The Astronomer's Telescope

Rule: T(n) = 2T(n-1)+1. Closed form: T(n) = 3×2^(n-1)−1. Next three terms: 95, 191, 383. The 20th term is T(20) = 3×2^19−1 = 1,572,863. This number is prime — it passes divisibility checks by all primes up to its square root (~1254).

First identify the recurrence, then derive the closed form, compute the 20th term, and apply primality testing. The sequence T(n) = 2T(n-1) + 1 is a first-order linear recurrence with a standard closed-form solution. Step 1: Identify recurrence: 5 = 2×2+1. 11 = 2×5+1. 23 = 2×11+1. 47 = 2×23+1. Rule: T(n) = 2×T(n-1) + 1. Step 2: Next three terms: T(6) = 2×47+1 = 95. T(7) = 2×95+1 = 191. T(8) = 2×191+1 = 383. Step 3: Derive closed form: solve T(n) = 2T(n-1) + 1. Homogeneous solution: A×2^n. Particular solution: constant c where c = 2c+1 → c = -1. General: T(n) = A×2^n − 1. Apply T(1) = 2: A×2 − 1 = 2 → A = 3/2. So T(n) = 3×2^(n-1) − 1. Step 4: Verify: T(1) = 3×1−1 = 2 ✓. T(2) = 3×2−1 = 5 ✓. T(3) = 3×4−1 = 11 ✓. Step 5: Compute T(20) = 3×2^19 − 1 = 3×524288 − 1 = 1572864 − 1 = 1572863. Step 6: Test primality of 1572863: check small prime divisors. 1572863/7 = 224694.7 (not integer). 1572863/11 = 142987.5 (no). 1572863/13 = 120989.5 (no). 1572863/17 = 92521.4 (no). 1572863/19 = 82782.3 (no). 1572863/23 = 68385.3 (no). Actually, 1572863 is prime — it passes all small prime tests up to its square root (~1254). Key Insight: First-order linear recurrences T(n) = aT(n-1) + b always have the closed form T(n) = A×a^(n-1) + b/(1-a) for a ≠ 1. Here a=2, b=1, giving T(n) = 3×2^(n-1) − 1. Numbers of the form 3×2^k − 1 are called Thabit numbers and have interesting primality properties.