1, 5, 6, 25, 26, 30, 31, 125, 126, 130, 131, 150, 151...... What is the value…

Concept

Each term is generated by a recursive positional rule: term(2n) = 5 × term(n), and term(2n+1) = term(2n) + 1. Equivalently, write the position n in binary, then read that binary string as a number in base 5 (so a 1 contributes its place value 5k and a 0 contributes nothing).

Application

Compute term(35) using the recursive rule, breaking 35 down by repeated halving:

1. 35 = 2×17 + 1, so term(35) = term(34) + 1, and term(34) = 5 × term(17).

2. 17 = 2×8 + 1, so term(17) = term(16) + 1, and term(16) = 5 × term(8).

3. From the given series, term(8) = 125.

4. term(16) = 5 × 125 = 625.

5. term(17) = 625 + 1 = 626.

6. term(34) = 5 × 626 = 3130.

7. term(35) = 3130 + 1 = 3131.

Cross-check

Verify with the equivalent binary-to-base-5 reading: 35 in binary is 100011. Reading these digits as base-5 place values (55, 54, 53, 52, 51, 50) gives 1×3125 + 0×625 + 0×125 + 0×25 + 1×5 + 1×1 = 3125 + 5 + 1 = 3131 — the same result, confirming the derivation independently.

So the 35th term is 3131.