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

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`%I #30 Mar 24 2017 00:47:54
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`%S 1,0,6,0,74,0,1140,0,19562,0,357756,0,6824684,0,134166696,0,
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`%T 2697855082,0,55213424556,0,1146078241284,0,24067465856088,0,
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`%U 510351502965548,0,10911807871502232,0,234970037988773560,0,5091149074269149520,0,110912377099411850090,0
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`%N The number of identity elements of length n in Z*Z^2.
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`%C 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).
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`%C 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.
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`%C Note that odd-numbered terms are zero.
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`%D Derek F. Holt, Sarah Rees, Claas E. RĂ¶ver, and Richard M. Thomas, Groups with Context-Free Co-Word Problem, J. London Math. Soc. (2005) 71 (3): 643-657. doi: 10.1112/S002461070500654X
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`%D Brough, Tara Rose, Groups with poly-context-free word problem, PhD thesis (2010), University of Warwick.
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`%H Nick Loughlin, <a href="/A199044/b199044.txt">Table of n, a(n) for n = 0..881</a>
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`%e 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.
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`%e 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(2i-1)=0 for all positive integers i.
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`%e The only word of length zero is the empty word, which vacuously represents the identity. Therefore, a_0=1.
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`%e 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.
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`%K nonn,word
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`%O 0,3
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`%A _Nick Loughlin_, Nov 02 2011
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`%E Edited by _Max Alekseyev_, Jan 24 2012
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`%E Edited by Nick Loughlin, Mar 12 2012
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