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Short communication Self-avoiding walks on Z f0, 1g Nikolai Nikolov Institute of Mathematics and Informatics, Bulgarian Academy of Sciences, Acad. G. Bonchev 8, 1113 Sofia, Bulgaria article info Article history: Received 25 May 2011 Accepted 22 July 2011 Available online 28 July 2011 Keywords: Self-avoiding walks abstract A short and simple proof of the formula for the number of the self-avoiding walks of length n on Z f0, 1g starting at the origin is given. & 2011 Elsevier B.V. All rights reserved. Let c n be the number of the self-avoiding walks of length n on Z f0, 1g starting at the origin. A formula for c n in terms of the Fibonacci numbers f n (f 1 ¼ f 2 ¼ 1 and f n ¼ f n1 þ f n2 for n Z3) can be found in Zeilberger (1996). The proof there uses generating functions. A combinatorial proof is given in Benjamin (2006). The purpose of this note is to obtain the formula for c n in a short and simple way. We shall need only the formulas for the first n odd and first n even Fibonacci numbers (which follow by induction): f 1 þ f 3 þþ f 2n1 ¼ f 2n , f 2 þ f 4 þþ f 2n ¼ f 2n þ 1 1: ð1Þ Proposition 1. Let c n be the number of the sequences A 0 , A 1 , ... , A n of pairwise different points in the plane with integer first coordinates and second coordinates 0 or 1 such that A 0 ¼ð0, 0Þ and 9x A i x A i1 9 þ 9y A i y A i1 9 ¼ 1 for 1 ri rn. Then c n ¼ 8f n e n for n Z2, where e n ¼ n for even n and e n ¼ 4 for odd n. Proof. Define the number d n similar to c n requiring the coordinates of the points to be non-negative; we shall call the respective sequences right. Denote by e n the number of the right sequences such that x A i x A i1 þ 9y A i y A i1 9 ¼ 1, 1 ri rn; call these sequences ultra-right and the other right sequences middle right. It is not difficult to see that e 0 ¼ 1, e 1 ¼ 2 and e i þ 2 ¼ e i þ 1 þ e i (to any ultra-right sequence A 0 , A 1 , ... , A i þ 2 associate A 0 , A 1 , ... , A i þ 1 if x A i þ 2 ¼ x A i þ 1 , and A 0 , A 1 , ... , A i otherwise). It follows that e n is ðn þ 2Þ-th Fibonacci number f n þ 2 . We associate to any middle right sequence A 0 , A 1 , ... , A n (n Z3) the ultra-right sequence A 0 , A 1 , ... , A k for which x k ¼ x n , k rn3. It is clear that x k þ 1 ¼ x k þ 1; we shall call such sequences super right. Note that the number of the super right sequences of length k is equal to e k1 ¼ f k þ 1 ð2Þ (this remains true fir k ¼ 0). There is an obvious bijection between the set of the middle right sequences of length n and the set of the super right sequences with lengths n3, n5, .... It follows by (2) and (1) that d n ¼ e n þ e n4 þ e n6 þ¼ 2f n þ 1 d n , ð3Þ Contents lists available at ScienceDirect journal homepage: www.elsevier.com/locate/jspi Journal of Statistical Planning and Inference 0378-3758/$ - see front matter & 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.jspi.2011.07.018 E-mail address: [email protected] Journal of Statistical Planning and Inference 142 (2012) 376–377

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Page 1: Self-avoiding walks on

Contents lists available at ScienceDirect

Journal of Statistical Planning and Inference

Journal of Statistical Planning and Inference 142 (2012) 376–377

0378-37

doi:10.1

E-m

journal homepage: www.elsevier.com/locate/jspi

Short communication

Self-avoiding walks on Z� f0,1g

Nikolai Nikolov

Institute of Mathematics and Informatics, Bulgarian Academy of Sciences, Acad. G. Bonchev 8, 1113 Sofia, Bulgaria

a r t i c l e i n f o

Article history:

Received 25 May 2011

Accepted 22 July 2011Available online 28 July 2011

Keywords:

Self-avoiding walks

58/$ - see front matter & 2011 Elsevier B.V. A

016/j.jspi.2011.07.018

ail address: [email protected]

a b s t r a c t

A short and simple proof of the formula for the number of the self-avoiding walks of

length n on Z� f0,1g starting at the origin is given.

& 2011 Elsevier B.V. All rights reserved.

Let cn be the number of the self-avoiding walks of length n on Z� f0,1g starting at the origin. A formula for cn in termsof the Fibonacci numbers fn (f1 ¼ f2 ¼ 1 and fn ¼ fn�1þ fn�2 for nZ3) can be found in Zeilberger (1996). The proof there usesgenerating functions. A combinatorial proof is given in Benjamin (2006). The purpose of this note is to obtain the formulafor cn in a short and simple way. We shall need only the formulas for the first n odd and first n even Fibonacci numbers(which follow by induction):

f1þ f3þ � � � þ f2n�1 ¼ f2n, f2þ f4þ � � � þ f2n ¼ f2nþ1�1: ð1Þ

Proposition 1. Let cn be the number of the sequences A0,A1, . . . ,An of pairwise different points in the plane with integer first

coordinates and second coordinates 0 or 1 such that A0 ¼ ð0,0Þ and 9xAi�xAi�1

9þ9yAi�yAi�1

9¼ 1 for 1r irn. Then cn ¼ 8fn�en for

nZ2, where en ¼ n for even n and en ¼ 4 for odd n.

Proof. Define the number dn similar to cn requiring the coordinates of the points to be non-negative; we shall call therespective sequences right. Denote by en the number of the right sequences such that

xAi�xAi�1

þ9yAi�yAi�1

9¼ 1, 1r irn;

call these sequences ultra-right and the other right sequences middle right. It is not difficult to see that e0 ¼ 1, e1 ¼ 2 andeiþ2 ¼ eiþ1þei (to any ultra-right sequence A0,A1, . . . ,Aiþ2 associate A0,A1, . . . ,Aiþ1 if xAiþ 2

¼ xAiþ 1, and A0,A1, . . . ,Ai

otherwise). It follows that en is ðnþ2Þ-th Fibonacci number fnþ2. We associate to any middle right sequenceA0,A1, . . . ,An (nZ3) the ultra-right sequence A0,A1, . . . ,Ak for which xk ¼ xn, krn�3. It is clear that xkþ1 ¼ xkþ1; we shallcall such sequences super right. Note that the number of the super right sequences of length k is equal to

ek�1 ¼ fkþ1 ð2Þ

(this remains true fir k¼0). There is an obvious bijection between the set of the middle right sequences of length n and theset of the super right sequences with lengths n�3,n�5, . . .. It follows by (2) and (1) that

dn ¼ enþen�4þen�6þ � � � ¼ 2fnþ1�dn, ð3Þ

ll rights reserved.

Page 2: Self-avoiding walks on

N. Nikolov / Journal of Statistical Planning and Inference 142 (2012) 376–377 377

where dn ¼ 1 for even n and dn ¼ 0 for odd n. Further, it is easy to see that

cn ¼ 2ðdnþgnÞ, nZ4, ð4Þ

where gn is the number of the sequences A0, . . . ,A2kþ1,A2kþ2, . . .An such that Aj ¼ ð�j,0Þ and Akþ1þ j ¼ ðj�k,1Þ for 0r jrk,and A2kþ2, . . . ,An is a ‘‘right’’ sequence starting at (1,1) (more precisely, the first coordinates are positive). Then

gn ¼ dn�4þdn�6þ � � � ,

and (3) and (1) imply that gn ¼ 2fn�2�n=2þ1 for even n and gn ¼ 2fn�2�2 for odd n. It remains to substitute this in (4), touse (3) again, and to check the formula for cn when n¼2,3 (note that c1 ¼ 3¼ 8f1�5). &

References

Benjamin, A.T., 2006. Self-avoiding walks and Fibonacci numbers. Fibonacci Quart. 44, 330–334.Zeilberger, D., 1996. Self-avoiding walks, the language of science, and Fibonacci numbers. J. Statist. Plann. Inference 54, 135–138.