Luce ETIENNE - cp's OEIS frontend

A338996 Numbers of squares and rectangles of all sizes in 3n(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.

0, 5, 27, 85, 205, 420, 770, 1302, 2070, 3135, 4565, 6435, 8827, 11830, 15540, 20060, 25500, 31977, 39615, 48545, 58905, 70840, 84502, 100050, 117650, 137475, 159705, 184527, 212135, 242730, 276520

Offset: 0

Luce ETIENNE, Nov 18 2020

			a(1) = 2*3-1 = 5, a(2) = 2*16-5 = 27, a(3) = 2*50-15 = 85, a(4) = 2*120-35 = 205, a(5) = 2*245-70 = 420, a(6) = 2*448-126 = 770.

Luce ETIENNE, Illustration of a(1), a(2), a(3) and a(4)
Wikipedia, Aztec diamond.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A000332, A002417, A004320, A045943, A258440, A330805.

CoefficientList[Series[x (2 x + 5)/(1 - x)^5, {x, 0, 30}], x] (* Michael De Vlieger, Dec 12 2020 *)

G.f.: x*(2*x + 5)/(1 - x)^5.
E.g.f.: exp(x)*x*(120 + 204*x + 76*x^2 + 7*x^3)/24. - Stefano Spezia, Nov 18 2020
a(n) = 5*(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = n*(n + 1)*(n + 2)*(7*n + 13)/24.
a(n) = 2*A004320(n) - A000332(n+3).
a(n) = 2*A000332(n+2) + 5*A000332(n+3).

A330805 Number of squares and rectangles in the interior of the square with vertices (n,0), (0,n), (-n,0) and (0,-n) in a square (x,y)-grid.

Original entry on oeis.org

0, 9, 51, 166, 410, 855, 1589, 2716, 4356, 6645, 9735, 13794, 19006, 25571, 33705, 43640, 55624, 69921, 86811, 106590, 129570, 156079, 186461, 221076, 260300, 304525, 354159, 409626, 471366, 539835, 615505, 698864, 790416, 890681, 1000195, 1119510, 1249194, 1389831

Offset: 0

Luce ETIENNE, Jan 01 2020

Collection: 2*n*(n+1)-ominoes.
Number of squares (all sizes): (8*n^3 + 24*n^2 + 22*n - 3*(-1)^n + 3)/12.
Number of rectangles (all sizes): (8*n^4 + 24*n^3 + 22*n^2 + 3*(-1)^n - 3)/12.

			a(1) = 4*1+5 = 9; a(2) = 4*5+31 = 51; a(3) = 4*15 + 106 = 166; a(4) = 4*36 + 270 = 410.

Teofil Bogdan and Mircea Dan Rus, Counting the lattice rectangles inside Aztec diamonds and square biscuits, arXiv:2007.13472 [math.CO], 2020.
Luce ETIENNE, Illustration of a(1), a(2) and a(3).
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000332, A004320, A046092, A111746, A212523.

LinearRecurrence[{5,-10,10,-5,1},{0,9,51,166,410},40] (* Harvey P. Dale, Jun 27 2020 *)

G.f.: x*(x + 3)^2/(1 - x)^5.
E.g.f.: (1/6)*exp(x)*x*(54 + 99*x + 40*x^2 + 4*x^3). - Stefano Spezia, Jan 01 2020
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = n*(n + 1)*(4*n^2 + 12*n + 11)/6.
a(n) = 4*A000332(n+3) + A212523(n+1).
a(n) = 9*A000332(n+3) + 6*A000332(n+2) + A000332(n+1). - Mircea Dan Rus, Aug 26 2020
a(n) = 3*A004320(n) + A004320(n-1). - Mircea Dan Rus, Aug 26 2020

A318054 a(n) = n(n + 1)(n^2 + n + 22)/24.

Original entry on oeis.org

0, 2, 7, 17, 35, 65, 112, 182, 282, 420, 605, 847, 1157, 1547, 2030, 2620, 3332, 4182, 5187, 6365, 7735, 9317, 11132, 13202, 15550, 18200, 21177, 24507, 28217, 32335, 36890, 41912, 47432, 53482, 60095, 67305, 75147, 83657, 92872, 102830, 113570, 125132, 137557

Offset: 0

Luce ETIENNE, Aug 14 2018

			a(1) = 2; a(2)= 5+2 = 7; a(3) = 10+5+2 = 17; a(4) = 18+10+5+2 = 35; a(5) = 30+18+10+5+2 = 65; a(6) = 47+30+18+10+5+2 = 112.

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Tenner, Bridget Eileen Reduced word manipulation: patterns and enumeration, J. Algebr. Comb. 46, No. 1, 189-217 (2017), w in S_n(231): l(w)=4.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Partial sums of A177787.

Cf. A089071, A022856, A152948.

List([0..30],n->n*(n+1)*(n^2+n+22)/24); # Muniru A Asiru, Aug 15 2018

seq(coeff(series(x*(2*x^2-3*x+2)/(1-x)^5, x,n+1),x,n),n=0..30); # Muniru A Asiru, Aug 15 2018

a(n) = n*(n+1)*(n^2+n+22)/24; \\ Michel Marcus, Aug 17 2018

G.f.: x*(2*x^2-3*x+2)/(1-x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = (1/6)*Sum_{i=1..n} (n-i)*((n-i)^2+11), for n >= 1.

A296636 Sequences n(n+1)(6n+1)/2 and n(n+1)(7n+1)/2 interleaved.

Original entry on oeis.org

0, 7, 8, 39, 45, 114, 132, 250, 290, 465, 540, 777, 903, 1204, 1400, 1764, 2052, 2475, 2880, 3355, 3905, 4422, 5148, 5694, 6630, 7189, 8372, 8925, 10395, 10920, 12720, 13192, 15368, 15759, 18360, 18639, 21717, 21850, 25460, 25410, 29610, 29337, 34188, 33649, 39215, 38364, 44712

Offset: 0

Luce ETIENNE, Dec 17 2017

Difference between these subsequences is A002411.

This sequence gives numbers of triangles all sizes in every n-th stage [of what? - N. J. A. Sloane, Feb 09 2018].

Colin Barker, Table of n, a(n) for n = 0..1000
Luce ETIENNE, Illustration
Eric Weisstein's World of Mathematics, Polyiamond
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

Cf. A002411, A002717, A006331, A033581, A033582, A255211.

List([0..50], n -> (2*n+1-(-1)^n)*(2*n+5-(-1)^n)*(13*n+10+(n-6)*(-1)^n)/128); # Bruno Berselli, Feb 12 2018

[(2*n+1-(-1)^n)*(2*n+5-(-1)^n)*(13*n+10+(n-6)*(-1)^n)/128: n in [0..50]]; // Bruno Berselli, Feb 12 2018

CoefficientList[Series[x (7 + 8 x + 11 x^2 + 13 x^3)/((1 - x)^4*(1 + x)^4), {x, 0, 46}], x] (* Michael De Vlieger, Dec 18 2017 *)
LinearRecurrence[{0,4,0,-6,0,4,0,-1},{0,7,8,39,45,114,132,250},50] (* Harvey P. Dale, May 01 2018 *)
Rest[Flatten[Table[With[{c=(n(n+1))/2},{c*(6n+1),c*(7n+1)}],{n,0,30}]]] (* Harvey P. Dale, Oct 11 2020 *)

concat(0, Vec(x*(7 + 8*x + 11*x^2 + 13*x^3) / ((1 - x)^4*(1 + x)^4) + O(x^80))) \\ Colin Barker, Dec 18 2017

a(n) = a(n-1)+4*a(n-2)-4*a(n-3)-6*a(n-4)+6*a(n-5)+4*a(n-6)-4*a(n-7)-a(n-8)+a(n-9).
a(n) = (2*n+1-(-1)^n)*(2*n+5-(-1)^n)*(13*n+10+(n-6)*(-1)^n)/128.
From Colin Barker, Dec 18 2017: (Start)
G.f.: x*(7 + 8*x + 11*x^2 + 13*x^3) / ((1 - x)^4*(1 + x)^4).
a(n) = 4*a(n-2) - 6*a(n-4) + 4*a(n-6) - a(n-8) for n>7.
(End)

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

A280304 a(n) = 3n(n^2 + 3*n + 4).

Original entry on oeis.org

0, 24, 84, 198, 384, 660, 1044, 1554, 2208, 3024, 4020, 5214, 6624, 8268, 10164, 12330, 14784, 17544, 20628, 24054, 27840, 32004, 36564, 41538, 46944, 52800, 59124, 65934, 73248, 81084, 89460, 98394, 107904, 118008, 128724, 140070, 152064, 164724, 178068, 192114, 206880, 222384, 238644, 255678, 273504, 292140, 311604, 331914, 353088, 375144, 398100

Offset: 0

Luce ETIENNE, Dec 31 2016

Numbers of unit triangles in a certain structure obtained from A006003.

			a(0) = 6*(1-1) = 0, a(1) = 6*(5-1) = 24, a(2) = 6*(15-1) = 84, a(3) = 6*(34-1) = 198, a(4) = 6*(65-1) = 384.

Colin Barker, Table of n, a(n) for n = 0..1000
Luce ETIENNE, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A003215, A005448, A006003, A033428, A213389, A269064.

[3*n*(n^2 + 3*n + 4) : n in [0..60]]; // Wesley Ivan Hurt, Dec 31 2016

A280304:=n->3*n*(n^2 + 3*n + 4): seq(A280304(n), n=0..60); # Wesley Ivan Hurt, Dec 31 2016

Table[3 n (n^2 + 3 n + 4), {n, 0, 50}] (* or *)
CoefficientList[Series[6 x (x^2 - 2 x + 4)/(1 - x)^4, {x, 0, 50}], x] (* Michael De Vlieger, Dec 31 2016 *)
LinearRecurrence[{4,-6,4,-1},{0,24,84,198},60] (* Harvey P. Dale, Feb 08 2023 *)

concat(0, Vec(6*x*(x^2-2*x+4) / (1-x)^4 + O(x^30))) \\ Colin Barker, Dec 31 2016

G.f.: 6*x*(x^2-2*x+4) / (1-x)^4.
a(n) = 6*(A006003(n+1)-1).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. - Colin Barker, Dec 31 2016

A274772 Zero together with the partial sums of A056640.

Original entry on oeis.org

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

I

			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)

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

A276071 n^3 followed by n^2 followed by n^4 followed by n.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 8, 4, 16, 2, 27, 9, 81, 3, 64, 16, 256, 4, 125, 25, 625, 5, 216, 36, 1296, 6, 343, 49, 2401, 7, 512, 64, 4096, 8, 729, 81, 6561, 9, 1000, 100, 10000, 10, 1331, 121, 14641, 11, 1728, 144, 20736, 12, 2197, 169, 28561, 13, 2744, 196, 38416, 14, 3375, 225, 50625, 15

Offset: 0

Luce ETIENNE, Aug 18 2016

			Rectangular array with four columns begins:
.   0,  0,    0, 0;
.   1,  1,    1, 1;
.   8,  4,   16, 2;
.  27,  9,   81, 3;
.  64, 16,  256, 4;
. 125, 25,  625, 5;
. 216, 36, 1296, 6; ...

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,5,0,0,0,-10,0,0,0,10,0,0,0,-5,0,0,0,1).

Cf. A110009.

&cat[[n^3,n^2,n^4,n]: n in [0..20]]; // Bruno Berselli, Aug 21 2016

Flatten[Table[{n^3, n^2, n^4, n}, {n, 0, 20}]] (* Bruno Berselli, Aug 21 2016 *)

concat(vector(4), Vec(x^4*(1+x+x^2+x^3+3*x^4-x^5+11*x^6-3*x^7-3*x^8-x^9+11*x^10+3*x^11-x^12+x^13+x^14-x^15) / ((1-x)^5*(1+x)^5*(1+x^2)^5) + O(x^60))) \\ Colin Barker, Aug 18 2016

a(n) = (2*n - 3 + (-1)^n + 2*(-1)^((2*n - 1 + (-1)^n)/4))*(n^3 - 2*n^2 + 28*n + 40 + (n^3 - 2*n^2 - 4*n - 56)*(-1)^n - (n^3 - 10*n^2 - 4*n + 72)*(-1)^((2*n - 1 + (-1)^n)/4) - (n^3 - 10*n^2 + 28*n - 88)*(-1)^((2*n + 1 - (-1)^n)/4))/2048.

G.f.: x^4*(1 + x + x^2 + x^3 + 3*x^4 - x^5 + 11*x^6 - 3*x^7 - 3*x^8 - x^9 + 11*x^10 + 3*x^11 - x^12 + x^13 + x^14 - x^15)/((1 - x)^5*(1 + x)^5*(1 + x^2)^5). - Colin Barker, Aug 18 2016

A275112 Zero together with the partial sums of A064412.

Original entry on oeis.org

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

cp's OEIS Frontend

User: Luce ETIENNE

A338996 Numbers of squares and rectangles of all sizes in 3*n*(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A330805 Number of squares and rectangles in the interior of the square with vertices (n,0), (0,n), (-n,0) and (0,-n) in a square (x,y)-grid.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A318054 a(n) = n*(n + 1)*(n^2 + n + 22)/24.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

GAP

Maple

PARI

Formula

A296636 Sequences n*(n+1)*(6*n+1)/2 and n*(n+1)*(7*n+1)/2 interleaved.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A280304 a(n) = 3*n*(n^2 + 3*n + 4).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A274772 Zero together with the partial sums of A056640.

Views

Author

Keywords

A338996 Numbers of squares and rectangles of all sizes in 3n(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.

A318054 a(n) = n(n + 1)(n^2 + n + 22)/24.

A296636 Sequences n(n+1)(6n+1)/2 and n(n+1)(7n+1)/2 interleaved.

A280304 a(n) = 3n(n^2 + 3*n + 4).

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