A143978 a(n) = floor(2*n*(n+1)/3).

1, 4, 8, 13, 20, 28, 37, 48, 60, 73, 88, 104, 121, 140, 160, 181, 204, 228, 253, 280, 308, 337, 368, 400, 433, 468, 504, 541, 580, 620, 661, 704, 748, 793, 840, 888, 937, 988, 1040, 1093, 1148, 1204, 1261, 1320, 1380, 1441, 1504, 1568, 1633, 1700, 1768, 1837

Offset: 1

Author

Clark Kimberling, Sep 06 2008

Keywords

• nonn
• easy

Comments

Second diagonal of array A143979, which counts certain unit squares in a lattice. First diagonal: A030511.

Convolution of A042965 with A000012, convolution of A131534 with A000027, and convolution of A106510 with A000217. - L. Edson Jeffery, Jan 24 2015

From Miquel A. Fiol, Aug 31 2024: (Start)

a(n+1) is the maximum number N of vertices of a circulant digraph with steps +-s1, s2, and diameter n.

Depending on the value of n, the following table shows the values of N, s1, and s2:

n | 3*r | 3*r-1 | 3*r-2 |

N | 6*r^2+6*r+1 | 6*r^2+2*r | 6*r^2-2*r |

s1 | 1 | r | r |

s2 | 6*r+3 | 3*r+1 | 3*r-1 |

(End)

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Bruno Berselli, Illustration of the initial terms.
• P. Morillo and Miquel A. Fiol, El diámetro de ciertos digrafos circulantes de triple paso, Stochastica X (1986), no. 3, 233-249.
• Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Crossrefs

Cf. A000217, A030511, A042965 (first differences), A106510, A131534, A143979.

Cf. A049347, A102283, A103115, A131713.

Programs

Maple

A143978:= n-> (6*n*(n+1) -1 + `mod`(n+2,3) - `mod`(n+1,3))/9;
seq(A143978(n), n=1..60); # G. C. Greubel, May 27 2020

Mathematica

Table[(6*n^2 +6*n -1  + Mod[n+2, 3] - Mod[n+1, 3])/9, {n, 60}] (* G. C. Greubel, May 27 2020 *)
Table[Floor[2n (n+1)/3],{n,60}] (* or *) LinearRecurrence[{2,-1,1,-2,1},{1,4,8,13,20},60] (* Harvey P. Dale, Aug 12 2025 *)

Formula

From R. J. Mathar, Oct 05 2009: (Start)

G.f.: x*(1 + x)^2/((1 + x + x^2)*(1-x)^3).

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5). (End)

a(n) = Sum_{k=1..(n+1)} A042965(k). - Klaus Purath, May 23 2020

From G. C. Greubel, May 27 2020: (Start)

a(n) = (ChebyshevU(n, -1/2) - ChebyshevU(n-1, -1/2) + (6*n^2 + 6*n -1))/9.

a(n) = (JacobiSymbol(n+1, 3) - JacobiSymbol(n, 3) + (6*n^2 + 6*n -1))/9.

a(n) = (A102283(n+1) - A102283(n) + A103115(n+1))/9

a(n) = (A131713(n) + A103115(n+1))/9. (End)

Sum_{n>=1} 1/a(n) = 3/2 + (tan(Pi/(2*sqrt(3)))-1)*Pi/(2*sqrt(3)). - Amiram Eldar, Sep 27 2022

E.g.f.: exp(-x/2)*(exp(3*x/2)*(6*x^2 + 12*x - 1) + cos(sqrt(3)*x/2) - sqrt(3)*sin(sqrt(3)*x/2))/9. - Stefano Spezia, Apr 05 2023