[seqfan] Re: Linear Recurrences T(n, k) version of A189418 rhombus count

[Ron Hardin]

Following http://oeis.org/A164897 continues for k=11..14
k=11 Empirical: a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>123
k=12 Empirical: a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>123
k=13 Empirical: a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>171
k=14 Empirical: a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>171

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



----- Original Message ----
> From: Ron Hardin <rhhardin at att.net>
> To: seqfan at list.seqfan.eu
> Sent: Sat, April 23, 2011 8:01:28 PM
> Subject: [seqfan] Linear Recurrences T(n, k) version of A189418 rhombus count
> 
> http://oeis.org/A189418 the  number of rhombi on a nXn grid, if you investigate 
>
> nXk grids, has identical  column recurrences
> 
> T(n,k)=Number of rhombuses on a (n+1)X(k+1)  grid
> Table  starts
> ..1..2...3...4...5...6....7....8....9...10...11...12...13...14...15...16...17
> ..2..6..10..15..20..26...32...39...46...54...62...71...80...90..100..111..122
> ..3.10..22..36..50..66...82..100..120..142..164..188..212..238..264..292..320
> ..4.15..36..66..96.130..164..204..248..296..344..396..448..504..560..620..680
> ..5.20..50..96.151.212..273..344..421..504..587..676..765..860..959.1064.1169
> ..6.26..66.130.212.312..412..527..650..782..914.1059.1204.1358.1520.1691.1862
> ..7.32..82.164.273.412..564..736..918.1112.1306.1520.1734.1960.2198.2448.2698
> ..8.39.100.204.344.527..736..984.1244.1520.1796.2104.2412.2736.3076.3438.3800
> ..9.46.120.248.421.650..918.1244.1601.1978.2355.2776.3197.3638.4103.4600.5097
> .10.54.142.296.504.782.1112.1520.1978.2478.2978.3535.4092.4674.5288.5945.6602
> 
> k=1  Empirical: a(n)=2*a(n-1)-a(n-2)
> k=2 Empirical:  a(n)=2*a(n-1)-2*a(n-3)+a(n-4)
> k=3 Empirical: a(n)=2*a(n-1)-2*a(n-3)+a(n-4)  for n>11
> k=4 Empirical: a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>11
> k=5  Empirical: a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>27
> k=6 Empirical:  a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>27
> k=7 Empirical:  a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>51
> k=8 Empirical:  a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>51
> k=9 Empirical:  a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>83
> k=10 Empirical:  a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>83
> 
> 
> if I've done it right,  they're all the same (after k=1) except for how far out 
>
> you have to go  before the recurrence applies, which seems to be given by 
> http://oeis.org/A164897
> 
> Is there an obvious  explanation?
> 
> 
> rhhardin at mindspring.com
> rhhardin at att.net (either)
> 
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