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From: Daniel V. <dan...@ya...> - 2025-11-23 16:33:52
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Hi All, I have the following lines: eq:(n+1)*f(n)-(n-1)*f(n-1)=(n+1)/(n-1);funcsolve(eq,f(n)); I get the following result: (n+1)*f(n)-f(n-1)*(n-1) = (n+1)/(n-1) f(n) = (%0*n^2+%1*n+%2)/((n-1)*(n+1)) What is the meaning of %0, %1, %2 in this result? Thanks, Daniel Volinski |