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Number of equivalence classes of compositions of n into parts of size 2 and 3 under the following equivalence relation: We make a "move" by changing three consecutive 2s into two consecutive 3s or vice versa. Two compositions are equivalent if we can reach one from the other by a series of moves.
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%I #18 Jan 08 2018 01:47:46

%S 1,0,1,1,1,2,1,3,2,3,4,3,6,4,7,7,7,11,8,14,12,15,19,16,26,21,30,32,32,

%T 46,38,57,54,63,79,71,104,93,121,134,135,184,165,226,228,257,319,301,

%U 411,394,484,548,559,731,696,896,943,1044,1280,1256,1628,1640,1941,2224,2301,2909

%N Number of equivalence classes of compositions of n into parts of size 2 and 3 under the following equivalence relation: We make a "move" by changing three consecutive 2s into two consecutive 3s or vice versa. Two compositions are equivalent if we can reach one from the other by a series of moves.

%C Sequence A133925 counts the equivalence classes with exactly one element.

%H Vincenzo Librandi, <a href="/A133926/b133926.txt">Table of n, a(n) for n = 0..1000</a>

%H Problem posed on the Art of Problem Solving forum, <a href="http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180422">String Replacement</a>

%F a(n) = a(n-2) + a(n-3) - a(n-6). - _Franklin T. Adams-Watters_, Oct 12 2013

%F G.f.: 1/(1-x^2-x^3+x^6). [_Joerg Arndt_, Oct 12 2013]

%e a(5) = 2 because the two compositions 23 and 32 are inequivalent. a(6) = 1 because the two compositions 222 and 33 are equivalent.

%t a=b=c=d=e=0;Delete[Table[z=a+b+c-e+1;a=b;b=c;c=d;d=e;e=z,{n,100}],{{1},{2}}] (* _Vladimir Joseph Stephan Orlovsky_, Feb 27 2011 *)

%o (PARI) Vec( 1/(1-x^2-x^3+x^6) +O(x^66) ) \\ _Joerg Arndt_, Oct 12 2013

%K easy,nonn

%O 0,6

%A _Joel B. Lewis_, Jan 07 2008, Jan 23 2008

%E More terms from _Joerg Arndt_, Oct 12 2013