41st Putnam 1980 Problem B3 Define a n by a 0 = α, a n+1 = 2a n - n 2 . For which α are all a n positive? Solution
: α ≥ 3. The trick is that we can solve the recurrence relation. A particular solution is obviously a polynomial with leading term n 2 , so we soon find a n = n 2 + 2n + 3. The general solution is then a n = n 2 + 2n + 3 + k2 n . The initial condition a 0 = α gives that k = α - 3. A necessary and sufficient condition for all a n to be positive is evidently k ≥ 0, or α ≥ 3. 41st Putnam 1980 © John Scholes jscholes@kalva.demon.co.uk 10 Dec 1999 Last corrected/updated 21 Nov 02