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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'.
5
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
1,2
COMMENTS
The sequence appears to be the interleaving of the four sequences A080856, A102083, A360416, A360417. This has been verified for values of k up to one million as of February 06 2023.
Above conjecture confirmed with more terms and linear recurrence. See supporting formula below. - Ray Chandler, Feb 10 2025
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).
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.
A080856(n) = A360417(n-1) + 2*A080856(n-1) - 2*A360417(n-2) - A080856(n-2) + A360417(n-3).
A102083(n) = A080856(n) + 2*A102083(n-1) - 2*A080856(n-1) - A102083(n-2) + A080856(n-2).
A360416(n) = A102083(n) + 2*A360416(n-1) - 2*A102083(n-1) - A360416(n-2) + A102083(n-2).
A360417(n) = A360416(n) + 2*A360417(n-1) - 2*A360416(n-1) - A360417(n-2) + A360416(n-2). (End)
EXAMPLE
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.
MATHEMATICA
mval[n_] := Floor[Sqrt[2 n] + 1/2]; (* A002024 *)
rval[n_] := (2 - 2 n + Round[Sqrt[2 n]] + Round[Sqrt[2 n]]^2)/2; (* A004736 *)
test[n_] := Mod[n - rval[n], mval[n]] == 0;
Select[Range[10000], test[#] &]
KEYWORD
nonn,easy
AUTHOR
James Propp, Feb 06 2023
STATUS
approved