[seqfan] Re: Looking for an interpretation

Ron Hardin rhhardin at att.net
Mon Dec 21 04:03:02 CET 2015


Here's a nX2 binary array consisting only of 2X2 squares with the following 7 NW NE SE SW patterns that has the same recurrence, if not the same series.
It came up at the the 553rd random pattern try.  

Its series though is  7 11 15 20 25 31 37 44 51 59 67 76 85 95 105 116 127 139 151 164 177 191 205 220 235 251 267 284 301 319

PAT="1100:1000:0101:1111:0000:1101:1010";

Some solutions for n=7:
..1..1....1..1....1..1....1..0....1..1....1..1....1..1....1..1....1..1....0..1
..1..1....1..1....1..1....0..1....1..1....1..1....1..1....1..1....1..1....1..0
..1..1....1..0....1..1....1..0....1..1....1..1....1..0....1..1....1..1....0..1
..1..0....0..0....1..1....0..0....1..1....1..1....0..1....1..1....1..1....1..0
..0..1....0..0....1..0....0..0....0..0....1..1....1..0....1..1....1..0....0..1
..1..0....0..0....0..1....0..0....0..0....1..1....0..0....1..1....0..1....1..0
..0..1....0..0....1..0....0..0....0..0....1..1....0..0....1..0....1..0....0..0
..1..0....0..0....0..0....0..0....0..0....0..0....0..0....0..1....0..1....0..0

Empirical a(n)=2*a(n-1)-2*a(n-3)+a(n-4)
 rhhardin at mindspring.com rhhardin at att.net (either)
 

 
      From: Ron Hardin <rhhardin at att.net>
 To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu> 
 Sent: Friday, December 18, 2015 6:48 AM
 Subject: [seqfan] Re: Looking for an interpretation
   
a(n)=2*a(n-1)-2*a(n-3)+a(n-4) for n>5, apparently.
 rhhardin at mindspring.com rhhardin at att.net (either) 

 
      From: Peter Luschny <peter.luschny at gmail.com>
 To: "seqfan at list.seqfan.eu" <seqfan at list.seqfan.eu> 
 Sent: Friday, December 18, 2015 5:06 AM
 Subject: [seqfan] Looking for an interpretation
  
Consider the algorithm:

x, y = 1, 2
repeat:
  print x
  x, y = x + y, x//y + 1

The assignment is simultaneous and x//y means floor(x/y).
This leads to:

1, 3, 4, 8, 10, 15, 18, 24, 28, 35, 40, 48, 54, 63, 70, ...

Can you contribute a combinatorial interpretation of this sequence?

Cheers, Peter

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