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A095708 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. 4

%I #34 Aug 21 2023 08:25:22

%S 1,1,1,1,2,5,24,409,16648,2590589,2837017232,14797643031281,

%T 589963307907379136,330879131533072568115765,

%U 1767380481751546168496112185408,144316081261864174415255171573551803529,331532406835855871870984528299440847142815328384

%N 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.

%C Leading order asymptotics of the sequence is log(a(n))~log(2)*n^3/18.

%C In general a(n) is a polynomial in q; here evaluated at the value q=2. For q=1 it is the Somos-4 sequence.

%D B. Grammaticos, F. Nijhoff and A. Ramani, Discrete Painlevé equations, CRM Series in Mathematical Physics, Ed. R. Conte, Springer-Verlag, New York (1999) 413.

%H Robin Visser, <a href="/A095708/b095708.txt">Table of n, a(n) for n = -2..39</a>

%H S. Fomin and A. Zelevinsky, <a href="http://dx.doi.org/10.1006/aama.2001.0770">The Laurent Phenomenon</a>, Advances in Applied Mathematics 28 (2002) 119-144.

%H A. N. W. Hone, <a href="http://dx.doi.org/10.1112/S0024609304004163">Elliptic curves and quadratic recurrence sequences</a>, Bull. Lond. Math. Soc. 37 (2005) 161-171.

%H A. N. W. Hone, <a href="http://arXiv.org/abs/0807.2538">Algebraic curves, integer sequences and a discrete Painlevé transcendent</a>, arXiv:0807.2538 [nlin.SI], 2008; Proceedings of SIDE 6, Helsinki, Finland, 2004.

%F a(n) = (2^(n-2)*a(n-1)*a(n-3) + a(n-2)^2)/a(n-4); a(-2)=a(-1)=a(0)=a(1)=1.

%F 0 = a(n+6)*a(n+2)*a(n+1) - 4*a(n+5)*a(n+4)*a(n) + 4*a(n+5)*a(n+2)*a(n+2) - a(n+4)*a(n+4)*a(n+1) for all n in Z. - _Michael Somos_, Jan 21 2014

%F 0 = a(n+5)*a(n+3)*a(n+1)*a(n+1) - 2*a(n+4)*a(n+4)*a(n+2)*a(n) + 2*a(n+4)*a(n+2)^3 + a(n+3)^3*a(n+1) for all n in Z. - _Michael Somos_, Jan 21 2014

%p t[0]:=1;t[1]:=1;t[ -2]:=1;t[ -1]:=1; alpha:=1;beta:=1; for n from 0 to 12 do t[n+2]:=simplify((alpha*2^n*t[n+1]*t[n-1]+beta*t[n]^2)/t[n-2]): od;

%t nmax = 12; t[-2] = t[-1] = t[0] = t[1] = 1;

%t Do[t[n+2] = (2^n*t[n+1]*t[n-1] + t[n]^2)/t[n-2], {n, 0, nmax}];

%t Table[t[n], {n, -2, nmax}] (* _Jean-François Alcover_, Aug 16 2018, from Maple *)

%Y Cf. A006720, A014125.

%K nonn

%O -2,5

%A _Andrew Hone_, Jul 07 2004

%E More terms from _Robin Visser_, Aug 20 2023

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Last modified April 26 21:53 EDT 2024. Contains 372004 sequences. (Running on oeis4.)