A093353 - OEIS

A093353

a(n) = (n + n mod 2)*(n + 1)/2.

2, 3, 8, 10, 18, 21, 32, 36, 50, 55, 72, 78, 98, 105, 128, 136, 162, 171, 200, 210, 242, 253, 288, 300, 338, 351, 392, 406, 450, 465, 512, 528, 578, 595, 648, 666, 722, 741, 800, 820, 882, 903, 968, 990, 1058, 1081, 1152, 1176, 1250, 1275, 1352, 1378, 1458

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Partial sums of A014682(n+1). - Paul Barry, Mar 31 2008

a(n) is the sum of all parts in the integer partitions of n+1 into two parts, see example. - Wesley Ivan Hurt, Jan 26 2013

Also the independence number of the n X n torus grid graph. - Eric W. Weisstein, Sep 06 2017

Also the number of circles we can draw on vertices of an (n+1)-sided regular polygon (using only a compass). - Matej Veselovac, Jan 21 2020

REFERENCES

W. R. Hare, S. T. Hedetniemi, R. Laskar, and J. Pfaff, Complete coloring parameters of graphs, Proceedings of the sixteenth Southeastern international conference on combinatorics, graph theory and computing (Boca Raton, Fla., 1985). Congr. Numer., Vol. 48 (1985), pp. 171-178. MR0830709 (87h:05088). See s_m on page 135. - N. J. A. Sloane, Apr 06 2012

LINKS

G. C. Greubel, Table of n, a(n) for n = 1..5000

Math StackExchange, Number of circles on vertices of a regular polygon, 2020.

Eric Weisstein's World of Mathematics, Independence Number.

Eric Weisstein's World of Mathematics, Torus Grid Graph.

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

FORMULA

a(2*n) = a(2*n-1) + n = A014105(n).

a(2*n+1) = a(2*n) + 3*n + 2 = A001105(n+1).

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

a(n) = (n+1)*(2*n+1-(-1)^n)/4. - Paul Barry, Mar 31 2008

a(n) = (n+1)*floor((n+1)/2). - Wesley Ivan Hurt, Jan 26 2013

a(n) = Sum_{i=1..floor((n+1)/2)} i + Sum_{i=ceiling((n+1)/2)..n} i. - Wesley Ivan Hurt, Jun 08 2013

From Amiram Eldar, Mar 10 2022: (Start)

Sum_{n>=1} 1/a(n) = Pi^2/12 + 2*(1-log(2)).

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/12 - 2*(1-log(2)). (End)

EXAMPLE

a(1) = 2 since 2 = (1+1) and the sum of the first and second parts in the partition is 2; a(2) = 3 since 3 = (1+2) and the sum of these parts is 3; a(3) = 8 since 4 = (1+3) = (2+2) and the sum of all the parts is 8. - Wesley Ivan Hurt, Jan 26 2013

MAPLE

a:=n->(n+1)*floor((n+1)/2); seq(a(k), k = 1..70);

MATHEMATICA

(* Contributions from Harvey P. Dale, Nov 15 2013: Start *)

Table[(n+Mod[n, 2])*(n+1)/2, {n, 60}]

LinearRecurrence[{1, 2, -2, -1, 1}, {2, 3, 8, 10, 18}, 60]

Module[{nn = 60, ab}, ab = Transpose[ Partition[ Accumulate[ Range[nn]], 2]]; Flatten[ Transpose[ {ab[[1]] + Range[nn/2], ab[[2]]}]]]

(* End *)

PROG

(PARI) a(n)=(n+1)\2*(n+1) \\ Charles R Greathouse IV, Jun 11 2015

(Magma) [(n+1)*(2*n+1-(-1)^n)/4: n in [1..60]]; // Vincenzo Librandi, Jan 23 2020

(SageMath) [(n+1)*int((n+1)//2) for n in range(1, 71)] # G. C. Greubel, Mar 14 2024