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[GH-ISSUE #4] an+1=an^2+1 ,a1=1(an^2表示an的平方） 求{an}的通项公式 Find the general term formula for {a_n} given that an+1=an^2+1 and ,a1=1 (where an^2 represents an squared） #1
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Originally created by @cdut007 on GitHub (Jan 20, 2025).
Original GitHub issue: https://github.com/deepseek-ai/DeepSeek-R1/issues/4
能给出答案，才是AGI ，目前只有最先进的AGI能给出答案