"Recurrence equations:" "I(N+1) = I(N)+J(N) +L(N)" "J(N+1) = I(N)" "K(N+1) = J(N)+K(N)" "L(N+1) = K(N)" "M(N+1) = L(N)+2M(N)" "" "Initial conditions:" "I(0) = 1" "J(0) = 0" "K(0) = 0" "L(0) = 0" "M(0) = 0" "" "Values of the equations for the first few N's:" " N = 0 1 2 3 4 5 6 7 8 ... to infinity" "I(N) = 1 1 2 3 6 11 21 39 73 ..." "J(N) = 0 1 1 2 3 6 11 21 39 ..." "K(N) = 0 0 1 2 4 7 13 24 45 ..." "L(N) = 0 0 0 1 2 4 7 13 24 ..." "M(N) = 0 0 0 0 1 4 12 31 75 ..." "" "This can be solved using a simple iterates function." "Here is an example of the of the iterates function." "NEWTON(u, x, x0, n) := ITERATES(x - u/DIF(u, x), x, x0, n)" "First a vector v is defined so v=[n,i,j,k,l,m]." "initial value of v is v0:=[0,1,0,0,0,0]." v0:=[0,1,0,0,0,0] "The recurrence equations:" FRED_I(v):=v SUB 2+v SUB 3+v SUB 5 FRED_J(v):=v SUB 2 FRED_K(v):=v SUB 3+v SUB 4 FRED_L(v):=v SUB 4 FRED_M(v):=v SUB 5+2*v SUB 6 "This can be one equation." FRED_AUX(v):=[v SUB 1+1,FRED_I(v),FRED_J(v),FRED_K(v),FRED_L(v),FRED_M(v)] "This is a solution:" FRED(n):=ITERATES(FRED_AUX(v_),v_,v0,n) "Test the it with n=8." FRED(8) ;Simp(#40) [[0,1,0,0,0,0],[1,1,1,0,0,0],[2,2,1,1,0,0],[3,3,2,2,1,0],[4,6,3,4,2,1],[5,11,~ 6,7,4,4],[6,21,11,13,7,12],[7,39,21,24,13,31],[8,73,39,45,24,75]] "There are simplier solutions. This is one example." FRED(16)` SUB 5 ;Simp(User) [0,0,0,1,2,4,7,13,24,45,84,157,293,547,1021,1906,3558] "LOOKUP 1 2 4 7 13 24 45 84 157 293 547 1021 1906 3558" FRED(22)` SUB 5 ;Simp(#46) [0,0,0,1,2,4,7,13,24,45,84,157,293,547,1021,1906,3558,6642,12399,23146,43208,~ 80659,150571] [6642,12399,23146,43208,80659,150571] "1+(x^4-x^3+2*x-1)*F(x) = 0" ;Simp(#36') FRED_AUX(v):=[v SUB 1+1,v SUB 5+v SUB 3+v SUB 2,v SUB 2,v SUB 4+v SUB 3,v SUB~ 4,2*v SUB 6+v SUB 5] FRED1_AUX(v):=[v SUB 4+v SUB 2+v SUB 1,v SUB 1,v SUB 3+v SUB 2,v SUB 3] FRED1(n):=ITERATES(FRED1_AUX(v_),v_,[1,0,0,0],n) FRED1(22)` SUB 4 ;Simp(#53) [0,0,0,1,2,4,7,13,24,45,84,157,293,547,1021,1906,3558,6642,12399,23146,43208,~ 80659,150571]