A064412 -id:A064412 - cp's OEIS frontend

A275112 Zero together with the partial sums of A064412.

0, 1, 6, 20, 52, 112, 215, 375, 613, 948, 1407, 2013, 2799, 3793, 5034, 6554, 8398, 10603, 13220, 16290, 19870, 24006, 28761, 34185, 40347, 47302, 55125, 63875, 73633, 84463, 96452, 109668, 124204, 140133, 157554, 176544, 197208, 219628, 243915, 270155, 298465, 328936, 361691, 396825, 434467, 474717, 517710, 563550, 612378, 664303, 719472

Offset: 0

Luce ETIENNE, Jul 17 2016

Colin Barker, Table of n, a(n) for n = 0..1000
Luce ETIENNE, Illustration of initial terms of this sequence and A064412
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,4,-4,2,2,-3,1).

Cf. A064412, A213389, A006003.

{0}~Join~Accumulate@ 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, 49}], x] (* Michael De Vlieger, Jul 18 2016, after Wesley Ivan Hurt at A064412, or *)
Table[(14 n^4 + 36 n^3 + 36 n^2 + 42 n + 11 + 3 (2 n - 1) (-1)^n - 8 (-1)^(((2 n - 1 + (-1)^n))/4))/128, {n, 50}] (* Michael De Vlieger, Jul 18 2016 *)
LinearRecurrence[{3,-2,-2,4,-4,2,2,-3,1},{0,1,6,20,52,112,215,375,613},60] (* Harvey P. Dale, Jun 19 2022 *)

concat(0, Vec(x*(1+x+x^2)*(1+2*x+x^2+3*x^3)/((1-x)^5*(1+x)^2*(1+x^2)) + O(x^50))) \\ Colin Barker, Jul 18 2016

a(n) = (28*n^4+36*n^3+18*n^2+12*n+(1-(-1)^n))/16 for n even.
a(n) = (28*n^4+92*n^3+114*n^2+68*n+17-(-1)^n)/16 for n odd.
a(n) = (14*n^4+36*n^3+36*n^2+42*n+11+3*(2*n-1)*(-1)^n-8*(-1)^(((2*n-1+(-1)^n))/4))/128.
G.f.: x*(1+x+x^2)*(1+2*x+x^2+3*x^3) / ((1-x)^5*(1+x)^2*(1+x^2)). - Colin Barker, Jul 18 2016

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

A282513 a(n) = floor((3*n + 2)^2/24 + 1/3).

Original entry on oeis.org

0, 1, 3, 5, 8, 12, 17, 22, 28, 35, 43, 51, 60, 70, 81, 92, 104, 117, 131, 145, 160, 176, 193, 210, 228, 247, 267, 287, 308, 330, 353, 376, 400, 425, 451, 477, 504, 532, 561, 590, 620, 651, 683, 715, 748, 782, 817, 852, 888, 925, 963

Offset: 0

Luce ETIENNE, Feb 17 2017

List of quadruples: 2*n*(3*n+1), (2*n+1)*(3*n+1), 6*n^2+8*n+3, (n+1)*(6*n+5). These terms belong to the sequences A033580, A033570, A126587 and A049452, respectively. See links for all the permutations.
After 0, subsequence of A025767.
It seems that a(n) is the smallest number of cells that need to be painted in a (n+1) X (n+1) grid, such that it has no unpainted hexominoes (see link to Kamenetsky and Pratt). - Rob Pratt, Dmitry Kamenetsky, Aug 30 2020

			Rectangular array with four columns:
.   0,   1,   3,   5;
.   8,  12,  17,  22;
.  28,  35,  43,  51;
.  60,  70,  81,  92;
. 104, 117, 131, 145, etc.
From _Rob Pratt_, Aug 30 2020: (Start)
For n = 3, painting only 2 cells would leave an unpainted hexomino, but painting the following 3 cells avoids all unpainted hexominoes:
    . . .
    . . X
    X X .
(End)

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Luce ETIENNE, Permutations
Dmitry Kamenetsky and Rob Pratt, 10x10 grid with no unpainted hexominoes, Puzzling StackExchange, October 2019.
Rob Pratt, Optimal solution for n=11.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Cf. A033436: floor((3*n)^2/24 + 1/3).
Cf. A000326, A000567, A025767, A033570, A033580, A049452, A064412, A126587, A222017, A269064, A274221.
Cf. A130519.
Minimum number of painted cells in other n-ominoes: A337501, A337502, A337503.

[(3*n^2+4*n+4) div 8: n in [0..50]]; // Bruno Berselli, Feb 17 2017

Table[Floor[(3 n + 2)^2/24 + 1/3], {n, 0, 50}] (* or *) CoefficientList[Series[x (1 + x + x^3)/((1 + x) (1 + x^2) (1 - x)^3), {x, 0, 50}], x] (* or *) Table[(6 n^2 + 8 n + 3 + Cos[n Pi] - 4 Cos[n Pi/2])/16, {n, 0, 50}] (* or *) Table[(3 n + 2)^2/24 + 1/3 + (-6 + (1 + (-1)^n) (1 + 2 I^((n + 1) (n + 2))))/16, {n, 0, 50}] (* Michael De Vlieger, Feb 17 2017 *)
LinearRecurrence[{2,-1,0,1,-2,1},{0,1,3,5,8,12},60] (* Harvey P. Dale, Aug 10 2024 *)

a(n)=(3*n^2 + 4*n + 4)\8 \\ Charles R Greathouse IV, Feb 17 2017

G.f.: x*(1 + x + x^3)/((1 + x)*(1 + x^2)*(1 - x)^3).
a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) for n>5.
a(n) = floor((3*n + 2)^2/24 + 2/3).
a(n) = (6*n^2 + 8*n + 3 + (-1)^n - 2*((-1)^((2*n - 1 + (-1)^n)/4) + (-1)^((2*n + 1 - (-1)^n)/4)))/16. Therefore:
a(2*k)   = (6*k^2 + 4*k + 1 - (-1)^k)/4,
a(2*k+1) = (k + 1)*(3*k + 2)/2.
a(n) = (6*n^2 + 8*n + 3 + cos(n*Pi) - 4*cos(n*Pi/2))/16.
a(n) = (3*n + 2)^2/24 + 1/3 + (-6 + (1 + (-1)^n)*(1 + 2*i^((n+1)*(n+2))))/16, where i=sqrt(-1).
a(n) = A130519(n+3)+A130519(n+2)+A130519(n). - R. J. Mathar, Jun 23 2021

Corrected and extended by Bruno Berselli, Feb 17 2017

A269064 At stage 1, start with a unit equilateral triangle. At each successive stage add 3*(n-1) new triangles around outside with vertex-to-vertex contacts. Sequence gives number of triangles at n-th stage.

Original entry on oeis.org

0, 1, 4, 10, 26, 48, 87, 135, 208, 293, 410, 542, 714, 904, 1141, 1399, 1712, 2049, 2448, 2874, 3370, 3896, 4499, 5135, 5856, 6613, 7462, 8350, 9338, 10368, 11505, 12687, 13984, 15329, 16796, 18314, 19962, 21664, 23503, 25399, 27440, 29541, 31794, 34110, 36586, 39128, 41837, 44615, 47568

Offset: 0

Luce ETIENNE, Feb 18 2016

At stage n, we count (6*n^2-6*n+5-3*(2*n-1)*(-1)^n)/8 unit up-pointing triangles and 3*(2*n^2-2*n+1+(2*n-1)*(-1)^n)/8 unit down-pointing triangles.
At stage n, the total number of unit triangles is (3*n^2-3*n+2)/2 = A005448(n). It is the same total as for A064412. Note also that A064412 gives number of triangles in a geometrical structure according to expansion side-side (mode S-S).
The edges of several unit triangles can form larger size triangles, and these are also up- or down-pointing. The number of all such larger is given by :(14*n^3-9*n^2+11*n+18-(9*n^2+3*n+14)*(-1)^n-4*((-1)^((2*n+1-(-1)^n)/4)))/64 up-pointing triangles and (14*n^3-15*n^2+35*n-6+(9*n^2+21*n+2)*(-1)^n+4*((-1)^((2*n-1+(-1)^n)/4)))/64 down-pointing triangles.
As for A265282 we observe that starting with n = 4 we can see and count hexagonal and dodecagonal forms for example in a reticular system (incomplete with hexagonal holes) by opposition to a compact shape A064412.

			a(0)= 0, a(1) = 1, a(2) = 4, a(3) = 7+3 = 10, a(4) = 19 + 6 + 1 = 26, a(5) = 31 + 12 + 4 + 1 = 48.

Colin Barker, Table of n, a(n) for n = 0..1000
Luce ETIENNE, Illustration of initial terms
Kival Ngaokrajang, Illustration of triangles expansion
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,2,-2,0,2,-1).

Cf. A003215, A005448, A005993, A033428, A064412, A061600, A006003, A039623, A102214, A265282.

[(14*n^3-12*n^2+23*n+6+3*(3*n-2)*(-1)^n+2*((-1)^((2*n-1+(-1)^n) div 4)-(-1)^((6*n-1+(-1)^n) div 4)))/32: n in [0..50]]; // Vincenzo Librandi, Feb 19 2016

Table[(14 n^3 - 12 n^2 + 23 n + 6 + 3 (3 n - 2) (-1)^n + 2 ((-1)^((2*n - 1 + (-1)^n) / 4) - (-1)^((6 n - 1 + (-1)^n) / 4))) / 32, {n, 0, 45}] (* Vincenzo Librandi, Feb 19 2016 *)

concat(0, Vec(x*(1+2*x+2*x^2+8*x^3+2*x^4+5*x^5+x^6)/((1-x)^4*(1+x)^2*(1+x^2)) + O(x^50))) \\ Colin Barker, Feb 24 2016

a(n) = (7*n^3-3*n^2+4*n)/2 for n even.
a(n) = (28*n^3+30*n^2+16*n+7+(-1)^n)/8 for n odd.
a(n) = (14*n^3-12*n^2+23*n+6+3*(3*n-2)*(-1)^n+2*((-1)^((2*n-1+(-1)^n)/4)-(-1)^((6*n-1+(-1)^n)/4)))/32.
G.f.: x*(1+2*x+2*x^2+8*x^3+2*x^4+5*x^5+x^6) / ((1-x)^4*(1+x)^2*(1+x^2)). - Colin Barker, Feb 24 2016

A274221 List of quadruples: 3n(3n-1), 3n(3n+1), (3n+1)^2, (3n+2)^2.

Original entry on oeis.org

0, 0, 1, 4, 6, 12, 16, 25, 30, 42, 49, 64, 72, 90, 100, 121, 132, 156, 169, 196, 210, 240, 256, 289, 306, 342, 361, 400, 420, 462, 484, 529, 552, 600, 625, 676, 702, 756, 784, 841, 870, 930, 961, 1024, 1056, 1122, 1156, 1225, 1260, 1332, 1369, 1444, 1482

Offset: 0

Luce ETIENNE, Sep 14 2016

For the formulae of the permutations of A152743, A045945, A016778 and A016790, see the link.

Luce ETIENNE, Permutations
Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1,-1,1).

Cf. A152743, A045945, A016778, A016790, A064412, A269064.

&cat [[3*n*(3*n-1), 3*n*(3*n+1), (3*n+1)^2, (3*n+2)^2]: n in [0..15]]; // Bruno Berselli, Sep 15 2016

Flatten[Table[{3 n (3 n - 1), 3 n (3 n + 1), (3 n + 1)^2, (3 n + 2)^2}, {n, 0, 15}]] (* Bruno Berselli, Sep 15 2016 *)

G.f.: x^2*(1+3*x+x^2+3*x^3+x^4)/((1-x)^3*(1+x)^2*(1+x^2)). - Robert Israel, Sep 15 2016
a(n) = (18*n^2-18*n+1-3*(2*n-1)*(-1)^n-4*(-1)^((2*n-1+(-1)^n)/4))/32. Therefore: a(2k) = (18*k^2-12*k+1-(-1)^k)/8, a(2k+1) = (18*k^2+12*k+1-(-1)^k)/8.
a(n) = A064412(n) - A269064(n) for n>0.
E.g.f.: ((9*x^2 - 3*x - 1)*sinh(x) + (9*x^2 + 3*x + 2)*cosh(x) - 2*(sin(x) + cos(x)))/16. - Stefano Spezia, Nov 07 2022

A213801 Number of 3 X 3 0..n symmetric arrays with all rows summing to floor(n*3/2).

Original entry on oeis.org

4, 13, 29, 57, 96, 153, 226, 323, 440, 587, 759, 967, 1204, 1483, 1796, 2157, 2556, 3009, 3505, 4061, 4664, 5333, 6054, 6847, 7696, 8623, 9611, 10683, 11820, 13047, 14344, 15737, 17204, 18773, 20421, 22177, 24016, 25969, 28010, 30171, 32424, 34803, 37279

Offset: 1

R. H. Hardin, Jun 20 2012

nonn

Row 3 of A213800.

Sequence is difference between numbers of triangles, regardless of size, in A064412 (a family of ((3*n^2+3*n+2)/2)-iamonds, see also illustration of initial terms there) and a quantity A077043 of triangles of dimension 1. - Luce ETIENNE, Aug 23 2014

			Some solutions for n=4:
..1..3..2....2..4..0....0..4..2....1..2..3....1..1..4....4..0..2....2..2..2
..3..1..2....4..0..2....4..0..2....2..2..2....1..3..2....0..2..4....2..2..2
..2..2..2....0..2..4....2..2..2....3..2..1....4..2..0....2..4..0....2..2..2
a(2)=5-1=4, a(3)=14-1=13, a(210)=4118206-8269=4109937. - _Luce ETIENNE_, Aug 23 2014

R. H. Hardin, Table of n, a(n) for n = 1..210

Cf. A064412, A077403.

Empirical: a(n) = 2*a(n-1) -2*a(n-3) +2*a(n-4) -2*a(n-5) +2*a(n-7) -a(n-8).
Empirical: G.f. -x*(-4-5*x-3*x^2-7*x^3-x^5-2*x^6+x^7) / ( (x^2+1)*(1+x)^2*(x-1)^4 ). - R. J. Mathar, Jul 04 2012
a(n) = (14*n^3+42*n^2+53*n+25+3*(n+1)*(-1)^n+2*((-1)^((2*n+1-(-1)^n)/4)-(-1)^((6*n+5-(-1)^n)/4)))/32. - Luce ETIENNE, Aug 23 2014
a(n) = A064412(n+1) - A077043((2*n+1-(-1)^n)/4). - Luce ETIENNE, Aug 23 2014

cp's OEIS Frontend

A275112 Zero together with the partial sums of A064412.

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

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A282513 a(n) = floor((3*n + 2)^2/24 + 1/3).

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A269064 At stage 1, start with a unit equilateral triangle. At each successive stage add 3*(n-1) new triangles around outside with vertex-to-vertex contacts. Sequence gives number of triangles at n-th stage.

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A274221 List of quadruples: 3*n*(3*n-1), 3*n*(3*n+1), (3*n+1)^2, (3*n+2)^2.

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A213801 Number of 3 X 3 0..n symmetric arrays with all rows summing to floor(n*3/2).

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A274221 List of quadruples: 3n(3n-1), 3n(3n+1), (3n+1)^2, (3n+2)^2.