MathDB
x_{n+1} = x_n^{s(n)} + 1, s(n)

Source: Iran MO Third Round 2021 N3

September 25, 2021
number theory

Problem Statement

x1x_1 is a natural constant. Prove that there does not exist any natural number m>2500m> 2500 such that the recursive sequence {xi}i=1∞\{x_i\} _{i=1} ^ \infty defined by xn+1=xns(n)+1x_{n+1} = x_n^{s(n)} + 1 becomes eventually periodic modulo mm. (That is there does not exist natural numbers NN and TT such that for each n≥Nn\geq N, m∣xn−xn+Tm\mid x_n - x_{n+T}). (s(n)s(n) is the sum of digits of nn.)
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