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A309407
a(n) = round(sqrt(3*n + 9/4)), with a(0) = 1.
6
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
0,2
COMMENTS
Diverges from A046693 at positions in A308766.
For the first 1750 terms, A046693(n)-a(n) is either 0 or 1.
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)
MATHEMATICA
Table[If[n == 0, 1, Round[Sqrt[3 n + 9/4]]], {n, 0, 88}]
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ed Pegg Jr, Jul 29 2019
STATUS
approved