S(n) = 1 if n = 1 2 if n = 2 S(n - 1) + S(n - 2) if n is even 2S(n - 1) - S(n - 2) if n is odd and n > 2 SOLUTION: def layup_s(n): # given base cases if n == 1: return 1 elif n == 2: return 2 s1, s2 = 1, 2 #first two sequence values for i in range(3, n + 1): if i % 2 == 0: #even case sn = s2 + s1 else: sn = 2 * s2 - s1 #odd case s1, s2 = s2, sn return s2 Can someone explain what is wrong with this solution? For smaller values it functions correctly: S(1) = 1, S(2) = 2, S(3) = 3, S(4) = 5, S(5) = 7 A recruiter is hell bent on telling me it is incorrect for S(10000) even though I had two LLMs generate the exact same answer as mine and I just do not understand the issue! EDIT : The task was to find S(10000) and am being told the value is incorrect.... Continue reading...
