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A008642
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Quarter-squares repeated.
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20
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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
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OFFSET
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0,3
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COMMENTS
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The area of the largest rectangle whose perimeter is not greater than n. - Dmitry Kamenetsky, Aug 30 2006
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
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
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REFERENCES
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D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 105.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 112, D(n).
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LINKS
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FORMULA
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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
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MAPLE
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seq((7/8+(-1)^k/8 + k + k^2/4)$2, k=0..100); # Robert Israel, Oct 08 2015
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MATHEMATICA
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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 *)
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PROG
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(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
(Magma) [Floor(((n+1)*((-1)^n+n+6)+9)/16): n in [0..70]]; // Vincenzo Librandi, Apr 02 2014
(Sage) [floor(floor(n/2+2)^2/2)/2 for n in (0..70)] # Bruno Berselli, Mar 03 2016
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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