Let's call sequences growing like x^n^2 "supergeometric". The Somos recurrence s[n]:=(s[n-1]*s[n-3]+s[n-2]^2)/s[n-4], for s[1],... = 1,1,1,1,..., grows supergeometrically. But if we initialize with 1,1,-1,-2/3,..., we get a supergeometrically *shrinking* sequence = a period 8 sequence times 3^-(n^2/16): s[n] = -2*%i^(n^2+1)*q^(n^2/64)*theta[1](%i*n*log(q)/8+n*%pi/2,q)/ theta[4](0,q)*3^(n^2/16) 2 2 n + 1 n /64 i n log(q) n pi 2 i q theta (---------- + ----, q) 1 8 2 s = - ------------------------------------------ n 2 n /16 theta (0, q) 3 4 = 1,1,-1,-2/3,1/3,1/9,-1/27,0,1/243,-1/729,-1/2187,2/3^9,... where the simplest equation I've found for q is 3^(1/4)*theta[4](0,q) = -theta[2](0,%i*q^(1/4))*%e^(7*i*%pi/8) 7 i pi ------ 1/4 1/4 8 3 theta (0, q) = - theta (0, i q ) e 4 2 Can we do better? Similarly, the period 8 sequence (1,1,-1,0,1,-1,-1,0)* comes from s[n] = theta[1](n*%pi/4,q)/(theta[2](0,q)*%i^(n^2/4-1)) n pi theta (----, q) 1 4 s = -------------------- n 2 n -- - 1 4 theta (0, q) i 2 with q a root of theta[2](0,q) = theta[2](0,%i*q)*sqrt(%i)/sqrt(2) theta (0, i q) sqrt(i) 2 theta (0, q) = ---------------------- 2 sqrt(2) And then there's the even simpler period 5 sequence (1,1,-1,-1,0)* from s[n] = theta[1](2*n*%pi/5,q)/theta[1](%pi/5,q) 2 n pi theta (------, q) 1 5 s = ----------------- n pi theta (--, q) 1 5 with q defined by theta[1](%pi/5,q) = theta[1](2*n*%pi/5,q) pi 2 n pi theta (--, q) = theta (------, q). 1 5 1 5 This theta is presumably algebraic? That would bode ill for closedform q. The only thetas I know how to invert are algebraic times gamma(s). Of course, all of these can be expressed as a pile of trigs But not when we generalize to 1,1,-1,x,... . Can anyone say why this makes polynomials in x? --rwg