A378185 a(n) = n + floor(n*r/s) + floor(n*r/t), where r=2^(1/4), s=2^(1/2), t=2^(3/4).

2, 5, 8, 11, 14, 18, 20, 23, 26, 29, 33, 36, 38, 41, 44, 48, 51, 54, 56, 59, 62, 66, 69, 72, 75, 77, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 112, 114, 117, 120, 123, 126, 130, 132, 135, 138, 141, 145, 148, 151, 153, 156, 160, 163, 166, 169, 171, 174, 178

Offset: 1

Author

Clark Kimberling, Jan 13 2025

Keywords

• nonn

Comments

The sequences A378142, A328185, A379510, partition the positive integers (A000027), as proved at A184812:

A378142: 3,6,10,13,17,21,24,28,32,35,...

A328185: 2,5,8,11,14,18,20,23,26,29,,...

A379510: 1,4,7,9,12,15,16,19,22,25,27,...

For each k in A000027, write "a" if k=A378142(n) for some n, "b" if k=A328185(n) for some n, and "c" if k=A379510(n) for some n. Concatenating these letters for k = 1,2,3,... spells the following infinite word:

cbacbacbcabcabccabcbacbacbcabcacbcabcbacbacbcacbacbcabcbacbcabcacbacbcabcabcbcacbacbacbcabcabccbacbacb...

Crossrefs

Cf. A000027, A184812, A378142, A379510.

Programs

Mathematica

r=2^(1/4); s=2^(1/2); t=2^(3/4);
a[n_]:=n+Floor[n*s/r]+Floor[n*t/r];
b[n_]:=n+Floor[n*r/s]+Floor[n*t/s];
c[n_]:=n+Floor[n*r/t]+Floor[n*s/t];
Table[a[n], {n, 1, 120}]  (* A378142 *)
Table[b[n], {n, 1, 120}]  (* A378185 *)
Table[c[n], {n, 1, 120}]  (* A379510 *)

Formula

a(n) = n + [w*n] + [n/w], where w = 2^(1/4) and [ ] = floor.