4 4 A Linear Recurrence Scheme for the Constant Term of , P[k](z) P[k](1/z) Where P[k](z) is the Rudin-Shapiro polynomial By Shalosh B. Ekhad Let , P[k](z), be the Rudin-Shapiro polynomial, that may be defined by the recurrence 2 2 P[k](z) = P[k - 1](z ) + z P[k - 1](-z ) and the initial condition P[0](z)=1 c1 c2 c3 c4 c5 Definition: For any monomial w, in a,b,A,B,z,, a A b B z , let c1 c2 c3 c4 c5 E[a A b B z ](k), be the coefficient of z^0 in the polynomial c1 c2 c3 c4 c5 P[k](z) P[k](1/z) P[k](-z) P[k](- 1/z) z 4 4 We are interested in getting a scheme for computing the sequence, E[a A ](k) 4 4 Let's express, E[a A ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 4 3 3 2 2 4 E[a A ](k) = -32 E[B A b a z](k - 1) - 12 E[B A a z](k - 1) 4 4 4 4 2 4 2 2 + 2 E[a A ](k - 1) + 2 E[B a z ](k - 1) + 12 E[A a b z](k - 1) 3 3 2 2 2 2 + 32 E[A B a b](k - 1) + 36 E[A B a b ](k - 1) 4 4 Note that, in the left side, the following monomials, a A , are already treated, 4 4 2 2 2 2 2 3 3 but we have to handle the new arrivals, B a z , A B a b , A B a b, 4 2 2 2 2 4 3 3 A a b z, B A a z, B A b a z, . Note that we still have to do handle the monomials in the set, 4 4 2 2 2 2 2 3 3 4 2 2 2 2 4 3 3 {B a z , A B a b , A B a b, A a b z, B A a z, B A b a z} 4 4 2 Let's express, E[B a z ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 4 2 3 3 3 2 3 E[B a z ](k) = -16 E[A B a b](k - 1) - 16 E[B A z b a ](k - 1) 4 4 3 4 4 2 2 4 - E[B a z ](k - 1) - E[a B z](k - 1) + 6 E[A B a ](k - 1) 4 2 2 4 2 2 2 2 2 2 4 + 6 E[A a b ](k - 1) + 6 E[A a b z ](k - 1) + 6 E[B A z a ](k - 1) 2 2 2 2 + 36 E[A B a b z](k - 1) 2 2 4 4 2 2 4 4 3 4 4 We have to handle the new arrivals, A B a , A a b , B a z , a B z, 3 3 4 2 2 2 2 2 2 4 2 2 2 2 3 2 3 A B a b, A a b z , B A z a , A B a b z, B A z b a 2 2 4 Note that we still have to do handle the monomials in the set, {A B a , 4 2 2 4 4 3 4 4 3 3 2 2 2 2 3 3 4 2 2 A a b , B a z , a B z, A B a b, A B a b , A B a b, A a b z, 4 2 2 2 2 2 4 2 2 2 4 2 2 2 2 3 3 A a b z , B A a z, B A z a , A B a b z, B A b a z, 3 2 3 B A z b a } 2 2 4 Let's express, E[A B a ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 2 2 4 2 2 2 2 2 2 4 E[A B a ](k) = -12 E[A B a b ](k - 1) - 4 E[B A a z](k - 1) 4 4 4 4 2 4 2 2 + 2 E[a A ](k - 1) + 2 E[B a z ](k - 1) + 4 E[A a b z](k - 1) 4 4 4 4 2 Note that, in the left side, the following monomials, a A , B a z , are already treated, 2 2 2 2 4 2 2 2 2 4 but we have to handle the new arrivals, A B a b , A a b z, B A a z, . 4 2 2 Note that we still have to do handle the monomials in the set, {A a b , 4 4 3 4 4 3 3 2 2 2 2 3 3 4 2 2 B a z , a B z, A B a b, A B a b , A B a b, A a b z, 4 2 2 2 2 2 4 2 2 2 4 2 2 2 2 3 3 A a b z , B A a z, B A z a , A B a b z, B A b a z, 3 2 3 B A z b a } 4 2 2 Let's express, E[A a b ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 2 2 2 2 2 2 2 2 4 E[A a b ](k) = -12 E[A B a b ](k - 1) - 4 E[B A a z](k - 1) 4 4 4 4 2 4 2 2 + 2 E[a A ](k - 1) + 2 E[B a z ](k - 1) + 4 E[A a b z](k - 1) 4 4 4 4 2 Note that, in the left side, the following monomials, a A , B a z , are already treated, 2 2 2 2 4 2 2 2 2 4 but we have to handle the new arrivals, A B a b , A a b z, B A a z, . 4 4 3 Note that we still have to do handle the monomials in the set, {B a z , 4 4 3 3 2 2 2 2 3 3 4 2 2 4 2 2 2 a B z, A B a b, A B a b , A B a b, A a b z, A a b z , 2 2 4 2 2 2 4 2 2 2 2 3 3 3 2 3 B A a z, B A z a , A B a b z, B A b a z, B A z b a } 4 4 3 Let's express, E[B a z ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 4 3 3 2 2 2 3 4 E[B a z ](k) = -24 E[A B a b z ](k - 1) - 4 E[A B a ](k - 1) 4 3 4 3 3 3 4 - 4 E[A a b z](k - 1) - 4 E[B b a z ](k - 1) - 4 E[z A B a ](k - 1) 2 3 4 4 3 4 3 2 - 4 E[z A B a ](k - 1) + 4 E[B a b](k - 1) + 4 E[A a b z ](k - 1) 3 3 4 3 2 2 + 4 E[B A z a ](k - 1) + 24 E[A B z a b ](k - 1) 2 2 3 2 2 2 3 + 24 E[z A B a b](k - 1) + 24 E[z A B a b](k - 1) 3 4 4 3 4 3 4 3 2 We have to handle the new arrivals, A B a , B a b, A a b z, A a b z , 3 3 4 4 3 3 3 4 2 3 4 3 2 2 2 3 2 2 B A z a , B b a z , z A B a , z A B a , A B a b z , A B z a b , 2 2 3 2 2 2 3 z A B a b, z A B a b 3 4 Note that we still have to do handle the monomials in the set, {A B a , 4 3 4 4 3 3 2 2 2 2 3 3 4 2 2 B a b, a B z, A B a b, A B a b , A B a b, A a b z, 4 2 2 2 4 3 4 3 2 2 2 4 2 2 2 4 3 3 4 A a b z , A a b z, A a b z , B A a z, B A z a , B A z a , 4 3 3 3 4 2 3 4 2 2 2 2 3 2 2 2 B b a z , z A B a , z A B a , A B a b z, A B a b z , 3 2 2 3 3 3 2 3 2 2 3 2 2 2 3 A B z a b , B A b a z, B A z b a , z A B a b, z A B a b} 3 4 Let's express, E[A B a ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 4 4 2 2 2 2 4 E[A B a ](k) = 6 E[A a b z](k - 1) + 6 E[B A a z](k - 1) 4 2 2 2 2 4 We have to handle the new arrivals, A a b z, B A a z 4 3 Note that we still have to do handle the monomials in the set, {B a b, 4 4 3 3 2 2 2 2 3 3 4 2 2 4 2 2 2 a B z, A B a b, A B a b , A B a b, A a b z, A a b z , 4 3 4 3 2 2 2 4 2 2 2 4 3 3 4 4 3 3 A a b z, A a b z , B A a z, B A z a , B A z a , B b a z , 3 4 2 3 4 2 2 2 2 3 2 2 2 3 2 2 z A B a , z A B a , A B a b z, A B a b z , A B z a b , 3 3 3 2 3 2 2 3 2 2 2 3 B A b a z, B A z b a , z A B a b, z A B a b} 4 3 Let's express, E[B a b](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 3 4 2 2 2 2 4 E[B a b](k) = 6 E[A a b z](k - 1) + 6 E[B A a z](k - 1) 4 2 2 2 2 4 We have to handle the new arrivals, A a b z, B A a z 4 4 Note that we still have to do handle the monomials in the set, {a B z, 3 3 2 2 2 2 3 3 4 2 2 4 2 2 2 4 3 A B a b, A B a b , A B a b, A a b z, A a b z , A a b z, 4 3 2 2 2 4 2 2 2 4 3 3 4 4 3 3 3 4 A a b z , B A a z, B A z a , B A z a , B b a z , z A B a , 2 3 4 2 2 2 2 3 2 2 2 3 2 2 3 3 z A B a , A B a b z, A B a b z , A B z a b , B A b a z, 3 2 3 2 2 3 2 2 2 3 B A z b a , z A B a b, z A B a b} 4 4 Let's express, E[a B z](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 4 3 2 2 3 2 2 E[a B z](k) = -24 E[A B a b ](k - 1) - 24 E[A B z a b ](k - 1) 2 2 3 3 4 3 4 2 - 24 E[z A B a b](k - 1) - 4 E[A B a ](k - 1) - 4 E[B A a z ](k - 1) 4 3 4 3 4 3 - 4 E[z a B A](k - 1) + 4 E[A a b](k - 1) + 4 E[A a b z](k - 1) 4 3 2 3 4 3 4 + 4 E[B b a z ](k - 1) + 4 E[z A B a ](k - 1) + 4 E[z b a B ](k - 1) 2 2 3 + 24 E[A B a b](k - 1) 3 4 4 3 2 2 3 3 2 2 We have to handle the new arrivals, A B a , A a b, A B a b, A B a b , 4 3 3 4 2 4 3 2 3 4 4 3 3 4 A a b z, B A a z , B b a z , z A B a , z a B A, z b a B , 3 2 2 2 2 3 A B z a b , z A B a b 3 4 Note that we still have to do handle the monomials in the set, {A B a , 4 3 3 3 2 2 2 2 2 2 3 3 2 2 3 3 A a b, A B a b, A B a b , A B a b, A B a b , A B a b, 4 2 2 4 2 2 2 4 3 4 3 2 2 2 4 2 2 2 4 A a b z, A a b z , A a b z, A a b z , B A a z, B A z a , 3 4 2 3 3 4 4 3 2 4 3 3 3 4 4 3 B A a z , B A z a , B b a z , B b a z , z A B a , z a B A, 3 4 2 3 4 2 2 2 2 3 2 2 2 3 2 2 z b a B , z A B a , A B a b z, A B a b z , A B z a b , 3 3 3 2 3 2 2 3 2 2 2 3 B A b a z, B A z b a , z A B a b, z A B a b} 3 4 Let's express, E[A B a ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 4 4 2 2 2 2 4 E[A B a ](k) = 6 E[A a b z](k - 1) + 6 E[B A a z](k - 1) 4 2 2 2 2 4 We have to handle the new arrivals, A a b z, B A a z 4 3 Note that we still have to do handle the monomials in the set, {A a b, 3 3 2 2 2 2 2 2 3 3 2 2 3 3 4 2 2 A B a b, A B a b , A B a b, A B a b , A B a b, A a b z, 4 2 2 2 4 3 4 3 2 2 2 4 2 2 2 4 3 4 2 A a b z , A a b z, A a b z , B A a z, B A z a , B A a z , 3 3 4 4 3 2 4 3 3 3 4 4 3 3 4 B A z a , B b a z , B b a z , z A B a , z a B A, z b a B , 2 3 4 2 2 2 2 3 2 2 2 3 2 2 3 3 z A B a , A B a b z, A B a b z , A B z a b , B A b a z, 3 2 3 2 2 3 2 2 2 3 B A z b a , z A B a b, z A B a b} 4 3 Let's express, E[A a b](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 3 4 2 2 2 2 4 E[A a b](k) = 6 E[A a b z](k - 1) + 6 E[B A a z](k - 1) 4 2 2 2 2 4 We have to handle the new arrivals, A a b z, B A a z 3 3 Note that we still have to do handle the monomials in the set, {A B a b, 2 2 2 2 2 2 3 3 2 2 3 3 4 2 2 4 2 2 2 A B a b , A B a b, A B a b , A B a b, A a b z, A a b z , 4 3 4 3 2 2 2 4 2 2 2 4 3 4 2 3 3 4 A a b z, A a b z , B A a z, B A z a , B A a z , B A z a , 4 3 2 4 3 3 3 4 4 3 3 4 2 3 4 B b a z , B b a z , z A B a , z a B A, z b a B , z A B a , 2 2 2 2 3 2 2 2 3 2 2 3 3 3 2 3 A B a b z, A B a b z , A B z a b , B A b a z, B A z b a , 2 2 3 2 2 2 3 z A B a b, z A B a b} 3 3 Let's express, E[A B a b](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 3 3 3 3 3 E[A B a b](k) = -8 E[A B a b](k - 1) - 8 E[B A b a z](k - 1) 4 4 2 4 4 - 2 E[B a z ](k - 1) + 2 E[a A ](k - 1) 4 4 4 4 2 Note that, in the left side, the following monomials, a A , B a z , are already treated, 3 3 3 3 but we have to handle the new arrivals, A B a b, B A b a z, . 2 2 2 2 Note that we still have to do handle the monomials in the set, {A B a b , 2 2 3 3 2 2 3 3 4 2 2 4 2 2 2 4 3 A B a b, A B a b , A B a b, A a b z, A a b z , A a b z, 4 3 2 2 2 4 2 2 2 4 3 4 2 3 3 4 4 3 2 A a b z , B A a z, B A z a , B A a z , B A z a , B b a z , 4 3 3 3 4 4 3 3 4 2 3 4 2 2 2 2 B b a z , z A B a , z a B A, z b a B , z A B a , A B a b z, 3 2 2 2 3 2 2 3 3 3 2 3 2 2 3 A B a b z , A B z a b , B A b a z, B A z b a , z A B a b, 2 2 2 3 z A B a b} 2 2 2 2 Let's express, E[A B a b ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 2 2 2 2 4 2 2 4 4 E[A B a b ](k) = -4 E[A a b z](k - 1) + 2 E[a A ](k - 1) 4 4 2 2 2 2 2 2 2 4 + 2 E[B a z ](k - 1) + 4 E[A B a b ](k - 1) + 4 E[B A a z](k - 1) 4 4 4 4 2 Note that, in the left side, the following monomials, a A , B a z , 2 2 2 2 A B a b , are already treated, 4 2 2 2 2 4 but we have to handle the new arrivals, A a b z, B A a z, . 2 2 3 Note that we still have to do handle the monomials in the set, {A B a b, 3 2 2 3 3 4 2 2 4 2 2 2 4 3 4 3 2 A B a b , A B a b, A a b z, A a b z , A a b z, A a b z , 2 2 4 2 2 2 4 3 4 2 3 3 4 4 3 2 4 3 3 B A a z, B A z a , B A a z , B A z a , B b a z , B b a z , 3 4 4 3 3 4 2 3 4 2 2 2 2 3 2 2 2 z A B a , z a B A, z b a B , z A B a , A B a b z, A B a b z , 3 2 2 3 3 3 2 3 2 2 3 2 2 2 3 A B z a b , B A b a z, B A z b a , z A B a b, z A B a b} 2 2 3 Let's express, E[A B a b](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 2 2 3 4 2 2 2 2 4 E[A B a b](k) = -2 E[A a b z](k - 1) - 2 E[B A a z](k - 1) 4 2 2 2 2 4 We have to handle the new arrivals, A a b z, B A a z 3 2 2 Note that we still have to do handle the monomials in the set, {A B a b , 3 3 4 2 2 4 2 2 2 4 3 4 3 2 2 2 4 A B a b, A a b z, A a b z , A a b z, A a b z , B A a z, 2 2 2 4 3 4 2 3 3 4 4 3 2 4 3 3 3 4 B A z a , B A a z , B A z a , B b a z , B b a z , z A B a , 4 3 3 4 2 3 4 2 2 2 2 3 2 2 2 z a B A, z b a B , z A B a , A B a b z, A B a b z , 3 2 2 3 3 3 2 3 2 2 3 2 2 2 3 A B z a b , B A b a z, B A z b a , z A B a b, z A B a b} 3 2 2 Let's express, E[A B a b ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 2 2 4 2 2 2 2 4 E[A B a b ](k) = -2 E[A a b z](k - 1) - 2 E[B A a z](k - 1) 4 2 2 2 2 4 We have to handle the new arrivals, A a b z, B A a z 3 3 Note that we still have to do handle the monomials in the set, {A B a b, 4 2 2 4 2 2 2 4 3 4 3 2 2 2 4 2 2 2 4 A a b z, A a b z , A a b z, A a b z , B A a z, B A z a , 3 4 2 3 3 4 4 3 2 4 3 3 3 4 4 3 B A a z , B A z a , B b a z , B b a z , z A B a , z a B A, 3 4 2 3 4 2 2 2 2 3 2 2 2 3 2 2 z b a B , z A B a , A B a b z, A B a b z , A B z a b , 3 3 3 2 3 2 2 3 2 2 2 3 B A b a z, B A z b a , z A B a b, z A B a b} 3 3 Let's express, E[A B a b](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 3 4 4 2 4 4 E[A B a b](k) = -2 E[B a z ](k - 1) + 2 E[a A ](k - 1) 3 3 3 3 + 8 E[A B a b](k - 1) + 8 E[B A b a z](k - 1) 4 4 4 4 2 Note that, in the left side, the following monomials, a A , B a z , 3 3 A B a b, are already treated, 3 3 but we have to handle the new arrivals, B A b a z, . 4 2 2 Note that we still have to do handle the monomials in the set, {A a b z, 4 2 2 2 4 3 4 3 2 2 2 4 2 2 2 4 3 4 2 A a b z , A a b z, A a b z , B A a z, B A z a , B A a z , 3 3 4 4 3 2 4 3 3 3 4 4 3 3 4 B A z a , B b a z , B b a z , z A B a , z a B A, z b a B , 2 3 4 2 2 2 2 3 2 2 2 3 2 2 3 3 z A B a , A B a b z, A B a b z , A B z a b , B A b a z, 3 2 3 2 2 3 2 2 2 3 B A z b a , z A B a b, z A B a b} 4 2 2 Let's express, E[A a b z](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 2 2 3 2 2 3 2 2 E[A a b z](k) = -8 E[A B a b ](k - 1) - 8 E[A B z a b ](k - 1) 3 4 3 4 3 4 - 4 E[z A B a ](k - 1) - 4 E[z b a B ](k - 1) + 4 E[A B a ](k - 1) 3 4 2 + 4 E[B A a z ](k - 1) 3 4 3 2 2 Note that, in the left side, the following monomials, A B a , A B a b , are already treated, 3 4 2 3 4 3 4 but we have to handle the new arrivals, B A a z , z A B a , z b a B , 3 2 2 A B z a b , . 4 2 2 2 Note that we still have to do handle the monomials in the set, {A a b z , 4 3 4 3 2 2 2 4 2 2 2 4 3 4 2 3 3 4 A a b z, A a b z , B A a z, B A z a , B A a z , B A z a , 4 3 2 4 3 3 3 4 4 3 3 4 2 3 4 B b a z , B b a z , z A B a , z a B A, z b a B , z A B a , 2 2 2 2 3 2 2 2 3 2 2 3 3 3 2 3 A B a b z, A B a b z , A B z a b , B A b a z, B A z b a , 2 2 3 2 2 2 3 z A B a b, z A B a b} 4 2 2 2 Let's express, E[A a b z ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 2 2 2 2 2 2 2 4 2 2 E[A a b z ](k) = -12 E[A B a b z](k - 1) - 2 E[A a b ](k - 1) 4 2 2 2 4 4 3 4 4 - 2 E[A a b z ](k - 1) - E[B a z ](k - 1) - E[a B z](k - 1) 2 2 4 2 2 2 4 + 6 E[A B a ](k - 1) + 6 E[B A z a ](k - 1) 2 2 4 4 2 2 Note that, in the left side, the following monomials, A B a , A a b , 4 4 3 4 4 4 2 2 2 B a z , a B z, A a b z , are already treated, 2 2 2 4 2 2 2 2 but we have to handle the new arrivals, B A z a , A B a b z, . 4 3 Note that we still have to do handle the monomials in the set, {A a b z, 4 3 2 2 2 4 2 2 2 4 3 4 2 3 3 4 4 3 2 A a b z , B A a z, B A z a , B A a z , B A z a , B b a z , 4 3 3 3 4 4 3 3 4 2 3 4 2 2 2 2 B b a z , z A B a , z a B A, z b a B , z A B a , A B a b z, 3 2 2 2 3 2 2 3 3 3 2 3 2 2 3 A B a b z , A B z a b , B A b a z, B A z b a , z A B a b, 2 2 2 3 z A B a b} 4 3 Let's express, E[A a b z](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 3 3 4 2 3 4 E[A a b z](k) = -4 E[B A a z ](k - 1) - 4 E[z b a B ](k - 1) 4 3 4 3 2 4 3 - 2 E[A a b](k - 1) - 2 E[B b a z ](k - 1) - 2 E[z a B A](k - 1) 4 3 3 4 3 4 + 2 E[A a b z](k - 1) + 4 E[A B a ](k - 1) + 4 E[z A B a ](k - 1) 2 2 3 2 2 3 + 12 E[A B a b](k - 1) + 12 E[z A B a b](k - 1) 3 4 4 3 Note that, in the left side, the following monomials, A B a , A a b, 2 2 3 4 3 A B a b, A a b z, are already treated, 3 4 2 4 3 2 3 4 but we have to handle the new arrivals, B A a z , B b a z , z A B a , 4 3 3 4 2 2 3 z a B A, z b a B , z A B a b, . 4 3 2 Note that we still have to do handle the monomials in the set, {A a b z , 2 2 4 2 2 2 4 3 4 2 3 3 4 4 3 2 4 3 3 B A a z, B A z a , B A a z , B A z a , B b a z , B b a z , 3 4 4 3 3 4 2 3 4 2 2 2 2 3 2 2 2 z A B a , z a B A, z b a B , z A B a , A B a b z, A B a b z , 3 2 2 3 3 3 2 3 2 2 3 2 2 2 3 A B z a b , B A b a z, B A z b a , z A B a b, z A B a b} 4 3 2 Let's express, E[A a b z ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 3 2 3 2 3 2 2 2 4 E[A a b z ](k) = -8 E[B A z b a ](k - 1) - 6 E[B A z a ](k - 1) 4 4 4 4 3 4 4 - E[a B z](k - 1) + E[B a z ](k - 1) + 2 E[A a z](k - 1) 2 2 4 3 3 3 3 + 6 E[A B a ](k - 1) + 8 E[A B a b](k - 1) + 16 E[A B a b z](k - 1) 2 2 4 4 4 3 Note that, in the left side, the following monomials, A B a , B a z , 4 4 3 3 a B z, A B a b, are already treated, 4 4 2 2 2 4 3 3 but we have to handle the new arrivals, A a z, B A z a , A B a b z, 3 2 3 B A z b a , . 4 4 Note that we still have to do handle the monomials in the set, {A a z, 2 2 4 2 2 2 4 3 4 2 3 3 4 4 3 2 4 3 3 B A a z, B A z a , B A a z , B A z a , B b a z , B b a z , 3 4 4 3 3 4 2 3 4 2 2 2 2 3 2 2 2 z A B a , z a B A, z b a B , z A B a , A B a b z, A B a b z , 3 3 3 2 2 3 3 3 2 3 2 2 3 A B a b z, A B z a b , B A b a z, B A z b a , z A B a b, 2 2 2 3 z A B a b} 4 4 Let's express, E[A a z](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 4 2 2 3 3 4 E[A a z](k) = -24 E[z A B a b](k - 1) - 4 E[z A B a ](k - 1) 4 3 3 4 3 4 - 4 E[z a B A](k - 1) - 4 E[z b a B ](k - 1) + 4 E[A B a ](k - 1) 4 3 4 3 3 4 2 + 4 E[A a b](k - 1) + 4 E[A a b z](k - 1) + 4 E[B A a z ](k - 1) 4 3 2 2 2 3 + 4 E[B b a z ](k - 1) + 24 E[A B a b](k - 1) 3 2 2 3 2 2 + 24 E[A B a b ](k - 1) + 24 E[A B z a b ](k - 1) 3 4 4 3 Note that, in the left side, the following monomials, A B a , A a b, 2 2 3 3 2 2 4 3 A B a b, A B a b , A a b z, are already treated, 3 4 2 4 3 2 3 4 but we have to handle the new arrivals, B A a z , B b a z , z A B a , 4 3 3 4 3 2 2 2 2 3 z a B A, z b a B , A B z a b , z A B a b, . 2 2 4 Note that we still have to do handle the monomials in the set, {B A a z, 2 2 2 4 3 4 2 3 3 4 4 3 2 4 3 3 3 4 B A z a , B A a z , B A z a , B b a z , B b a z , z A B a , 4 3 3 4 2 3 4 2 2 2 2 3 2 2 2 3 3 z a B A, z b a B , z A B a , A B a b z, A B a b z , A B a b z, 3 2 2 3 3 3 2 3 2 2 3 2 2 2 3 A B z a b , B A b a z, B A z b a , z A B a b, z A B a b} 2 2 4 Let's express, E[B A a z](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 2 2 4 2 2 3 4 3 E[B A a z](k) = -8 E[A B a b](k - 1) - 4 E[z a B A](k - 1) 4 3 4 3 4 3 2 + 4 E[A a b](k - 1) + 4 E[A a b z](k - 1) + 4 E[B b a z ](k - 1) 2 2 3 + 8 E[z A B a b](k - 1) 4 3 2 2 3 Note that, in the left side, the following monomials, A a b, A B a b, 4 3 A a b z, are already treated, 4 3 2 4 3 2 2 3 but we have to handle the new arrivals, B b a z , z a B A, z A B a b, . 2 2 2 4 Note that we still have to do handle the monomials in the set, {B A z a , 3 4 2 3 3 4 4 3 2 4 3 3 3 4 4 3 B A a z , B A z a , B b a z , B b a z , z A B a , z a B A, 3 4 2 3 4 2 2 2 2 3 2 2 2 3 3 z b a B , z A B a , A B a b z, A B a b z , A B a b z, 3 2 2 3 3 3 2 3 2 2 3 2 2 2 3 A B z a b , B A b a z, B A z b a , z A B a b, z A B a b} 2 2 2 4 Let's express, E[B A z a ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 2 2 2 4 2 2 2 2 2 2 4 E[B A z a ](k) = -12 E[A B a b z](k - 1) - 2 E[A B a ](k - 1) 2 2 2 4 4 4 3 4 4 - 2 E[B A z a ](k - 1) - E[B a z ](k - 1) - E[a B z](k - 1) 4 2 2 4 2 2 2 + 6 E[A a b ](k - 1) + 6 E[A a b z ](k - 1) 2 2 4 4 2 2 Note that, in the left side, the following monomials, A B a , A a b , 4 4 3 4 4 4 2 2 2 2 2 2 4 B a z , a B z, A a b z , B A z a , are already treated, 2 2 2 2 but we have to handle the new arrivals, A B a b z, . 3 4 2 Note that we still have to do handle the monomials in the set, {B A a z , 3 3 4 4 3 2 4 3 3 3 4 4 3 3 4 B A z a , B b a z , B b a z , z A B a , z a B A, z b a B , 2 3 4 2 2 2 2 3 2 2 2 3 3 3 2 2 z A B a , A B a b z, A B a b z , A B a b z, A B z a b , 3 3 3 2 3 2 2 3 2 2 2 3 B A b a z, B A z b a , z A B a b, z A B a b} 3 4 2 Let's express, E[B A a z ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 4 2 3 3 3 2 3 E[B A a z ](k) = -16 E[A B a b z](k - 1) - 8 E[B A z b a ](k - 1) 4 2 2 4 4 3 4 4 - 6 E[A a b ](k - 1) - E[B a z ](k - 1) + E[a B z](k - 1) 4 4 4 2 2 2 3 3 + 2 E[A a z](k - 1) + 6 E[A a b z ](k - 1) + 8 E[A B a b](k - 1) 4 2 2 4 4 Note that, in the left side, the following monomials, A a b , A a z, 4 4 3 4 4 3 3 4 2 2 2 B a z , a B z, A B a b, A a b z , are already treated, 3 3 3 2 3 but we have to handle the new arrivals, A B a b z, B A z b a , . 3 3 4 Note that we still have to do handle the monomials in the set, {B A z a , 4 3 2 4 3 3 3 4 4 3 3 4 2 3 4 B b a z , B b a z , z A B a , z a B A, z b a B , z A B a , 2 2 2 2 3 2 2 2 3 3 3 2 2 3 3 A B a b z, A B a b z , A B a b z, A B z a b , B A b a z, 3 2 3 2 2 3 2 2 2 3 B A z b a , z A B a b, z A B a b} 3 3 4 Let's express, E[B A z a ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 3 4 3 2 2 2 3 2 2 E[B A z a ](k) = -12 E[A B a b z ](k - 1) - 12 E[A B z a b ](k - 1) 4 3 4 3 3 3 4 - 4 E[B a b](k - 1) - 4 E[B b a z ](k - 1) - 2 E[z A B a ](k - 1) 3 4 3 3 4 2 3 4 + 2 E[A B a ](k - 1) + 2 E[B A z a ](k - 1) + 2 E[z A B a ](k - 1) 4 3 4 3 2 + 4 E[A a b z](k - 1) + 4 E[A a b z ](k - 1) 3 4 4 3 Note that, in the left side, the following monomials, A B a , B a b, 4 3 4 3 2 3 3 4 A a b z, A a b z , B A z a , are already treated, 4 3 3 3 4 2 3 4 but we have to handle the new arrivals, B b a z , z A B a , z A B a , 3 2 2 2 3 2 2 A B a b z , A B z a b , . 4 3 2 Note that we still have to do handle the monomials in the set, {B b a z , 4 3 3 3 4 4 3 3 4 2 3 4 2 2 2 2 B b a z , z A B a , z a B A, z b a B , z A B a , A B a b z, 3 2 2 2 3 3 3 2 2 3 3 3 2 3 A B a b z , A B a b z, A B z a b , B A b a z, B A z b a , 2 2 3 2 2 2 3 z A B a b, z A B a b} 4 3 2 Let's express, E[B b a z ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 3 2 3 3 3 3 E[B b a z ](k) = -16 E[A B a b z](k - 1) - 8 E[A B a b](k - 1) 2 2 2 4 4 4 4 4 3 - 6 E[B A z a ](k - 1) - E[a B z](k - 1) + E[B a z ](k - 1) 4 4 2 2 4 3 2 3 + 2 E[A a z](k - 1) + 6 E[A B a ](k - 1) + 8 E[B A z b a ](k - 1) 2 2 4 4 4 Note that, in the left side, the following monomials, A B a , A a z, 4 4 3 4 4 3 3 2 2 2 4 B a z , a B z, A B a b, B A z a , are already treated, 3 3 3 2 3 but we have to handle the new arrivals, A B a b z, B A z b a , . 4 3 3 Note that we still have to do handle the monomials in the set, {B b a z , 3 4 4 3 3 4 2 3 4 2 2 2 2 3 2 2 2 z A B a , z a B A, z b a B , z A B a , A B a b z, A B a b z , 3 3 3 2 2 3 3 3 2 3 2 2 3 A B a b z, A B z a b , B A b a z, B A z b a , z A B a b, 2 2 2 3 z A B a b} 4 3 3 Let's express, E[B b a z ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 3 3 2 2 2 3 3 4 E[B b a z ](k) = -12 E[z A B a b](k - 1) - 4 E[A B a ](k - 1) 3 3 4 3 4 4 3 - 4 E[B A z a ](k - 1) - 4 E[z A B a ](k - 1) + 2 E[B a b](k - 1) 4 3 4 3 2 4 3 3 + 2 E[A a b z](k - 1) + 2 E[A a b z ](k - 1) + 2 E[B b a z ](k - 1) 2 3 4 2 2 3 + 4 E[z A B a ](k - 1) + 12 E[z A B a b](k - 1) 3 4 4 3 Note that, in the left side, the following monomials, A B a , B a b, 4 3 4 3 2 3 3 4 4 3 3 A a b z, A a b z , B A z a , B b a z , are already treated, 3 4 2 3 4 2 2 3 but we have to handle the new arrivals, z A B a , z A B a , z A B a b, 2 2 2 3 z A B a b, . 3 4 Note that we still have to do handle the monomials in the set, {z A B a , 4 3 3 4 2 3 4 2 2 2 2 3 2 2 2 3 3 z a B A, z b a B , z A B a , A B a b z, A B a b z , A B a b z, 3 2 2 3 3 3 2 3 2 2 3 2 2 2 3 A B z a b , B A b a z, B A z b a , z A B a b, z A B a b} 3 4 Let's express, E[z A B a ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 4 3 2 2 4 3 E[z A B a ](k) = -12 E[A B a b ](k - 1) - 4 E[A a b](k - 1) 3 4 3 4 2 3 4 + 2 E[A B a ](k - 1) + 2 E[B A a z ](k - 1) + 2 E[z A B a ](k - 1) 3 4 4 3 4 3 2 + 2 E[z b a B ](k - 1) + 4 E[A a b z](k - 1) + 4 E[B b a z ](k - 1) 4 3 3 2 2 + 4 E[z a B A](k - 1) + 12 E[A B z a b ](k - 1) 3 4 4 3 Note that, in the left side, the following monomials, A B a , A a b, 3 2 2 4 3 3 4 2 4 3 2 3 4 A B a b , A a b z, B A a z , B b a z , z A B a , are already treated, 4 3 3 4 3 2 2 but we have to handle the new arrivals, z a B A, z b a B , A B z a b , . 4 3 Note that we still have to do handle the monomials in the set, {z a B A, 3 4 2 3 4 2 2 2 2 3 2 2 2 3 3 z b a B , z A B a , A B a b z, A B a b z , A B a b z, 3 2 2 3 3 3 2 3 2 2 3 2 2 2 3 A B z a b , B A b a z, B A z b a , z A B a b, z A B a b} 4 3 Let's express, E[z a B A](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 4 3 3 2 2 4 3 E[z a B A](k) = -12 E[A B z a b ](k - 1) - 4 E[A a b](k - 1) 3 4 3 4 2 3 4 - 2 E[A B a ](k - 1) - 2 E[B A a z ](k - 1) - 2 E[z A B a ](k - 1) 3 4 4 3 4 3 2 - 2 E[z b a B ](k - 1) + 4 E[A a b z](k - 1) + 4 E[B b a z ](k - 1) 4 3 3 2 2 + 4 E[z a B A](k - 1) + 12 E[A B a b ](k - 1) 3 4 4 3 Note that, in the left side, the following monomials, A B a , A a b, 3 2 2 4 3 3 4 2 4 3 2 3 4 4 3 A B a b , A a b z, B A a z , B b a z , z A B a , z a B A, are already treated, 3 4 3 2 2 but we have to handle the new arrivals, z b a B , A B z a b , . 3 4 Note that we still have to do handle the monomials in the set, {z b a B , 2 3 4 2 2 2 2 3 2 2 2 3 3 3 2 2 z A B a , A B a b z, A B a b z , A B a b z, A B z a b , 3 3 3 2 3 2 2 3 2 2 2 3 B A b a z, B A z b a , z A B a b, z A B a b} 3 4 Let's express, E[z b a B ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 4 3 4 3 4 E[z b a B ](k) = -4 E[A B a ](k - 1) - 4 E[z A B a ](k - 1) 4 3 4 3 2 4 3 - 2 E[A a b](k - 1) - 2 E[B b a z ](k - 1) - 2 E[z a B A](k - 1) 4 3 3 4 2 3 4 + 2 E[A a b z](k - 1) + 4 E[B A a z ](k - 1) + 4 E[z b a B ](k - 1) 2 2 3 2 2 3 + 12 E[A B a b](k - 1) + 12 E[z A B a b](k - 1) 3 4 4 3 Note that, in the left side, the following monomials, A B a , A a b, 2 2 3 4 3 3 4 2 4 3 2 3 4 4 3 A B a b, A a b z, B A a z , B b a z , z A B a , z a B A, 3 4 z b a B , are already treated, 2 2 3 but we have to handle the new arrivals, z A B a b, . 2 3 4 Note that we still have to do handle the monomials in the set, {z A B a , 2 2 2 2 3 2 2 2 3 3 3 2 2 3 3 A B a b z, A B a b z , A B a b z, A B z a b , B A b a z, 3 2 3 2 2 3 2 2 2 3 B A z b a , z A B a b, z A B a b} 2 3 4 Let's express, E[z A B a ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 2 3 4 3 3 4 2 2 E[z A B a ](k) = -8 E[A B a b](k - 1) - 6 E[A a b ](k - 1) 4 4 3 4 4 4 4 - E[B a z ](k - 1) + E[a B z](k - 1) + 2 E[A a z](k - 1) 4 2 2 2 3 2 3 + 6 E[A a b z ](k - 1) + 8 E[B A z b a ](k - 1) 3 3 + 16 E[A B a b z](k - 1) 4 2 2 4 4 Note that, in the left side, the following monomials, A a b , A a z, 4 4 3 4 4 3 3 4 2 2 2 B a z , a B z, A B a b, A a b z , are already treated, 3 3 3 2 3 but we have to handle the new arrivals, A B a b z, B A z b a , . 2 2 2 2 Note that we still have to do handle the monomials in the set, {A B a b z, 3 2 2 2 3 3 3 2 2 3 3 3 2 3 A B a b z , A B a b z, A B z a b , B A b a z, B A z b a , 2 2 3 2 2 2 3 z A B a b, z A B a b} 2 2 2 2 Let's express, E[A B a b z](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 2 2 2 2 E[A B a b z](k) = 0 We have to handle the new arrivals, 1 Note that we still have to do handle the monomials in the set, {1, 3 2 2 2 3 3 3 2 2 3 3 3 2 3 A B a b z , A B a b z, A B z a b , B A b a z, B A z b a , 2 2 3 2 2 2 3 z A B a b, z A B a b} Let's express, E[1](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences E[1](k) = E[1](k - 1) Note that, in the left side, the following monomials, 1, are already treated, 3 2 2 2 Note that we still have to do handle the monomials in the set, {A B a b z , 3 3 3 2 2 3 3 3 2 3 2 2 3 A B a b z, A B z a b , B A b a z, B A z b a , z A B a b, 2 2 2 3 z A B a b} 3 2 2 2 Let's express, E[A B a b z ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 2 2 2 4 2 2 2 4 4 3 E[A B a b z ](k) = -2 E[A a b z ](k - 1) - E[B a z ](k - 1) 4 4 4 2 2 4 4 + E[a B z](k - 1) + 2 E[A a b ](k - 1) + 2 E[A a z](k - 1) 4 2 2 4 4 Note that, in the left side, the following monomials, A a b , A a z, 4 4 3 4 4 4 2 2 2 B a z , a B z, A a b z , are already treated, 3 3 Note that we still have to do handle the monomials in the set, {A B a b z, 3 2 2 3 3 3 2 3 2 2 3 2 2 2 3 A B z a b , B A b a z, B A z b a , z A B a b, z A B a b} 3 3 Let's express, E[A B a b z](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 3 3 4 2 4 3 2 E[A B a b z](k) = -2 E[B A a z ](k - 1) - 2 E[B b a z ](k - 1) 3 4 3 4 4 3 - 2 E[z A B a ](k - 1) + 2 E[A B a ](k - 1) + 2 E[A a b](k - 1) 4 3 4 3 3 4 + 2 E[A a b z](k - 1) + 2 E[z a B A](k - 1) + 2 E[z b a B ](k - 1) 3 4 4 3 Note that, in the left side, the following monomials, A B a , A a b, 4 3 3 4 2 4 3 2 3 4 4 3 3 4 A a b z, B A a z , B b a z , z A B a , z a B A, z b a B , are already treated, Note that we still have to do handle the monomials in the set, 3 2 2 3 3 3 2 3 2 2 3 2 2 2 3 {A B z a b , B A b a z, B A z b a , z A B a b, z A B a b} 3 2 2 Let's express, E[A B z a b ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 2 2 3 2 2 3 4 E[A B z a b ](k) = -4 E[A B z a b ](k - 1) + 2 E[A B a ](k - 1) 3 4 2 3 4 3 4 + 2 E[B A a z ](k - 1) + 2 E[z A B a ](k - 1) + 2 E[z b a B ](k - 1) 3 2 2 + 4 E[A B a b ](k - 1) 3 4 3 2 2 Note that, in the left side, the following monomials, A B a , A B a b , 3 4 2 3 4 3 4 3 2 2 B A a z , z A B a , z b a B , A B z a b , are already treated, Note that we still have to do handle the monomials in the set, 3 3 3 2 3 2 2 3 2 2 2 3 {B A b a z, B A z b a , z A B a b, z A B a b} 3 3 Let's express, E[B A b a z](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 3 3 4 4 3 2 E[B A b a z](k) = -2 E[A B a ](k - 1) - 2 E[B b a z ](k - 1) 3 4 4 3 4 3 - 2 E[z b a B ](k - 1) + 2 E[A a b](k - 1) + 2 E[A a b z](k - 1) 3 4 2 3 4 4 3 + 2 E[B A a z ](k - 1) + 2 E[z A B a ](k - 1) + 2 E[z a B A](k - 1) 3 4 4 3 Note that, in the left side, the following monomials, A B a , A a b, 4 3 3 4 2 4 3 2 3 4 4 3 3 4 A a b z, B A a z , B b a z , z A B a , z a B A, z b a B , are already treated, Note that we still have to do handle the monomials in the set, 3 2 3 2 2 3 2 2 2 3 {B A z b a , z A B a b, z A B a b} 3 2 3 Let's express, E[B A z b a ](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 3 2 3 4 4 3 4 4 E[B A z b a ](k) = E[B a z ](k - 1) + E[a B z](k - 1) 3 3 3 2 3 + 4 E[A B a b](k - 1) + 4 E[B A z b a ](k - 1) 4 4 3 4 4 Note that, in the left side, the following monomials, B a z , a B z, 3 3 3 2 3 A B a b, B A z b a , are already treated, Note that we still have to do handle the monomials in the set, 2 2 3 2 2 2 3 {z A B a b, z A B a b} 2 2 3 Let's express, E[z A B a b](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 2 2 3 2 2 3 2 2 3 E[z A B a b](k) = -4 E[A B a b](k - 1) - 4 E[z A B a b](k - 1) 4 3 4 3 2 4 3 - 2 E[A a b](k - 1) - 2 E[B b a z ](k - 1) - 2 E[z a B A](k - 1) 4 3 + 2 E[A a b z](k - 1) 4 3 2 2 3 Note that, in the left side, the following monomials, A a b, A B a b, 4 3 4 3 2 4 3 2 2 3 A a b z, B b a z , z a B A, z A B a b, are already treated, 2 2 2 3 Note that we still have to do handle the monomials in the set, {z A B a b} 2 2 2 3 Let's express, E[z A B a b](k), in terms of the values of other, related, sequences at, k - 1 Applying the defining recurrence, and using trivial equivalences 2 2 2 3 2 2 4 4 4 E[z A B a b](k) = -2 E[A B a ](k - 1) - E[a B z](k - 1) 4 4 3 4 4 2 2 2 4 + E[B a z ](k - 1) + 2 E[A a z](k - 1) + 2 E[B A z a ](k - 1) 2 2 4 4 4 Note that, in the left side, the following monomials, A B a , A a z, 4 4 3 4 4 2 2 2 4 B a z , a B z, B A z a , are already treated, Nothing left to do. Summing up we found the following scheme for the sequences 4 4 3 4 2 2 4 3 4 E[1](k), E[a A ](k), E[A B a ](k), E[A B a ](k), E[A B a ](k), 4 2 2 4 3 4 4 4 3 4 4 2 E[A a b ](k), E[A a b](k), E[A a z](k), E[B a b](k), E[B a z ](k), 4 4 3 4 4 3 3 2 2 2 2 E[B a z ](k), E[a B z](k), E[A B a b](k), E[A B a b ](k), 2 2 3 3 2 2 3 3 4 2 2 E[A B a b](k), E[A B a b ](k), E[A B a b](k), E[A a b z](k), 4 2 2 2 4 3 4 3 2 2 2 4 E[A a b z ](k), E[A a b z](k), E[A a b z ](k), E[B A a z](k), 2 2 2 4 3 4 2 3 3 4 4 3 2 E[B A z a ](k), E[B A a z ](k), E[B A z a ](k), E[B b a z ](k), 4 3 3 3 4 4 3 3 4 E[B b a z ](k), E[z A B a ](k), E[z a B A](k), E[z b a B ](k), 2 3 4 2 2 2 2 3 2 2 2 E[z A B a ](k), E[A B a b z](k), E[A B a b z ](k), 3 3 3 2 2 3 3 E[A B a b z](k), E[A B z a b ](k), E[B A b a z](k), 3 2 3 2 2 3 2 2 2 3 E[B A z b a ](k), E[z A B a b](k), E[z A B a b](k) as follows E[1](k) = E[1](k - 1) 4 4 3 3 2 2 4 E[a A ](k) = -32 E[B A b a z](k - 1) - 12 E[B A a z](k - 1) 4 4 4 4 2 4 2 2 + 2 E[a A ](k - 1) + 2 E[B a z ](k - 1) + 12 E[A a b z](k - 1) 3 3 2 2 2 2 + 32 E[A B a b](k - 1) + 36 E[A B a b ](k - 1) 3 4 4 2 2 2 2 4 E[A B a ](k) = 6 E[A a b z](k - 1) + 6 E[B A a z](k - 1) 2 2 4 2 2 2 2 2 2 4 E[A B a ](k) = -12 E[A B a b ](k - 1) - 4 E[B A a z](k - 1) 4 4 4 4 2 4 2 2 + 2 E[a A ](k - 1) + 2 E[B a z ](k - 1) + 4 E[A a b z](k - 1) 3 4 4 2 2 2 2 4 E[A B a ](k) = 6 E[A a b z](k - 1) + 6 E[B A a z](k - 1) 4 2 2 2 2 2 2 2 2 4 E[A a b ](k) = -12 E[A B a b ](k - 1) - 4 E[B A a z](k - 1) 4 4 4 4 2 4 2 2 + 2 E[a A ](k - 1) + 2 E[B a z ](k - 1) + 4 E[A a b z](k - 1) 4 3 4 2 2 2 2 4 E[A a b](k) = 6 E[A a b z](k - 1) + 6 E[B A a z](k - 1) 4 4 2 2 3 3 4 E[A a z](k) = -24 E[z A B a b](k - 1) - 4 E[z A B a ](k - 1) 4 3 3 4 3 4 - 4 E[z a B A](k - 1) - 4 E[z b a B ](k - 1) + 4 E[A B a ](k - 1) 4 3 4 3 3 4 2 + 4 E[A a b](k - 1) + 4 E[A a b z](k - 1) + 4 E[B A a z ](k - 1) 4 3 2 2 2 3 + 4 E[B b a z ](k - 1) + 24 E[A B a b](k - 1) 3 2 2 3 2 2 + 24 E[A B a b ](k - 1) + 24 E[A B z a b ](k - 1) 4 3 4 2 2 2 2 4 E[B a b](k) = 6 E[A a b z](k - 1) + 6 E[B A a z](k - 1) 4 4 2 3 3 3 2 3 E[B a z ](k) = -16 E[A B a b](k - 1) - 16 E[B A z b a ](k - 1) 4 4 3 4 4 2 2 4 - E[B a z ](k - 1) - E[a B z](k - 1) + 6 E[A B a ](k - 1) 4 2 2 4 2 2 2 2 2 2 4 + 6 E[A a b ](k - 1) + 6 E[A a b z ](k - 1) + 6 E[B A z a ](k - 1) 2 2 2 2 + 36 E[A B a b z](k - 1) 4 4 3 3 2 2 2 3 4 E[B a z ](k) = -24 E[A B a b z ](k - 1) - 4 E[A B a ](k - 1) 4 3 4 3 3 3 4 - 4 E[A a b z](k - 1) - 4 E[B b a z ](k - 1) - 4 E[z A B a ](k - 1) 2 3 4 4 3 4 3 2 - 4 E[z A B a ](k - 1) + 4 E[B a b](k - 1) + 4 E[A a b z ](k - 1) 3 3 4 3 2 2 + 4 E[B A z a ](k - 1) + 24 E[A B z a b ](k - 1) 2 2 3 2 2 2 3 + 24 E[z A B a b](k - 1) + 24 E[z A B a b](k - 1) 4 4 3 2 2 3 2 2 E[a B z](k) = -24 E[A B a b ](k - 1) - 24 E[A B z a b ](k - 1) 2 2 3 3 4 3 4 2 - 24 E[z A B a b](k - 1) - 4 E[A B a ](k - 1) - 4 E[B A a z ](k - 1) 4 3 4 3 4 3 - 4 E[z a B A](k - 1) + 4 E[A a b](k - 1) + 4 E[A a b z](k - 1) 4 3 2 3 4 3 4 + 4 E[B b a z ](k - 1) + 4 E[z A B a ](k - 1) + 4 E[z b a B ](k - 1) 2 2 3 + 24 E[A B a b](k - 1) 3 3 3 3 3 3 E[A B a b](k) = -8 E[A B a b](k - 1) - 8 E[B A b a z](k - 1) 4 4 2 4 4 - 2 E[B a z ](k - 1) + 2 E[a A ](k - 1) 2 2 2 2 4 2 2 4 4 E[A B a b ](k) = -4 E[A a b z](k - 1) + 2 E[a A ](k - 1) 4 4 2 2 2 2 2 2 2 4 + 2 E[B a z ](k - 1) + 4 E[A B a b ](k - 1) + 4 E[B A a z](k - 1) 2 2 3 4 2 2 2 2 4 E[A B a b](k) = -2 E[A a b z](k - 1) - 2 E[B A a z](k - 1) 3 2 2 4 2 2 2 2 4 E[A B a b ](k) = -2 E[A a b z](k - 1) - 2 E[B A a z](k - 1) 3 3 4 4 2 4 4 E[A B a b](k) = -2 E[B a z ](k - 1) + 2 E[a A ](k - 1) 3 3 3 3 + 8 E[A B a b](k - 1) + 8 E[B A b a z](k - 1) 4 2 2 3 2 2 3 2 2 E[A a b z](k) = -8 E[A B a b ](k - 1) - 8 E[A B z a b ](k - 1) 3 4 3 4 3 4 - 4 E[z A B a ](k - 1) - 4 E[z b a B ](k - 1) + 4 E[A B a ](k - 1) 3 4 2 + 4 E[B A a z ](k - 1) 4 2 2 2 2 2 2 2 4 2 2 E[A a b z ](k) = -12 E[A B a b z](k - 1) - 2 E[A a b ](k - 1) 4 2 2 2 4 4 3 4 4 - 2 E[A a b z ](k - 1) - E[B a z ](k - 1) - E[a B z](k - 1) 2 2 4 2 2 2 4 + 6 E[A B a ](k - 1) + 6 E[B A z a ](k - 1) 4 3 3 4 2 3 4 E[A a b z](k) = -4 E[B A a z ](k - 1) - 4 E[z b a B ](k - 1) 4 3 4 3 2 4 3 - 2 E[A a b](k - 1) - 2 E[B b a z ](k - 1) - 2 E[z a B A](k - 1) 4 3 3 4 3 4 + 2 E[A a b z](k - 1) + 4 E[A B a ](k - 1) + 4 E[z A B a ](k - 1) 2 2 3 2 2 3 + 12 E[A B a b](k - 1) + 12 E[z A B a b](k - 1) 4 3 2 3 2 3 2 2 2 4 E[A a b z ](k) = -8 E[B A z b a ](k - 1) - 6 E[B A z a ](k - 1) 4 4 4 4 3 4 4 - E[a B z](k - 1) + E[B a z ](k - 1) + 2 E[A a z](k - 1) 2 2 4 3 3 3 3 + 6 E[A B a ](k - 1) + 8 E[A B a b](k - 1) + 16 E[A B a b z](k - 1) 2 2 4 2 2 3 4 3 E[B A a z](k) = -8 E[A B a b](k - 1) - 4 E[z a B A](k - 1) 4 3 4 3 4 3 2 + 4 E[A a b](k - 1) + 4 E[A a b z](k - 1) + 4 E[B b a z ](k - 1) 2 2 3 + 8 E[z A B a b](k - 1) 2 2 2 4 2 2 2 2 2 2 4 E[B A z a ](k) = -12 E[A B a b z](k - 1) - 2 E[A B a ](k - 1) 2 2 2 4 4 4 3 4 4 - 2 E[B A z a ](k - 1) - E[B a z ](k - 1) - E[a B z](k - 1) 4 2 2 4 2 2 2 + 6 E[A a b ](k - 1) + 6 E[A a b z ](k - 1) 3 4 2 3 3 3 2 3 E[B A a z ](k) = -16 E[A B a b z](k - 1) - 8 E[B A z b a ](k - 1) 4 2 2 4 4 3 4 4 - 6 E[A a b ](k - 1) - E[B a z ](k - 1) + E[a B z](k - 1) 4 4 4 2 2 2 3 3 + 2 E[A a z](k - 1) + 6 E[A a b z ](k - 1) + 8 E[A B a b](k - 1) 3 3 4 3 2 2 2 3 2 2 E[B A z a ](k) = -12 E[A B a b z ](k - 1) - 12 E[A B z a b ](k - 1) 4 3 4 3 3 3 4 - 4 E[B a b](k - 1) - 4 E[B b a z ](k - 1) - 2 E[z A B a ](k - 1) 3 4 3 3 4 2 3 4 + 2 E[A B a ](k - 1) + 2 E[B A z a ](k - 1) + 2 E[z A B a ](k - 1) 4 3 4 3 2 + 4 E[A a b z](k - 1) + 4 E[A a b z ](k - 1) 4 3 2 3 3 3 3 E[B b a z ](k) = -16 E[A B a b z](k - 1) - 8 E[A B a b](k - 1) 2 2 2 4 4 4 4 4 3 - 6 E[B A z a ](k - 1) - E[a B z](k - 1) + E[B a z ](k - 1) 4 4 2 2 4 3 2 3 + 2 E[A a z](k - 1) + 6 E[A B a ](k - 1) + 8 E[B A z b a ](k - 1) 4 3 3 2 2 2 3 3 4 E[B b a z ](k) = -12 E[z A B a b](k - 1) - 4 E[A B a ](k - 1) 3 3 4 3 4 4 3 - 4 E[B A z a ](k - 1) - 4 E[z A B a ](k - 1) + 2 E[B a b](k - 1) 4 3 4 3 2 4 3 3 + 2 E[A a b z](k - 1) + 2 E[A a b z ](k - 1) + 2 E[B b a z ](k - 1) 2 3 4 2 2 3 + 4 E[z A B a ](k - 1) + 12 E[z A B a b](k - 1) 3 4 3 2 2 4 3 E[z A B a ](k) = -12 E[A B a b ](k - 1) - 4 E[A a b](k - 1) 3 4 3 4 2 3 4 + 2 E[A B a ](k - 1) + 2 E[B A a z ](k - 1) + 2 E[z A B a ](k - 1) 3 4 4 3 4 3 2 + 2 E[z b a B ](k - 1) + 4 E[A a b z](k - 1) + 4 E[B b a z ](k - 1) 4 3 3 2 2 + 4 E[z a B A](k - 1) + 12 E[A B z a b ](k - 1) 4 3 3 2 2 4 3 E[z a B A](k) = -12 E[A B z a b ](k - 1) - 4 E[A a b](k - 1) 3 4 3 4 2 3 4 - 2 E[A B a ](k - 1) - 2 E[B A a z ](k - 1) - 2 E[z A B a ](k - 1) 3 4 4 3 4 3 2 - 2 E[z b a B ](k - 1) + 4 E[A a b z](k - 1) + 4 E[B b a z ](k - 1) 4 3 3 2 2 + 4 E[z a B A](k - 1) + 12 E[A B a b ](k - 1) 3 4 3 4 3 4 E[z b a B ](k) = -4 E[A B a ](k - 1) - 4 E[z A B a ](k - 1) 4 3 4 3 2 4 3 - 2 E[A a b](k - 1) - 2 E[B b a z ](k - 1) - 2 E[z a B A](k - 1) 4 3 3 4 2 3 4 + 2 E[A a b z](k - 1) + 4 E[B A a z ](k - 1) + 4 E[z b a B ](k - 1) 2 2 3 2 2 3 + 12 E[A B a b](k - 1) + 12 E[z A B a b](k - 1) 2 3 4 3 3 4 2 2 E[z A B a ](k) = -8 E[A B a b](k - 1) - 6 E[A a b ](k - 1) 4 4 3 4 4 4 4 - E[B a z ](k - 1) + E[a B z](k - 1) + 2 E[A a z](k - 1) 4 2 2 2 3 2 3 + 6 E[A a b z ](k - 1) + 8 E[B A z b a ](k - 1) 3 3 + 16 E[A B a b z](k - 1) 2 2 2 2 E[A B a b z](k) = 0 3 2 2 2 4 2 2 2 4 4 3 E[A B a b z ](k) = -2 E[A a b z ](k - 1) - E[B a z ](k - 1) 4 4 4 2 2 4 4 + E[a B z](k - 1) + 2 E[A a b ](k - 1) + 2 E[A a z](k - 1) 3 3 3 4 2 4 3 2 E[A B a b z](k) = -2 E[B A a z ](k - 1) - 2 E[B b a z ](k - 1) 3 4 3 4 4 3 - 2 E[z A B a ](k - 1) + 2 E[A B a ](k - 1) + 2 E[A a b](k - 1) 4 3 4 3 3 4 + 2 E[A a b z](k - 1) + 2 E[z a B A](k - 1) + 2 E[z b a B ](k - 1) 3 2 2 3 2 2 3 4 E[A B z a b ](k) = -4 E[A B z a b ](k - 1) + 2 E[A B a ](k - 1) 3 4 2 3 4 3 4 + 2 E[B A a z ](k - 1) + 2 E[z A B a ](k - 1) + 2 E[z b a B ](k - 1) 3 2 2 + 4 E[A B a b ](k - 1) 3 3 3 4 4 3 2 E[B A b a z](k) = -2 E[A B a ](k - 1) - 2 E[B b a z ](k - 1) 3 4 4 3 4 3 - 2 E[z b a B ](k - 1) + 2 E[A a b](k - 1) + 2 E[A a b z](k - 1) 3 4 2 3 4 4 3 + 2 E[B A a z ](k - 1) + 2 E[z A B a ](k - 1) + 2 E[z a B A](k - 1) 3 2 3 4 4 3 4 4 E[B A z b a ](k) = E[B a z ](k - 1) + E[a B z](k - 1) 3 3 3 2 3 + 4 E[A B a b](k - 1) + 4 E[B A z b a ](k - 1) 2 2 3 2 2 3 2 2 3 E[z A B a b](k) = -4 E[A B a b](k - 1) - 4 E[z A B a b](k - 1) 4 3 4 3 2 4 3 - 2 E[A a b](k - 1) - 2 E[B b a z ](k - 1) - 2 E[z a B A](k - 1) 4 3 + 2 E[A a b z](k - 1) 2 2 2 3 2 2 4 4 4 E[z A B a b](k) = -2 E[A B a ](k - 1) - E[a B z](k - 1) 4 4 3 4 4 2 2 2 4 + E[B a z ](k - 1) + 2 E[A a z](k - 1) + 2 E[B A z a ](k - 1) In order to simplify notation, let 4 4 4 4 2 2 2 4 E[1](k) = E[a A ](k), E[2](k) = E[B a z ](k), E[3](k) = E[A B a ](k), 4 2 2 4 4 3 3 4 E[4](k) = E[A a b ](k), E[5](k) = E[B a z ](k), E[6](k) = E[A B a ](k), 4 3 4 4 3 4 E[7](k) = E[B a b](k), E[8](k) = E[a B z](k), E[9](k) = E[A B a ](k), 4 3 3 3 E[10](k) = E[A a b](k), E[11](k) = E[A B a b](k), 2 2 2 2 2 2 3 E[12](k) = E[A B a b ](k), E[13](k) = E[A B a b](k), 3 2 2 3 3 E[14](k) = E[A B a b ](k), E[15](k) = E[A B a b](k), 4 2 2 4 2 2 2 E[16](k) = E[A a b z](k), E[17](k) = E[A a b z ](k), 4 3 4 3 2 E[18](k) = E[A a b z](k), E[19](k) = E[A a b z ](k), 4 4 2 2 4 E[20](k) = E[A a z](k), E[21](k) = E[B A a z](k), 2 2 2 4 3 4 2 E[22](k) = E[B A z a ](k), E[23](k) = E[B A a z ](k), 3 3 4 4 3 2 E[24](k) = E[B A z a ](k), E[25](k) = E[B b a z ](k), 4 3 3 3 4 E[26](k) = E[B b a z ](k), E[27](k) = E[z A B a ](k), 4 3 3 4 E[28](k) = E[z a B A](k), E[29](k) = E[z b a B ](k), 2 3 4 2 2 2 2 E[30](k) = E[z A B a ](k), E[31](k) = E[A B a b z](k), 3 2 2 2 E[32](k) = E[1](k), E[33](k) = E[A B a b z ](k), 3 3 3 2 2 E[34](k) = E[A B a b z](k), E[35](k) = E[A B z a b ](k), 3 3 3 2 3 E[36](k) = E[B A b a z](k), E[37](k) = E[B A z b a ](k), 2 2 3 2 2 2 3 E[38](k) = E[z A B a b](k), E[39](k) = E[z A B a b](k) Our scheme becomes E[1](k) = -32 E[36](k - 1) - 12 E[21](k - 1) + 2 E[1](k - 1) + 2 E[2](k - 1) + 12 E[16](k - 1) + 32 E[15](k - 1) + 36 E[12](k - 1) E[2](k) = -16 E[11](k - 1) - 16 E[37](k - 1) - E[5](k - 1) - E[8](k - 1) + 6 E[3](k - 1) + 6 E[4](k - 1) + 6 E[17](k - 1) + 6 E[22](k - 1) + 36 E[31](k - 1) E[3](k) = -12 E[12](k - 1) - 4 E[21](k - 1) + 2 E[1](k - 1) + 2 E[2](k - 1) + 4 E[16](k - 1) E[4](k) = -12 E[12](k - 1) - 4 E[21](k - 1) + 2 E[1](k - 1) + 2 E[2](k - 1) + 4 E[16](k - 1) E[5](k) = -24 E[33](k - 1) - 4 E[6](k - 1) - 4 E[18](k - 1) - 4 E[26](k - 1) - 4 E[27](k - 1) - 4 E[30](k - 1) + 4 E[7](k - 1) + 4 E[19](k - 1) + 4 E[24](k - 1) + 24 E[35](k - 1) + 24 E[38](k - 1) + 24 E[39](k - 1) E[6](k) = 6 E[16](k - 1) + 6 E[21](k - 1) E[7](k) = 6 E[16](k - 1) + 6 E[21](k - 1) E[8](k) = -24 E[14](k - 1) - 24 E[35](k - 1) - 24 E[38](k - 1) - 4 E[9](k - 1) - 4 E[23](k - 1) - 4 E[28](k - 1) + 4 E[10](k - 1) + 4 E[18](k - 1) + 4 E[25](k - 1) + 4 E[27](k - 1) + 4 E[29](k - 1) + 24 E[13](k - 1) E[9](k) = 6 E[16](k - 1) + 6 E[21](k - 1) E[10](k) = 6 E[16](k - 1) + 6 E[21](k - 1) E[11](k) = -8 E[15](k - 1) - 8 E[36](k - 1) - 2 E[2](k - 1) + 2 E[1](k - 1) E[12](k) = -4 E[16](k - 1) + 2 E[1](k - 1) + 2 E[2](k - 1) + 4 E[12](k - 1) + 4 E[21](k - 1) E[13](k) = -2 E[16](k - 1) - 2 E[21](k - 1) E[14](k) = -2 E[16](k - 1) - 2 E[21](k - 1) E[15](k) = -2 E[2](k - 1) + 2 E[1](k - 1) + 8 E[15](k - 1) + 8 E[36](k - 1) E[16](k) = -8 E[14](k - 1) - 8 E[35](k - 1) - 4 E[27](k - 1) - 4 E[29](k - 1) + 4 E[9](k - 1) + 4 E[23](k - 1) E[17](k) = -12 E[31](k - 1) - 2 E[4](k - 1) - 2 E[17](k - 1) - E[5](k - 1) - E[8](k - 1) + 6 E[3](k - 1) + 6 E[22](k - 1) E[18](k) = -4 E[23](k - 1) - 4 E[29](k - 1) - 2 E[10](k - 1) - 2 E[25](k - 1) - 2 E[28](k - 1) + 2 E[18](k - 1) + 4 E[9](k - 1) + 4 E[27](k - 1) + 12 E[13](k - 1) + 12 E[38](k - 1) E[19](k) = -8 E[37](k - 1) - 6 E[22](k - 1) - E[8](k - 1) + E[5](k - 1) + 2 E[20](k - 1) + 6 E[3](k - 1) + 8 E[11](k - 1) + 16 E[34](k - 1) E[20](k) = -24 E[38](k - 1) - 4 E[27](k - 1) - 4 E[28](k - 1) - 4 E[29](k - 1) + 4 E[9](k - 1) + 4 E[10](k - 1) + 4 E[18](k - 1) + 4 E[23](k - 1) + 4 E[25](k - 1) + 24 E[13](k - 1) + 24 E[14](k - 1) + 24 E[35](k - 1) E[21](k) = -8 E[13](k - 1) - 4 E[28](k - 1) + 4 E[10](k - 1) + 4 E[18](k - 1) + 4 E[25](k - 1) + 8 E[38](k - 1) E[22](k) = -12 E[31](k - 1) - 2 E[3](k - 1) - 2 E[22](k - 1) - E[5](k - 1) - E[8](k - 1) + 6 E[4](k - 1) + 6 E[17](k - 1) E[23](k) = -16 E[34](k - 1) - 8 E[37](k - 1) - 6 E[4](k - 1) - E[5](k - 1) + E[8](k - 1) + 2 E[20](k - 1) + 6 E[17](k - 1) + 8 E[11](k - 1) E[24](k) = -12 E[33](k - 1) - 12 E[35](k - 1) - 4 E[7](k - 1) - 4 E[26](k - 1) - 2 E[27](k - 1) + 2 E[6](k - 1) + 2 E[24](k - 1) + 2 E[30](k - 1) + 4 E[18](k - 1) + 4 E[19](k - 1) E[25](k) = -16 E[34](k - 1) - 8 E[11](k - 1) - 6 E[22](k - 1) - E[8](k - 1) + E[5](k - 1) + 2 E[20](k - 1) + 6 E[3](k - 1) + 8 E[37](k - 1) E[26](k) = -12 E[39](k - 1) - 4 E[6](k - 1) - 4 E[24](k - 1) - 4 E[27](k - 1) + 2 E[7](k - 1) + 2 E[18](k - 1) + 2 E[19](k - 1) + 2 E[26](k - 1) + 4 E[30](k - 1) + 12 E[38](k - 1) E[27](k) = -12 E[14](k - 1) - 4 E[10](k - 1) + 2 E[9](k - 1) + 2 E[23](k - 1) + 2 E[27](k - 1) + 2 E[29](k - 1) + 4 E[18](k - 1) + 4 E[25](k - 1) + 4 E[28](k - 1) + 12 E[35](k - 1) E[28](k) = -12 E[35](k - 1) - 4 E[10](k - 1) - 2 E[9](k - 1) - 2 E[23](k - 1) - 2 E[27](k - 1) - 2 E[29](k - 1) + 4 E[18](k - 1) + 4 E[25](k - 1) + 4 E[28](k - 1) + 12 E[14](k - 1) E[29](k) = -4 E[9](k - 1) - 4 E[27](k - 1) - 2 E[10](k - 1) - 2 E[25](k - 1) - 2 E[28](k - 1) + 2 E[18](k - 1) + 4 E[23](k - 1) + 4 E[29](k - 1) + 12 E[13](k - 1) + 12 E[38](k - 1) E[30](k) = -8 E[11](k - 1) - 6 E[4](k - 1) - E[5](k - 1) + E[8](k - 1) + 2 E[20](k - 1) + 6 E[17](k - 1) + 8 E[37](k - 1) + 16 E[34](k - 1) E[31](k) = 0 E[32](k) = E[32](k - 1) E[33](k) = -2 E[17](k - 1) - E[5](k - 1) + E[8](k - 1) + 2 E[4](k - 1) + 2 E[20](k - 1) E[34](k) = -2 E[23](k - 1) - 2 E[25](k - 1) - 2 E[27](k - 1) + 2 E[9](k - 1) + 2 E[10](k - 1) + 2 E[18](k - 1) + 2 E[28](k - 1) + 2 E[29](k - 1) E[35](k) = -4 E[35](k - 1) + 2 E[9](k - 1) + 2 E[23](k - 1) + 2 E[27](k - 1) + 2 E[29](k - 1) + 4 E[14](k - 1) E[36](k) = -2 E[9](k - 1) - 2 E[25](k - 1) - 2 E[29](k - 1) + 2 E[10](k - 1) + 2 E[18](k - 1) + 2 E[23](k - 1) + 2 E[27](k - 1) + 2 E[28](k - 1) E[37](k) = E[5](k - 1) + E[8](k - 1) + 4 E[11](k - 1) + 4 E[37](k - 1) E[38](k) = -4 E[13](k - 1) - 4 E[38](k - 1) - 2 E[10](k - 1) - 2 E[25](k - 1) - 2 E[28](k - 1) + 2 E[18](k - 1) E[39](k) = -2 E[3](k - 1) - E[8](k - 1) + E[5](k - 1) + 2 E[20](k - 1) + 2 E[22](k - 1) Subject to the obvious initial conditions E[1](0) = 1, E[2](0) = 0, E[3](0) = 1, E[4](0) = 1, E[5](0) = 0, E[6](0) = 1, E[7](0) = 1, E[8](0) = 0, E[9](0) = 1, E[10](0) = 1, E[11](0) = 1, E[12](0) = 1, E[13](0) = 1, E[14](0) = 1, E[15](0) = 1, E[16](0) = 0, E[17](0) = 0, E[18](0) = 0, E[19](0) = 0, E[20](0) = 0, E[21](0) = 0, E[22](0) = 0, E[23](0) = 0, E[24](0) = 0, E[25](0) = 0, E[26](0) = 0, E[27](0) = 0, E[28](0) = 0, E[29](0) = 0, E[30](0) = 0, E[31](0) = 0, E[32](0) = 1, E[33](0) = 0, E[34](0) = 0, E[35](0) = 0, E[36](0) = 0, E[37](0) = 0, E[38](0) = 0, E[39](0) = 0 Now let's try to find explicit expressions for the (ordinary) generating fun\ ctions, in the variable infinity ----- \ k F[i](t) = ) E[i](k) t / ----- k = 0 For i from 1 to, 39 The above recurrences for the sequences, E[i](k), translate to the following system of linear equations for F[i](t), for i from 1 to, 39 F[1](t) = 1 + t (-32 F[36](t) - 12 F[21](t) + 2 F[1](t) + 2 F[2](t) + 12 F[16](t) + 32 F[15](t) + 36 F[12](t)) F[2](t) = t (-16 F[11](t) - 16 F[37](t) - F[5](t) - F[8](t) + 6 F[3](t) + 6 F[4](t) + 6 F[17](t) + 6 F[22](t) + 36 F[31](t)) F[3](t) = 1 + t (-12 F[12](t) - 4 F[21](t) + 2 F[1](t) + 2 F[2](t) + 4 F[16](t)) F[4](t) = 1 + t (-12 F[12](t) - 4 F[21](t) + 2 F[1](t) + 2 F[2](t) + 4 F[16](t)) F[5](t) = t (-24 F[33](t) - 4 F[6](t) - 4 F[18](t) - 4 F[26](t) - 4 F[27](t) - 4 F[30](t) + 4 F[7](t) + 4 F[19](t) + 4 F[24](t) + 24 F[35](t) + 24 F[38](t) + 24 F[39](t)) F[6](t) = 1 + t (6 F[16](t) + 6 F[21](t)) F[7](t) = 1 + t (6 F[16](t) + 6 F[21](t)) F[8](t) = t (-24 F[14](t) - 24 F[35](t) - 24 F[38](t) - 4 F[9](t) - 4 F[23](t) - 4 F[28](t) + 4 F[10](t) + 4 F[18](t) + 4 F[25](t) + 4 F[27](t) + 4 F[29](t) + 24 F[13](t)) F[9](t) = 1 + t (6 F[16](t) + 6 F[21](t)) F[10](t) = 1 + t (6 F[16](t) + 6 F[21](t)) F[11](t) = 1 + t (-8 F[15](t) - 8 F[36](t) - 2 F[2](t) + 2 F[1](t)) F[12](t) = 1 + t (-4 F[16](t) + 2 F[1](t) + 2 F[2](t) + 4 F[12](t) + 4 F[21](t)) F[13](t) = 1 + t (-2 F[16](t) - 2 F[21](t)) F[14](t) = 1 + t (-2 F[16](t) - 2 F[21](t)) F[15](t) = 1 + t (-2 F[2](t) + 2 F[1](t) + 8 F[15](t) + 8 F[36](t)) F[16](t) = t ( -8 F[14](t) - 8 F[35](t) - 4 F[27](t) - 4 F[29](t) + 4 F[9](t) + 4 F[23](t) ) F[17](t) = t (-12 F[31](t) - 2 F[4](t) - 2 F[17](t) - F[5](t) - F[8](t) + 6 F[3](t) + 6 F[22](t)) F[18](t) = t (-4 F[23](t) - 4 F[29](t) - 2 F[10](t) - 2 F[25](t) - 2 F[28](t) + 2 F[18](t) + 4 F[9](t) + 4 F[27](t) + 12 F[13](t) + 12 F[38](t)) F[19](t) = t (-8 F[37](t) - 6 F[22](t) - F[8](t) + F[5](t) + 2 F[20](t) + 6 F[3](t) + 8 F[11](t) + 16 F[34](t)) F[20](t) = t (-24 F[38](t) - 4 F[27](t) - 4 F[28](t) - 4 F[29](t) + 4 F[9](t) + 4 F[10](t) + 4 F[18](t) + 4 F[23](t) + 4 F[25](t) + 24 F[13](t) + 24 F[14](t) + 24 F[35](t)) F[21](t) = t (-8 F[13](t) - 4 F[28](t) + 4 F[10](t) + 4 F[18](t) + 4 F[25](t) + 8 F[38](t)) F[22](t) = t (-12 F[31](t) - 2 F[3](t) - 2 F[22](t) - F[5](t) - F[8](t) + 6 F[4](t) + 6 F[17](t)) F[23](t) = t (-16 F[34](t) - 8 F[37](t) - 6 F[4](t) - F[5](t) + F[8](t) + 2 F[20](t) + 6 F[17](t) + 8 F[11](t)) F[24](t) = t (-12 F[33](t) - 12 F[35](t) - 4 F[7](t) - 4 F[26](t) - 2 F[27](t) + 2 F[6](t) + 2 F[24](t) + 2 F[30](t) + 4 F[18](t) + 4 F[19](t)) F[25](t) = t (-16 F[34](t) - 8 F[11](t) - 6 F[22](t) - F[8](t) + F[5](t) + 2 F[20](t) + 6 F[3](t) + 8 F[37](t)) F[26](t) = t (-12 F[39](t) - 4 F[6](t) - 4 F[24](t) - 4 F[27](t) + 2 F[7](t) + 2 F[18](t) + 2 F[19](t) + 2 F[26](t) + 4 F[30](t) + 12 F[38](t)) F[27](t) = t (-12 F[14](t) - 4 F[10](t) + 2 F[9](t) + 2 F[23](t) + 2 F[27](t) + 2 F[29](t) + 4 F[18](t) + 4 F[25](t) + 4 F[28](t) + 12 F[35](t)) F[28](t) = t (-12 F[35](t) - 4 F[10](t) - 2 F[9](t) - 2 F[23](t) - 2 F[27](t) - 2 F[29](t) + 4 F[18](t) + 4 F[25](t) + 4 F[28](t) + 12 F[14](t)) F[29](t) = t (-4 F[9](t) - 4 F[27](t) - 2 F[10](t) - 2 F[25](t) - 2 F[28](t) + 2 F[18](t) + 4 F[23](t) + 4 F[29](t) + 12 F[13](t) + 12 F[38](t)) F[30](t) = t (-8 F[11](t) - 6 F[4](t) - F[5](t) + F[8](t) + 2 F[20](t) + 6 F[17](t) + 8 F[37](t) + 16 F[34](t)) F[31](t) = 0 F[32](t) = 1 + t F[32](t) F[33](t) = t (-2 F[17](t) - F[5](t) + F[8](t) + 2 F[4](t) + 2 F[20](t)) F[34](t) = t (-2 F[23](t) - 2 F[25](t) - 2 F[27](t) + 2 F[9](t) + 2 F[10](t) + 2 F[18](t) + 2 F[28](t) + 2 F[29](t)) F[35](t) = t ( -4 F[35](t) + 2 F[9](t) + 2 F[23](t) + 2 F[27](t) + 2 F[29](t) + 4 F[14](t) ) F[36](t) = t (-2 F[9](t) - 2 F[25](t) - 2 F[29](t) + 2 F[10](t) + 2 F[18](t) + 2 F[23](t) + 2 F[27](t) + 2 F[28](t)) F[37](t) = t (F[5](t) + F[8](t) + 4 F[11](t) + 4 F[37](t)) F[38](t) = t (-4 F[13](t) - 4 F[38](t) - 2 F[10](t) - 2 F[25](t) - 2 F[28](t) + 2 F[18](t)) F[39](t) = t (-2 F[3](t) - F[8](t) + F[5](t) + 2 F[20](t) + 2 F[22](t)) Solving this system gives explicit expressions for each of, F[i](t), and in particular, we find that our object of desire 11 10 9 8 F[1](t) = - (90194313216 t - 15300820992 t - 1979711488 t - 292552704 t 7 6 5 4 3 2 - 22216704 t + 10649600 t - 1024 t - 144384 t + 7008 t + 664 t / 10 9 - 54 t - 1) / ((8 t + 1) (16 t - 1) (1409286144 t - 264241152 t / 8 7 6 5 4 3 - 25690112 t - 4128768 t - 311296 t + 170496 t - 2624 t - 2208 t 2 + 148 t + 8 t - 1)) This concludes the article that took, 0.150, seconds to generate.