

A199044


The number of identity elements of length n in Z*Z^2.


1



1, 0, 6, 0, 74, 0, 1140, 0, 19562, 0, 357756, 0, 6824684, 0, 134166696, 0, 2697855082, 0, 55213424556, 0, 1146078241284, 0, 24067465856088, 0, 510351502965548, 0, 10911807871502232, 0, 234970037988773560, 0, 5091149074269149520, 0, 110912377099411850090, 0
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OFFSET

0,3


COMMENTS

Z*Z^2 is the free product of the free group on one letter (say, x) and the free abelian group on two letters (say, y and z).
Viewed as the quotient of the free group F on three letters {x,y,z} by the normal subgroup generated by the commutator [y,z], the sequence gives the number of words in F of length n that are sent to the identity in Z*Z^2 under the quotient map.
Note that oddnumbered terms are zero.


REFERENCES

Derek F. Holt, Sarah Rees, Claas E. Röver, and Richard M. Thomas, Groups with ContextFree CoWord Problem, J. London Math. Soc. (2005) 71 (3): 643657. doi: 10.1112/S002461070500654X
Brough, Tara Rose, Groups with polycontextfree word problem, PhD thesis (2010), University of Warwick.


LINKS



EXAMPLE

The identity from the free group F maps to the identity in Z*Z^2, and is the only word of length zero in F, so a(0)=1.
The group Z*Z^2 maps onto the direct product C_2^3, the group of exponent 2 with 8 elements. Therefore no elements of odd length are sent to the identity and thus a(2i1)=0 for all positive integers i.
The only word of length zero is the empty word, which vacuously represents the identity. Therefore, a_0=1.
For n=2, there are a_2=6 identities; each is a (positive or negative) generator x,y, or z, followed or preceded by its inverse. We have the words x*x^1, y*y^1, z*z^1, plus the reverse of each.


CROSSREFS



KEYWORD

nonn,word


AUTHOR



EXTENSIONS

Edited by Nick Loughlin, Mar 12 2012


STATUS

approved



