A056640 -id:A056640 - cp's OEIS frontend

A274772 Zero together with the partial sums of A056640.

0, 1, 6, 24, 66, 149, 292, 520, 860, 1345, 2010, 2896, 4046, 5509, 7336, 9584, 12312, 15585, 19470, 24040, 29370, 35541, 42636, 50744, 59956, 70369, 82082, 95200, 109830, 126085, 144080, 163936, 185776, 209729, 235926, 264504, 295602, 329365, 365940, 405480, 448140, 494081, 543466, 596464, 653246, 713989, 778872, 848080, 921800, 1000225, 1083550

Offset: 0

Luce ETIENNE, Nov 11 2016

			a(0) = 0, a(1) = 1, a(2) = 6, a(3) = 24, a(4) = 66.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

Cf. A001844, A005900, A056640.

LinearRecurrence[{4,-5,0,5,-4,1},{0,1,6,24,66,149},60] (* Harvey P. Dale, Jun 19 2021 *)

concat(0, Vec(x*(1 + 2*x + 5*x^2) / ((1 - x)^5 * (1 + x)) + O(x^50))) \\ Colin Barker, Nov 11 2016

a(n) = (4*n^4+8*n^3+2*n^2+4*n+3*(1-(-1)^n))/24. Therefore :
a(2*k) = k*(k+1)*(8*k^2+1)/3, a(2*k+1) = (k+1)*(8*k^3+16*k^2+9*k+3)/3.
From Colin Barker, Nov 11 2016: (Start)
G.f.: x*(1 + 2*x + 5*x^2) / ((1 - x)^5 * (1 + x)).
a(n) = 4*a(n-1) - 5*a(n-2) + 5*a(n-4) - 4*a(n-5) + a(n-6) for n>5.
(End)

A085601 a(n) = 2 * (4^n + 2^n) + 1.

Original entry on oeis.org

5, 13, 41, 145, 545, 2113, 8321, 33025, 131585, 525313, 2099201, 8392705, 33562625, 134234113, 536903681, 2147549185, 8590065665, 34360000513, 137439477761, 549756862465, 2199025352705, 8796097216513, 35184380477441

Offset: 0

Jun Mizuki (suzuki32(AT)sanken.osaka-u.ac.jp), Jul 07 2003

Begin with a square tile.
Place square tiles on each edge to form a diamond shape.
Count the tiles: a(0) = 5.
Add tiles to fill the enclosing square.
Go to step 2.

Iain Fox, Table of n, a(n) for n = 0..1660
Index entries for linear recurrences with constant coefficients, signature (7,-14,8)

Cf. A028403, A046142, A056640, A064412.

Cf. A343175 (essentially the same).

Table[2(4^n+2^n)+1,{n,0,30}] (* or *) LinearRecurrence[{7,-14,8},{5,13,41},30] (* Harvey P. Dale, Dec 30 2017 *)

first(n) = Vec((5 - 22*x + 20*x^2)/(1 - 7*x + 14*x^2 - 8*x^3) + O(x^n)) \\ Iain Fox, Dec 30 2017

From R. J. Mathar, Apr 20 2009: (Start)
a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3).
G.f.: -(5 - 22*x + 20*x^2)/((x - 1)*(2*x - 1)*(4*x - 1)).
(End)
E.g.f.: e^x + 2*(e^(2*x) + e^(4*x)). - Iain Fox, Dec 30 2017

Edited by Franklin T. Adams-Watters and Don Reble, Aug 15 2006

A064412 At stage 1, start with a unit equilateral equiangular triangle. At each successive stage add 3*(n-1) new triangles around outside with edge-to-edge contacts. Sequence gives number of triangles (regardless of size) at n-th stage.

Original entry on oeis.org

1, 5, 14, 32, 60, 103, 160, 238, 335, 459, 606, 786, 994, 1241, 1520, 1844, 2205, 2617, 3070, 3580, 4136, 4755, 5424, 6162, 6955, 7823, 8750, 9758, 10830, 11989, 13216, 14536, 15929, 17421, 18990, 20664, 22420, 24287, 26240, 28310, 30471, 32755, 35134, 37642

Offset: 1

Robert G. Wilson v, Sep 29 2001

Number of unit triangles at n-th stage = 3n(n-1)/2 + 1, A005448.

			a(4) = 32: 19 triangles of side 1, 10 of side 2 and 3 of side 3.

Anthony Gardiner, "Mathematical Puzzling," Dover Publications, Inc., Mineola, NY., 1987, page 88.

N. J. A. Sloane, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,2,-2,0,2,-1).

Cf. A056640.

A064412:=n->(14*n^3+6*n^2+5*n+7+3*(n-1)*(-1)^n-2*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((6*n-1+(-1)^n)/4)))/32; seq(A064412(n), n=1..30); # Wesley Ivan Hurt, Jun 27 2014

CoefficientList[Series[(1 + x + x^2) (1 + 2 x + x^2 + 3 x^3)/((1 - x)^2 (1 - x^2) (1 - x^4)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Jun 27 2014 *)
LinearRecurrence[{2,0,-2,2,-2,0,2,-1},{1,5,14,32,60,103,160,238},50] (* Harvey P. Dale, Apr 12 2016 *)

a(n)=polcoeff(x*(1+x+x^2)*(1+2*x+x^2+3*x^3)/((1-x)^2*(1-x^2)*(1-x^4))+x*O(x^n),n)

G.f.: (1+x+x^2)(1+2x+x^2+3x^3)/((1-x)^2(1-x^2)(1-x^4)).
a(2n+1) = (7n^3+12n^2+7n+2)/2; a(2n) = (28n^3+6n^2+4n+1+(-1)^(n+1))/8. - Len Smiley, Oct 07 2001
a(n) = (14*n^3+6*n^2+5*n+7+3*(n-1)*(-1)^n-2*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((6*n-1+(-1)^n)/4)))/32. - Luce ETIENNE, Jun 27 2014

A255840 a(n) = (4n^2 - 4n + 1 - (-1)^n)/2.

Original entry on oeis.org

0, 1, 4, 13, 24, 41, 60, 85, 112, 145, 180, 221, 264, 313, 364, 421, 480, 545, 612, 685, 760, 841, 924, 1013, 1104, 1201, 1300, 1405, 1512, 1625, 1740, 1861, 1984, 2113, 2244, 2381, 2520, 2665, 2812, 2965, 3120, 3281, 3444, 3613, 3784, 3961, 4140, 4325, 4512

Offset: 0

Wesley Ivan Hurt, Mar 07 2015

Take an n X n square grid and add unit squares along each side except for the corners --> do this repeatedly along each side with the same restriction until no squares can be added. a(n) is the total area of each figure. The perimeter, P, of each figure is given by P(n) = 4*A042963(n), n>0 (see example).

For n>0, partial sums of a(n) are in A056640.

			                                                                 _
                                                               _|_|_
                            _              _ _               _|_|_|_|_
                          _|_|_          _|_|_|_           _|_|_|_|_|_|_
              _ _       _|_|_|_|_      _|_|_|_|_|_       _|_|_|_|_|_|_|_|_
    _        |_|_|     |_|_|_|_|_|    |_|_|_|_|_|_|     |_|_|_|_|_|_|_|_|_|
   |_|       |_|_|       |_|_|_|      |_|_|_|_|_|_|       |_|_|_|_|_|_|_|
                           |_|          |_|_|_|_|           |_|_|_|_|_|
                                          |_|_|               |_|_|_|
                                                                |_|
   n=1        n=2          n=3             n=4                  n=5

Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000290 (squares), A002620 (quarter-squares), A042963.

Cf. A056640, A085046, A255876.

[(4*n^2 - 4*n + 1 - (-1)^n)/2 : n in [0..100]];

A255840:=n->(4*n^2 - 4*n + 1 - (-1)^n)/2: seq(A255840(n), n=0..100);

CoefficientList[Series[x (1 + 2 x + 5 x^2)/((1 + x) (1 - x)^3), {x, 0, 50}], x]

vector(100,n,(4*(n-1)^2 - 4*(n-1) + 1 + (-1)^n)/2) \\ Derek Orr, Mar 09 2015

G.f.: x*(1+2*x+5*x^2)/((1+x)*(1-x)^3).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
a(n) = A000290(n) + 4*A002620(n).
a(n) - a(n-1) = A047471(n). - Wesley Ivan Hurt, Apr 28 2017

cp's OEIS Frontend

A274772 Zero together with the partial sums of A056640.

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A085601 a(n) = 2 * (4^n + 2^n) + 1.

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A064412 At stage 1, start with a unit equilateral equiangular triangle. At each successive stage add 3*(n-1) new triangles around outside with edge-to-edge contacts. Sequence gives number of triangles (regardless of size) at n-th stage.

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Maple

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A255840 a(n) = (4*n^2 - 4*n + 1 - (-1)^n)/2.

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Maple

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A255840 a(n) = (4n^2 - 4n + 1 - (-1)^n)/2.