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a(n) = round(sqrt(3*n + 9/4)), with a(0) = 1.
6

%I #33 Jun 19 2024 14:12:03

%S 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,

%T 10,10,10,10,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,13,13,13,13,

%U 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

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

%C Diverges from A046693 at positions in A308766.

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

%H Ed Pegg Jr, <a href="http://demonstrations.wolfram.com/SparseRulers/">Sparse Rulers</a>.

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

%F From _Michael Chu_, Jan 17 2022: (Start)

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

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

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

%o (PARI) a(n) = if (n, round(sqrt(3*n + 9/4)), 1); \\ _Michel Marcus_, Jan 18 2022

%o (Python)

%o from math import isqrt

%o from sympy import integer_nthroot

%o def A309407(n):

%o if n == 0: return 1

%o a, b = integer_nthroot(12*n+9,2)

%o return a-(c:=isqrt(3*n+2))-(b&(c&1^1)) # _Chai Wah Wu_, Jun 19 2024

%Y Cf. A046693, A308766, A326499.

%K nonn,easy

%O 0,2

%A _Ed Pegg Jr_, Jul 29 2019