*a(n-2)+544*a(n-3)-1568*a(n-4)+17920*a(n-5)-16384*a(n-6) for n>9
k=5: a(n)=101*a(n-1)-2532*a(n-2)+10624*a(n-3)-8192*a(n-4) for n>7
k=6: a(n)=193*a(n-1)-8384*a(n-2)+8320*a(n-3)-24704*a(n-4)+1073152*a(n-5)-1048576*a(n-6) for n>9
k=7: a(n)=385*a(n-1)-33152*a(n-2)+33024*a(n-3)-98560*a(n-4)+8486912*a(n-5)-8388608*a(n-6) for n>9
k=8: a(n)=777*a(n-1)-137992*a(n-2)+1185792*a(n-3)-1048576*a(n-4) for n>7
k=9: a(n)=1537*a(n-1)-525824*a(n-2)+525312*a(n-3)-1573888*a(n-4)+538443776*a(n-5)-536870912*a(n-6) for n>9
k=10: a(n)=3073*a(n-1)-2100224*a(n-2)+2099200*a(n-3)-6293504*a(n-4)+4301258752*a(n-5)-4294967296*a(n-6) for n>9
k=11: a(n)=6161*a(n-1)-8493072*a(n-2)+142704640*a(n-3)-134217728*a(n-4) for n>7



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