[seqfan] Re: Linear recurrences with coefficients 1

[Ron Hardin]

Putting together a computation of length n+4 arrays with no 3*two terms=2*three terms gives
30 58 112 216 416 802 1546 2980 5744 11072 21342 41138 79296 152848 294624 567906 1094674
which matches http://oeis.org/search?q=id:A135492 which has recurrence
a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4)

I wonder what it is to get an order 5 recurrence....

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


>________________________________
> From: Ron Hardin <rhhardin at att.net>
>To: "seqfan at list.seqfan.eu" <seqfan at list.seqfan.eu> 
>Sent: Monday, October 20, 2014 3:56 PM
>Subject: [seqfan] Linear recurrences with coefficients 1
> 
>
>This just came up.  Without looking into it yet, I wonder if there's a class of combinatorial problems varied from this for various length recurrences
>
>/tmp/ekf
>T(n,k)=Number of length n+6 0..k arrays with no seven consecutive terms having four times the sum of any three elements equal to three times the sum of the remaining four
>
>Table starts
>...126....1792......14126.......66948.......240870........708106........1809164
>...250....4586......49730......290032......1249890.......4264120.......12523980
>...496...11874.....175616.....1257084......6487112......25682744.......86727506
>...984...30876.....620648.....5448778.....33673760.....154701784......600707602
>..1952...80354....2194096....23617364....174811880.....931904308.....4161333538
>..3872..208876....7758744...102366154....907556600....5613806736....28830275130
>..7680..541624...27446576...443680650...4711821902...33817920836...199757196902
>.15234.1400008...97142002..1922960684..24462953912..203721491688..1384167006986
>.30218.3618986..343823110..8334332160.127008089170.1227234165196..9592120648562
>.59940.9363890.1217189036.36123881064.659412923000.7392963064454.66472910762904
>
>Empirical for column k:
>k=1: a(n)=a(n-1)+a(n-2)+a(n-3)+a(n-4)+a(n-5)+a(n-6)
>
>
>
>rhhardin at mindspring.com
>rhhardin at att.net (either)
>
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