c is a positive integer. The sequence f1, f2, f3, ... is defined by f1 = 1, f2 = c, fn+1 = 2 fn - fn-1 + 2. Show that for each k there is an r such that fk fk+1 = fr.
Answer
k2 + (c-3)k + (c-4)
Solution
We have fn+1 - fn = fn - fn-1 + 2, so fn+1 - fn = 2n+c-3. Hence fn = (n-2)2 + (n-1)c = n2+bn-b, where b = c-4. Hence fkfk+1 = (k2+bk-b)(k2+bk+2k+1). Put r = k2+(b+1)k-b. Then fkfk+1 = (r-k)(r+k+b+1) = r2+br-b = fr. So r = k2+(b+1)k-b = k2 + (c-3)k + (c-4).
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
12 February 2004
Last corrected/updated 12 Feb 04