A359698 Least k > 0 such that the first n digits of 2^k and 3^k are identical.
1, 17, 193, 619, 2016, 91958, 91958, 8186278, 45392361, 977982331, 26450915298, 91600221212, 196425900073, 14810317269038, 44430951807114, 626642721222487, 626642721222487, 102882886570917135, 874191214492184404, 3830977578643912683, 86801197487071715103
Offset: 0
Examples
n k = a(n) 1st n digits -- ----------- ------------- 0 1 1 17 1... 2 193 12... 3 619 217... 4 2016 7524... 5 91958 13071... 6 91958 130719... 7 8186278 1701547... 8 45392361 17179395... 9 977982331 725070476... 10 26450915298 2919267309... a(3) = 619 because 2^619 = 2.175...*10^186 and 3^619 = 2.177...*10^295 both begin with the same three digits (in base ten), and this is not true of any smaller positive integer than 619. a(0) = 1 because 2^1 and 3^1 share zero leading digits.
Links
- Zhao Hui Du, Table of n, a(n) for n = 0..1000
Extensions
a(11)-a(20) from Jon E. Schoenfield, May 21 2023