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Part of 1999 Putnam

Problems(2)

Putnam 1999 A3

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12/22/2012
Consider the power series expansion 11−2x−x2=∑n=0∞anxn.\dfrac{1}{1-2x-x^2}=\sum_{n=0}^\infty a_nx^n. Prove that, for each integer n≥0n\geq 0, there is an integer mm such that an2+an+12=am.a_n^2+a_{n+1}^2=a_m.
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Putnam 1999 B3

Source:

10/28/2012
Let A={(x,y):0≤x,y<1}.A=\{(x,y): 0\le x,y < 1\}. For (x,y)∈A,(x,y)\in A, let S(x,y)=∑12≤mn≤2xmyn,S(x,y)=\sum_{\frac12\le\frac mn\le2}x^my^n, where the sum ranges over all pairs (m,n)(m,n) of positive integers satisfying the indicated inequalities. Evaluate lim⁡(x,y)→(1,1),(x,y)∈A(1−xy2)(1−x2y)S(x,y).\lim_{(x,y)\to(1,1),(x,y)\in A}(1-xy^2)(1-x^2y)S(x,y).
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