A129317 The second of the pair of consecutive integers k and k+1 such that sopfr(k) divides sopfr(k+1), where sopfr(k) is the sum of the prime factors of k, counting multiplicity.
6, 9, 16, 78, 126, 161, 253, 497, 715, 949, 981, 1046, 1054, 1261, 1331, 1379, 1405, 1431, 1509, 1521, 1611, 1751, 1863, 1891, 2171, 2492, 2681, 2822, 3095, 3101, 3249, 3401, 3592, 3611, 3653, 3809, 4186, 4192, 4385, 4453, 4501, 4599, 4907, 5121, 5146
Offset: 1
Examples
a(6)=161 since sopfr(160)=sopfr(2^5*5)=10+5=15 and sopfr(161)=sopfr(7*23)=30.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Formula
sopfr(k+1) mod sopfr(k) = 0.
a(n) = A129316(n+1). - Amiram Eldar, Oct 26 2019
Comments