A360418 Numbers k such that, in a listing of all congruence classes of positive integers, the k-th congruence class contains k. Here the class r mod m (with r in {1,...,m}) precedes the class a' mod b' (with r' in {1,...,m'}) iff m < m' or r > r'.
1, 2, 3, 5, 13, 17, 20, 25, 41, 48, 53, 61, 85, 95, 102, 113, 145, 158, 167, 181, 221, 237, 248, 265, 313, 332, 345, 365, 421, 443, 458, 481, 545, 570, 587, 613, 685, 713, 732, 761, 841, 872, 893, 925, 1013, 1047, 1070, 1105, 1201, 1238, 1263, 1301, 1405, 1445, 1472, 1513, 1625, 1668, 1697, 1741, 1861
Offset: 1
Examples
The 1st congruence class in the list (with m=1 and r=1) is {1,2,3,...} which contains 1, so 1 is in the sequence. The 2nd congruence class (with m=2 and r=2) is {2,4,6,...} which contains 2, so 2 is in the sequence. The 3rd congruence class (with m=2 and r=1) is {1,3,5,...} which contains 3, so 3 is in the sequence. The 4th congruence class (with m=3 and r=3) is {3,6,9,...} which does not contain 4, so 4 is not in the sequence.
Links
- Ray Chandler, Table of n, a(n) for n = 1..14142 (terms up to 10^8)
- Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 2, -2, 0, 0, -1, 1).
Programs
Formula
From Ray Chandler, Feb 10 2025: (Start)
a(n) = a(n-1) + 2*a(n-4) - 2*a(n-5) - a(n-8) + a(n-9) for n > 8.
Comments