Question

make a sequence of integers that starts with 43. each new term is the sum of the squares of the digits of the previous term. the sequence would be a₁= 43, a₂= 4²+3² =25, and so on. keep going until you find the first integer that appears twice in the sequence

Answer 1

Let's create the sequence of integers based on the rules you provided:

a₁ = 43

a₂ = 4² + 3² = 16 + 9 = 25

a₃ = 2² + 5² = 4 + 25 = 29

a₄ = 2² + 9² = 4 + 81 = 85

a₅ = 8² + 5² = 64 + 25 = 89

a₆ = 8² + 9² = 64 + 81 = 145

a₇ = 1² + 4² + 5² = 1 + 16 + 25 = 42

a₈ = 4² + 2² = 16 + 4 = 20

a₉ = 2² + 0² = 4 + 0 = 4

The sequence continues from here. Let's check if any integer appears twice in the sequence:

a₁ = 43

a₂ = 25

a₃ = 29

a₄ = 85

a₅ = 89

a₆ = 145

a₇ = 42

a₈ = 20

a₉ = 4

The first integer that appears twice in the sequence is 4, which repeats at a₉. So, the first integer that appears twice in the sequence is 4.

Answer 2

a₁ = 43
a₂ = 4² + 3² = 16 + 9 = 25
a₃ = 2² + 5² = 4 + 25 = 29
a₄ = 2² + 9² = 4 + 81 = 85
a₅ = 8² + 5² = 64 + 25 = 89
a₆ = 8² + 9² = 64 + 81 = 145
a₇ = 1² + 4² + 5² = 1 + 16 + 25 = 42
a₈ = 4² + 2² = 16 + 4 = 20
a₉ = 2² + 0² = 4 + 0 = 4
a₁₀ = 4² = 16
a₁₁ = 1² + 6² = 1 + 36 = 37
a₁₂ = 3² + 7² = 9 + 49 = 58
a₁₃ = 5² + 8² = 25 + 64 = 89

As we can see, the first integer that appears twice in the sequence is 89.