Fomin, Sergey; Zelevinsky, Andrei The Laurent phenomenon. (English) Zbl 1012.05012 Adv. Appl. Math. 28, No. 2, 119-144 (2002). Summary: A composition of birational maps given by Laurent polynomials need not be given by Laurent polynomials; however, sometimes – quite unexpectedly – it does. We suggest a unified treatment of this phenomenon, which covers a large class of applications. In particular, we settle in the affirmative a conjecture of D. Gale and R. Robinson on integrality of generalized Somos sequences, and prove the Laurent property for several multidimensional recurrences, confirming conjectures by J. Propp, N. Elkies, and M. Kleber. Cited in 9 ReviewsCited in 146 Documents MSC: 05A15 Exact enumeration problems, generating functions Keywords:Gale-Robinson conjecture; birational maps; Laurent polynomials; Somos sequences × Cite Format Result Cite Review PDF Full Text: DOI arXiv Online Encyclopedia of Integer Sequences: Somos-4 sequence: a(0)=a(1)=a(2)=a(3)=1; for n >= 4, a(n) = (a(n-1) * a(n-3) + a(n-2)^2) / a(n-4). Somos-5 sequence: a(n) = (a(n-1) * a(n-4) + a(n-2) * a(n-3)) / a(n-5), with a(0) = a(1) = a(2) = a(3) = a(4) = 1. Somos-6 sequence: a(n) = (a(n-1) * a(n-5) + a(n-2) * a(n-4) + a(n-3)^2) / a(n-6), a(0) = ... = a(5) = 1. Somos-7 sequence: a(n) = (a(n-1) * a(n-6) + a(n-2) * a(n-5) + a(n-3) * a(n-4)) / a(n-7), a(0) = ... = a(6) = 1. Bisection of A001400. a(n+1) = (a(n)^2 + a(n-1)^2)/a(n-2), with a(1) = a(2) = a(3) = 1. a(1) = a(2) = a(3) = a(4) = 1; a(n) = (a(n-1)*a(n-3) + a(n-2)^4)/a(n-4). Tau-functions of the q-discrete Painlevé I equation, f(n+1) = (A*q^n*f(n) + B)/(f(n)^2*f(n-1)), for q=2 and A=B=1, with f(n) = a(n+1)*a(n-1)/a(n)^2. Generalized Somos-4 sequence with a(n-2)^2 replaced by a(n-2)^5. a(n+2) = (a(n+1)^3 + a(n+1)^2)/a(n) with a(0)=1, a(1)=1. a(n+2) = (a(n+1)^3 + a(n+1))/a(n) with a(0)=1, a(1)=1. a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)*a(n-2)+a(n-1)*a(n-3)+a(n-2)*a(n-3))/a(n-4) with four initial ones. a(n) where a(n) * a(n-5) * a(n-10) = a(n-1) * a(n-6) * a(n-8) + a(n-2) * a(n-4) * a(n-9), with a(1) = ... = a(10) = 1. a(n) = (a(n-1)*a(n-2)^2+1)/a(n-3) with a(0)=a(1)=a(2)=1. a(n)=(a(n-1)*a(n-2)^5+1)/a(n-3) with a(0)=a(1)=a(2)=1. a(n)=(a(n-1)^2*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1. a(n)=(a(n-1)^3*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1. a(n)=(a(n-1)^4*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1. a(n) = (a(n-1)^2*a(n-2)^2 + 1)/a(n-3) with a(0)=a(1)=a(2)=1. a(n)=(a(n-1)^2*a(n-2)^4+1)/a(n-3) with a(0)=a(1)=a(2)=1. a(n) = (a(n-1)^2*a(n-2)^5+1)/a(n-3) with a(0)=a(1)=a(2)=1. a(n)=(a(n-1)^3*a(n-2)^2+1)/a(n-3) with a(0)=a(1)=a(2)=1. a(n)=(a(n-1)^3*a(n-2)^3+1)/a(n-3) with a(0)=a(1)=a(2)=1. a(n)=(a(n-1)^2*a(n-3)+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n)=(a(n-1)^3*a(n-3)+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n)=(a(n-1)*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n)=(a(n-1)^2*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n) = (a(n-1)^3*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n) = (a(n-1)*a(n-3)^3+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n)=(a(n-1)^2*a(n-3)^3+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n)=(a(n-1)^3*a(n-3)^3+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n)=(a(n-1)*a(n-3)^4+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n) = (a(n-1)^2*a(n-3)^4+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n)=(a(n-1)^3*a(n-3)^4+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1. a(n) = (a(n-1)+1)*(a(n-3)+1)/a(n-4) for n > 3, a(0) = a(1) = a(2) = a(3) = 1. 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