A056829 -id:A056829 - cp's OEIS frontend

A069813 Maximum number of triangles in polyiamond with perimeter n.

1, 2, 3, 6, 7, 10, 13, 16, 19, 24, 27, 32, 37, 42, 47, 54, 59, 66, 73, 80, 87, 96, 103, 112, 121, 130, 139, 150, 159, 170, 181, 192, 203, 216, 227, 240, 253, 266, 279, 294, 307, 322, 337, 352, 367, 384, 399, 416, 433, 450, 467, 486, 503, 522, 541, 560, 579

Offset: 3

Winston C. Yang (winston(AT)cs.wisc.edu), Apr 30 2002

			a(10) = 16 because the maximum number of triangles in a polyiamond of perimeter 10 is 16.

Colin Barker, Table of n, a(n) for n = 3..1000
W. C. Yang, R. R. Meyer, Maximal and minimal polyiamonds, 2002.
Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1,-1,1).

Cf. A000105, A000577, A027709, A030511 (bisection), A057729, A067628.

R:=PowerSeriesRing(Integers(), 65); Coefficients(R!( x^3*(x^2-x-1)*(x^2+1)/((x-1)^3*(x+1)*(x^2+x+1)))); // Marius A. Burtea, Jan 03 2020

A069813 := proc(n)
    round(n^2/6) ;
    if modp(n,6) <> 0 then
        %-1 ;
    else
        % ;
    end if;
end proc: # R. J. Mathar, Jul 14 2015

LinearRecurrence[{1, 1, 0, -1, -1, 1}, {1, 2, 3, 6, 7, 10}, 60] (* Jean-François Alcover, Jan 03 2020 *)

a(n) = round(n^2/6) - (n % 6 != 0) \\ Michel Marcus, Jul 17 2013

Vec(x^3*(x^2-x-1)*(x^2+1)/((x-1)^3*(x+1)*(x^2+x+1)) + O(x^60)) \\ Colin Barker, Jan 19 2015

a(n) = round(n^2/6) - (0 if n = 0 mod 6, 1 else) = A056829(n)-A097325(n).
From Colin Barker, Jan 18 2015: (Start)
a(n) = round((-25 + 9*(-1)^n + 8*exp(-2/3*i*n*Pi) + 8*exp((2*i*n*Pi)/3) + 6*n^2)/36), where i=sqrt(-1).
G.f.: x^3*(1+x-x^2)*(1+x^2) / ((1-x)^3*(1+x)*(1+x+x^2)). (End)
a(n) = A001399(n-3) + A001399(n-4) + A001399(n-6) - A001399(n-7). - R. J. Mathar, Jul 14 2015

A173691 Partial sums of round(n^2/6).

Original entry on oeis.org

0, 0, 1, 3, 6, 10, 16, 24, 35, 49, 66, 86, 110, 138, 171, 209, 252, 300, 354, 414, 481, 555, 636, 724, 820, 924, 1037, 1159, 1290, 1430, 1580, 1740, 1911, 2093, 2286, 2490, 2706, 2934, 3175, 3429, 3696, 3976, 4270, 4578, 4901, 5239, 5592, 5960, 6344, 6744, 7161

Offset: 0

Mircea Merca, Nov 25 2010

Partial sums of A056829.

			a(5) = round(1/6) + round(4/6) + round(9/6) + round(16/6) + round(25/6) = 0 + 1 + 2 + 3 + 4.
Note that 9/6 = 1.5 is rounded up.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,0,1,-3,3,-1).

Cf. A056829.

[Floor((2*n^3+3*n^2+6*n+16)/36): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011

Maple
```
a(n):=round((2*n^(3)+3*n^(2)+6*n)/(36))
```

Accumulate[Round[Range[0, 50]^2/6]] (* or *) LinearRecurrence[{3,-3,1,0, 0,1,-3,3,-1}, {0,0,1,3,6,10,16,24,35}, 60] (* Harvey P. Dale, Jan 08 2014 *)
CoefficientList[Series[x^2(1+x^4)/((1+x)(1-x+x^2)(1+x+x^2)(1-x)^4), {x, 0, 60}], x] (* Vincenzo Librandi, Mar 26 2014 *)

vector(60, n, n--; (16+6*n+3*n^2+2*n^3)\36) \\ G. C. Greubel, Jul 02 2019

[floor((16+6*n+3*n^2+2*n^3)/36) for n in (0..60)] # G. C. Greubel, Jul 02 2019

a(n) = Sum_{k=0..n} round(k^2/6).
a(n) = round((2*n^3 + 3*n^2 + 6*n)/36).
a(n) = round((4*n^3 + 6*n^2 + 12*n + 5)/72).
a(n) = floor((2*n^3 + 3*n^2 + 6*n + 16)/36).
a(n) = ceiling((2*n^3 + 3*n^2 + 6*n - 11)/36).
a(n) = a(n-6) + n^2 - 5*n + 10, n > 5.
G.f.: x^2*(1+x^4)/((1+x)*(1-x+x^2)*(1+x+x^2)*(1-x)^4). - Bruno Berselli, Jan 12 2011

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A069813 Maximum number of triangles in polyiamond with perimeter n.

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A173691 Partial sums of round(n^2/6).

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