by Michael Somos 11 Apr 1999 Given integers { k , p , q , r } such that 6 <= k , 0 < p < q < r <= k/2 . Given variables { x , y , z } , define a sequence by a(1) = a(2) = ... = a(k) = 1 , and x * a(n-p)*a(n-k+p) + y * a(n-q)*a(n-k+q) + z * a(n-r)*a(n-k+r) a(n) = --------------------------------------------------------------- . a(n-k) By definition, a(n) is a rational function in variables { x , y , z } with non-negative integer coefficients. Conjecture: a(n) is always a polynomial if and only if p+q = r or k-r . It first fails to be a polynomial at a(n) when the denominator a(n-k) = x+y+z or x(x+y+z)+y+z , and when k+1 <= n-k <= k+1+p . end. References: Richard K. Guy, Unsolved Problems in Number Theory, 2nd edition, page 215. "E15 ... Raphael Robinson has observed ... x_nx_{n-k}=ax_{n-p}x_{n-k+p}+ bx_{n-q}x_{n-k+q}+cx_{n-r}x_{n-k+r} ... appears to generate integers from the starting values ... a>=0,b>=0,c>=0, p>=1,q>=1,r>=1, p+q+r=k." David Gale, The Mathematical Intelligencer, Vol 13, No. 1, 1991, page 42. "... Conjecture: For any p, q, r < k the recursion a_n a_{n-k} = xa_{n-p}a_{n-k+p} + ya_{n-q}a_{n-k+q} + za_{n-r}a_{n-k+r} (7) generates integers if and only if p, q, r can be chosen so that p+q+r=k. (Robinson's evidence is only for the case x = y = z = 1. The arbitrary x, y, z are my responsibility.)"

Further Conjecture by Michael Somos 12 Apr 1999 : Given integers { k, p, q, r, s } such that 8 <= k , 0 < p < q < r < s <= k/2 . Given variables { x, y, z, w } , define a sequence by a(1) = a(2) = ... = a(k) = 1 , and x*a(n-p)*a(n-k+p)+y*a(n-q)*a(n-k+q)+z*a(n-r)*a(n-k+r)+w*a(n-s)*a(n-k+s) a(n)=-----------------------------------------------------------------------. a(n-k) Conjecture: a(n) always fails to be a polynomial at a(n) when the denominator a(n-k) = x+y+z+w or x^2+(y+z+w)*x+y+z+w and when k+1 <= n-k <= k+1+p . end. Comment: The Somos-4, Somos-5, Somos-6, Somos-7 sequences are all special cases of the Robinson sequence conjecture. After that, his sequences produce integers and mine don't.

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