A309407 a(n) = round(sqrt(3*n + 9/4)), with a(0) = 1.

1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16

Offset: 0

Author

Ed Pegg Jr, Jul 29 2019

Keywords

• nonn
• easy

Comments

Diverges from A046693 at positions in A308766.

For the first 1750 terms, A046693(n)-a(n) is either 0 or 1.

Links

• Ed Pegg Jr, Sparse Rulers.

Crossrefs

Cf. A046693, A308766, A326499.

Programs

Mathematica

Table[If[n == 0, 1, Round[Sqrt[3 n + 9/4]]], {n, 0, 88}]

PARI

a(n) = if (n, round(sqrt(3*n + 9/4)), 1); \\ Michel Marcus, Jan 18 2022

Python

from math import isqrt
from sympy import integer_nthroot
def A309407(n):
    if n == 0: return 1
    a, b = integer_nthroot(12*n+9,2)
    return a-(c:=isqrt(3*n+2))-(b&(c&1^1)) # Chai Wah Wu, Jun 19 2024

Formula

a(n) = round(sqrt(3*n + 9/4)).

From Michael Chu, Jan 17 2022: (Start)

a(12*k^2 - 6*k) = 6*k - 2 for k>0.

a(12*k^2 + 6*k) = 6*k + 2 for k>0. (End)