If the input is: gu:=SymDel(N^2-N-1,n,N,[1,1],[0,1],[1,0,1,1],10,A,B); SymToDelSy(gu,N,A,B, {\ }); the output is [(2*ln(A)-ln(B))/ln(A), N^2-N-1, N+1] the irrationality measure is: 2 -------------------- ------------------------------ for log(b/a) the input is: gu:=SymDel( -(a-b)^2*(1+n)+(3+2*n)*(b+a)*N+(-2-n)*N^2,n,N,[0,2*(b-a)],[1,a+\ b],[1,1,1,1],10,A,B); SymToDelSy(gu,N,A,B,{a,b}); the output is [(2*ln(A)-ln(B))/(ln(A)+1), -N^2+2*N*a+2*N*b-a^2+2*a*b-b^2, -a^2+2*a*b-b^2+N] the irrationality measure is: -2*(-ln(abs(a+b+2*a^(1/2)*b^(1/2)))+ln(abs(a-b)))/(ln(abs(a+b+2*a^(1/2)*b^(1/2) ))+1) -------------------- ------------------------------ for log((x+1)/x) the input is: gu:=SymDel(n+1-(2*x+1)*(3+2*n)*N+(2+n)*N^2,n,N,[0,2],[1,2*x+1],[1,1,1,1],10,\ A,B); SymToDelSy(gu,N,A,B,{x}); the output is [(2*ln(A)-ln(B))/(ln(A)+1), N^2-4*N*x-2*N+1, N-1] the irrationality measure is: 2*ln(abs(2*x+1+2*x^(1/2)*(x+1)^(1/2)))/(ln(abs(2*x+1+2*x^(1/2)*(x+1)^(1/2)))+1) -------------------- ------------------------------ For Arctan(1/(2*k+1)), 2*k+1 odd: the input is: ope:=AZd(2^(3*n)*(2*k+1)^(2*n)*x^(2*n)*(1-x)^(2*n)/((2*k+1)^2+x^2)^(2*n+1),x\ ,n,N)[1]: SymToDelSy(SymDel(ope,n,N,[0,-3*(2*k+1)],[1,3*(2*k+1)^2+1],[2,1,1,1],10,A,B\ ),N,A,B,{k}); the output is the recurrence operator is 4*(4*n+7)*(2*n+1)*(1+n)-4*(5+4*n)*(16*k^2*n^2+40*k^2*n+16*k*n^2+21*k^2+40*k*n+6 *n^2+21*k+15*n+8)*N+(4*n+3)*(3+2*n)*(n+2)*N^2 the irrationality measure is: 2*(ln(abs(8*2^(1/2)*(2*k^2+2*k+1)^(1/2)*k+4*2^(1/2)*(2*k^2+2*k+1)^(1/2)+16*k^2+ 16*k+6))-ln(2))/(ln(abs(8*2^(1/2)*(2*k^2+2*k+1)^(1/2)*k+4*2^(1/2)*(2*k^2+2*k+1) ^(1/2)+16*k^2+16*k+6))+2) ------------------------------ For Arctan(1/(2*k), 2*k even: the input is: SymToDelSy(SymDel( 4+4*n-2*(2*n+3)*N*2*k+(-n-2)*N^2 ,n,N,[0,-2],[1,-4*k],[1\ ,1,1,1],10,A,B),N,A,B,{k}); the irrationality measure is: 2*(ln(abs(4*k+2*(4*k^2+1)^(1/2)))-ln(2))/(ln(abs(4*k+2*(4*k^2+1)^(1/2)))+1) -------------------- -------------------- the whole thing took, 1.340, seconds.