A008642 Quarter-squares repeated.

1, 1, 2, 2, 4, 4, 6, 6, 9, 9, 12, 12, 16, 16, 20, 20, 25, 25, 30, 30, 36, 36, 42, 42, 49, 49, 56, 56, 64, 64, 72, 72, 81, 81, 90, 90, 100, 100, 110, 110, 121, 121, 132, 132, 144, 144, 156, 156, 169, 169, 182, 182, 196, 196, 210, 210, 225, 225

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

The area of the largest rectangle whose perimeter is not greater than n. - Dmitry Kamenetsky, Aug 30 2006

Also number of partitions of n into parts 1, 2 or 4. - Reinhard Zumkeller, Aug 12 2011

Let us consider a rectangle composed of unit squares. Then count how many squares are necessary to surround this rectangle by a layer whose width is 1 unit. And repeat this surrounding ad libitum. This sequence, prepended by 4 zeros and with offset 0, gives the number of rectangles that need 2*n unit squares in one of their surrounding layers. - Michel Marcus, Sep 19 2015

a(n) is the number of nonnegative integer solutions (x,y,z) for n-2 <= 2*x + 3*y + 4*z <= n. For example, the two solutions for 1 <= 2*x + 3*y + 4*z <= 3 are (1,0,0) and (0,1,0). - Ran Pan, Oct 07 2015

Conjecture: Consider the number of compositions of n>=4*k+8 into odd parts, where the order of the parts 1,3,..,2k+1 does not count. Then, as k approaches infinity, a(n-4*k-8) is equal to the number of these restricted compositions minus A000009(n), the number of strict partitions of n. - Gregory L. Simay, Aug 12 2016

From Gus Wiseman, May 17 2019: (Start)

Also the number of length-3 integer partitions of n + 4 whose largest part is greater than the sum of the other two. These are unordered triples that cannot be the sides of a triangle. For example, the a(1) = 1 through a(10) = 9 partitions are (A = 10, B = 11, C = 12):

(311) (411) (421) (521) (522) (622) (632) (732) (733) (833)

(511) (611) (531) (631) (641) (741) (742) (842)

(621) (721) (722) (822) (751) (851)

(711) (811) (731) (831) (832) (932)

(821) (921) (841) (941)

(911) (A11) (922) (A22)

(931) (A31)

(A21) (B21)

(B11) (C11)

(End)

This sequence, prepended by four 0's and with offset 0, is the number of partitions of n into four parts whose smallest two parts are equal. - Wesley Ivan Hurt, Jan 05 2021

This sequence, prepended by four 0's and with offset 0, is the number of incongruent obtuse triangles formed from the vertices of a regular n-gon. - Frank M Jackson, Nov 27 2022

References

• D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 105.
• L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 112, D(n).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Sara C. Billey, Matjaž Konvalinka, and Joshua P. Swanson, Asymptotic normality of the major index on standard tableaux, arXiv:1905.00975 [math.CO], 2019.
• Ran Pan, Exercise U, Project P.
• Harold N. Ward, A Normal Graph Algebra, arXiv:2201.00389 [math.CO], 2022.
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1,-1,1).

Crossrefs

Cf. A002620.

Cf. A001399, A005044 (triangles without self-intersections), A069905, A124278, A266223, A325686, A325689, A325690, A325691, A325695.

Programs

Magma

[Floor(((n+1)*((-1)^n+n+6)+9)/16): n in [0..70]]; // Vincenzo Librandi, Apr 02 2014

Maple

seq((7/8+(-1)^k/8 + k + k^2/4)$2, k=0..100); # Robert Israel, Oct 08 2015

Mathematica

CoefficientList[Series[1/((1-x)(1-x^2)(1-x^4)), {x, 0, 70}], x] (* Vincenzo Librandi, Apr 02 2014 *)
LinearRecurrence[{1,1,-1,1,-1,-1,1},{1,1,2,2,4,4,6}, 70] (* Harvey P. Dale, Jun 03 2015 *)
Table[Floor[((n + 1) ((-1)^n + n + 6) + 9)/16], {n, 0, 70}] (* Michael De Vlieger, Aug 14 2016 *)

PARI

Vec(1/((1-x)*(1-x^2)*(1-x^4)) + O(x^70)) \\ Michel Marcus, Mar 31 2014

PARI

vector(70, n, n--; floor(((n+1)*((-1)^n+n+6)+9)/16)) \\ Altug Alkan, Oct 08 2015

Sage

[floor(floor(n/2+2)^2/2)/2 for n in (0..70)] # Bruno Berselli, Mar 03 2016

Formula

G.f.: 1/((1-x)*(1-x^2)*(1-x^4)).

a(n) = (2*n^2 + 14*n + 21 + (2*n + 7)*(-1)^n)/32 + ((1 + (-1)^n)/2 - (1 - (-1)^n)*i/2)*i^n/8, with i = sqrt(-1).

a(n) = floor(((n+1)*((-1)^n+n+6)+9)/16). - Tani Akinari, Jun 16 2013

a(n) = Sum_{i=1..floor((n+6)/2)} floor((n+6-2*i-(n mod 2))/4). - Wesley Ivan Hurt, Mar 31 2014

a(0)=1, a(1)=1, a(2)=2, a(3)=2, a(4)=4, a(5)=4, a(6)=6; for n>6, a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-4) - a(n-5) - a(n-6) + a(n-7). - Harvey P. Dale, Jun 03 2015

a(n) = floor(floor(n/2+2)^2/4) = floor(floor(n/2+2)^2/2)/2. - Bruno Berselli, Mar 03 2016

E.g.f.: ((14 + 7*x + x^2)*cosh(x) + 2*(cos(x) + sin(x)) + (7 + 9*x + x^2)*sinh(x))/16. - Stefano Spezia, Mar 05 2023

a(n) = floor((n + 4)/4)*floor((n + 6)/4). - Ridouane Oudra, Apr 01 2023