A163433 -id:A163433 - cp's OEIS frontend

A163437 Number of different fixed (possibly) disconnected polyominoes (of any area) bounded tightly by an n X n square.

1, 7, 322, 51472, 29671936, 64588152832, 545697103347712, 18161310923858378752, 2399054119350722118025216, 1262710910458264839283982467072, 2653270028014955753823799266500411392

Offset: 1

David Bevan, Jul 28 2009

nonn

			a(2)=7: 2 rotations of the strictly disconnected domino consisting of two squares connected at a vertex, 4 rotations of the L tromino, and the square tetromino.

G. C. Greubel, Table of n, a(n) for n = 1..55

Cf. A162677 (bound not necessarily tight), A163433 (fixed disconnected trominoes), A163434 (fixed disconnected tetrominoes), A163435 (fixed disconnected pentominoes), A163436 (fixed disconnected n-ominoes).

Table[2^(n^2) - 4*2^((n - 1)*n) + 4*2^((n - 1)^2) + 2*2^((n - 2)*n) -
  4*2^((n - 2)*(n - 1)) + 2^((n - 2)^2), {n, 1, 25}] (* G. C. Greubel, Dec 23 2016 *)

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

A077958 Expansion of 1/(1-2*x^3).

Original entry on oeis.org

1, 0, 0, 2, 0, 0, 4, 0, 0, 8, 0, 0, 16, 0, 0, 32, 0, 0, 64, 0, 0, 128, 0, 0, 256, 0, 0, 512, 0, 0, 1024, 0, 0, 2048, 0, 0, 4096, 0, 0, 8192, 0, 0, 16384, 0, 0, 32768, 0, 0, 65536, 0, 0, 131072, 0, 0, 262144, 0, 0, 524288, 0, 0, 1048576, 0, 0, 2097152, 0, 0, 4194304, 0, 0, 8388608, 0

Offset: 0

N. J. A. Sloane, Nov 17 2002

a(n) is the number of L-tromino tilings of the n X 2 rectangle (see Exercise 2 in Grinberg). - Stefano Spezia, Nov 26 2019

G. C. Greubel, Table of n, a(n) for n = 0..1002
Darij Grinberg, Math 222: Enumerative Combinatorics, Fall 2019: Homework 1, Drexel University, Department of Mathematics, 2019.
Index entries for linear recurrences with constant coefficients, signature (0,0,2).

Cf. A006801, A162673, A163433, A329185.

R:=PowerSeriesRing(Integers(), 80); Coefficients(R!( 1/(1-2*x^3) )); // G. C. Greubel, Jun 23 2019

CoefficientList[Series[1/(1-2*x^3),{x,0,80}],x] (* Vladimir Joseph Stephan Orlovsky, Jan 30 2012 *)
Riffle[Riffle[2^Range[0,30],0],0,{3,-1,3}] (* Harvey P. Dale, Dec 18 2012 *)

Vec(1/(1-2*x^3)+O(x^80)) \\ Charles R Greathouse IV, Sep 27 2012

a(n)= !(n%3)<<(n\3) \\ Ruud H.G. van Tol, Mar 28 2025

(1/(1-2*x^3)).series(x, 80).coefficients(x, sparse=False) # G. C. Greubel, Jun 23 2019

From Stefano Spezia, Nov 26 2019: (Start)
a(n) = 2^(n/3) if 3 divides n, otherwise a(n) = 0 (see Exercise 2 in Grinberg).
E.g.f.: (1/3)*(exp(-(-2)^(1/3)*x) + exp(2^(1/3)*x) + exp((-1)^(2/3)*2^(1/3)*x)). (End)

A163434 Number of different fixed (possibly) disconnected tetrominoes bounded tightly by an n X n square.

Original entry on oeis.org

0, 1, 70, 425, 1426, 3577, 7526, 14065, 24130, 38801, 59302, 87001, 123410, 170185, 229126, 302177, 391426, 499105, 627590, 779401, 957202, 1163801, 1402150, 1675345, 1986626, 2339377, 2737126, 3183545, 3682450, 4237801, 4853702

Offset: 1

David Bevan, Jul 28 2009

			a(2)=1: the (connected) square tetromino.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A162674, A163433, A163435, A163437.

Join[{0}, Table[(2 n^2 - 4 n + 1)*(3 n^2 - 6 n + 1), {n, 2, 50}]] (* or *) Join[{0}, LinearRecurrence[{5,-10,10,-5,1}, {1, 70, 425, 1426, 3577}, 50]] (* G. C. Greubel, Dec 23 2016 *)

concat([0], Vec(x^2*(1+65*x+85*x^2-9*x^3+2*x^4)/(1-x)^5 + O(x^50))) \\ G. C. Greubel, Dec 23 2016

a(n) = (2n^2 -4n +1)*(3n^2 -6n +1), n>1.
G.f.: x^2*(1+65*x+85*x^2-9*x^3+2*x^4)/(1-x)^5. - Colin Barker, Apr 25 2012
E.g.f.: (6*x^4 + 12*x^3 - x^2 + x + 1)*exp(x) - 2 x - 1. - G. C. Greubel, Dec 23 2016

A103115 a(n) = 6n(n-1) - 1.

Original entry on oeis.org

-1, -1, 11, 35, 71, 119, 179, 251, 335, 431, 539, 659, 791, 935, 1091, 1259, 1439, 1631, 1835, 2051, 2279, 2519, 2771, 3035, 3311, 3599, 3899, 4211, 4535, 4871, 5219, 5579, 5951, 6335, 6731, 7139, 7559, 7991, 8435, 8891, 9359, 9839, 10331, 10835, 11351, 11879, 12419

Offset: 0

Jacob Landon (jacoblandon(AT)aol.com), May 09 2009

Star numbers A003154 minus 2.
What A163433 does for a triangle, this sequence is doing for a square but giving one-half the results. Take a square with vertices n, n+1, n+2, and n+3 and find the sum of the four products of each four vertices times the sum of the other three; at n you have n((n+1)+(n+2)+(n+3)) and so on for the other three vertices. The result of all four is 12*n^2 + 36*n + 22; half this is 6*n^2 + 18*n + 11 and gives the numbers in this sequence starting with n = 0. - J. M. Bergot, May 23 2012
Multiplying a(n) by 16 gives the sum of the convolution with itself of each of the 24 permutations of four consecutive numbers. - J. M. Bergot, May 15 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000914, A003154, A163433.

[6*n*(n-1)-1: n in [0..50]]; // Vincenzo Librandi, May 16 2017

Table[6n(n-1)-1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{-1,-1,11},50] (* Harvey P. Dale, Nov 14 2011 *)
CoefficientList[Series[(1 - 2 x - 11 x^2) / (x - 1)^3, {x, 0, 50}], x] (* Vincenzo Librandi, May 16 2017 *)

a(n)=6*n*(n-1)-1 \\ Charles R Greathouse IV, Jun 16 2017

a(n) = A003154(n) - 2.
G.f.: (1-2*x-11*x^2)/(x-1)^3. - R. J. Mathar, May 11 2009 (adapted by Vincenzo Librandi, May 16 2017).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=-1, a(1)=-1, a(2)=11. - Harvey P. Dale, Nov 14 2011
a(n) = (n-2)*(n-1 + n + n+1) + (n-1)*(n + n+1) + n*(n+1), which is applying A000914 to four consecutive numbers. - J. M. Bergot, May 15 2017
Sum_{n>=1} 1/a(n) = tan(sqrt(5/3)*Pi/2)*Pi/(2*sqrt(15)). Amiram Eldar, Aug 20 2022
E.g.f.: exp(x)*(6*x^2 - 1). - Elmo R. Oliveira, Jan 16 2025

Edited and extended by R. J. Mathar, May 11 2009

Entries rechecked by N. J. A. Sloane, Jul 18 2009

A162245 Triangle T(n,m) = 6mn + 3m + 3n + 1 read by rows.

Original entry on oeis.org

13, 22, 37, 31, 52, 73, 40, 67, 94, 121, 49, 82, 115, 148, 181, 58, 97, 136, 175, 214, 253, 67, 112, 157, 202, 247, 292, 337, 76, 127, 178, 229, 280, 331, 382, 433, 85, 142, 199, 256, 313, 370, 427, 484, 541, 94, 157, 220, 283, 346, 409, 472, 535, 598, 661

Offset: 1

Vincenzo Librandi, Jun 28 2009

If h belongs to the main diagonal of the triangle then 6*h+3 is a square since T(n,n) = (3/2)*(2*n+1)^2-1/2 and 6*T(n,n)+3 = 9*(2*n+1)^2. Also, the first column is A017209 (after 4). - Vincenzo Librandi, Nov 20 2012

			Triangle begins:
13;
22, 37;
31, 52,  73;
40, 67,  94,  121;
49, 82,  115, 148, 181;
58, 97,  136, 175, 214, 253;
67, 112, 157, 202, 247, 292, 337;
76, 127, 178, 229, 280, 331, 382, 433; etc.

Vincenzo Librandi, Rows n = 1..100, flattened

Cf. A003154, A017209, A083487, A163433.

[6*n*k + 3*n + 3*k + 1:  k in [1..n],  n in [1..11]]; // Vincenzo Librandi, Nov 20 2012

Flatten@Table[6*m*n + 3*m + 3*n + 1, {n, 20}, {m, n}] (* Vincenzo Librandi, Mar 03 2012 *)

Row sums: Sum_{m=1..n} T(n,m) = n*(5+6*n^2+15*n)/2. - R. J. Mathar, Jul 26 2009
T(n,m) = 3*A083487(n,m)+1. - R. J. Mathar, Jul 26 2009
T(k,k) = A003154(k+1) and T(k+1,k) = A163433(k+2). - Avi Friedlich, May 22 2015

Edited by R. J. Mathar, Jul 26 2009

A289181 Number of 6-cycles in the n X n knight graph.

Original entry on oeis.org

0, 0, 0, 20, 164, 616, 1348, 2352, 3628, 5176, 6996, 9088, 11452, 14088, 16996, 20176, 23628, 27352, 31348, 35616, 40156, 44968, 50052, 55408, 61036, 66936, 73108, 79552, 86268, 93256, 100516, 108048, 115852, 123928, 132276, 140896, 149788, 158952, 168388, 178096

Offset: 1

Eric W. Weisstein, Jun 27 2017

Eric Weisstein's World of Mathematics, Graph Cycle
Eric Weisstein's World of Mathematics, Knight Graph
Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).

Cf. A163433 (4-cycles).

Table[Length[FindCycle[KnightTourGraph[n, n], {6}, All]], {n, 20}]
Table[Piecewise[{{0, n < 4}, {20, n == 4}, {164, n == 5}}, 4 (34 n^2 - 259 n + 484)], {n, 20}]
Join[{0, 0, 0, 20, 164}, LinearRecurrence[{3, -3, 1}, {1036, 408, 52}, {6, 20}]]
CoefficientList[Series[(4 x^3 (-5 - 26 x - 46 x^2 + 7 x^3 + 2 x^4))/(-1 + x)^3, {x, 0, 20}], x]

a(n)=4*if(n>5,34*n^2 - 259*n + 484, max(36*n-139, 0)) \\ Charles R Greathouse IV, Oct 21 2022

For n > 5, a(n) = 4 (34 n^2 - 259 n + 484).

G.f.: (4*x^4*(-5 - 26*x - 46*x^2 + 7*x^3 + 2*x^4))/(-1 + x)^3.

cp's OEIS Frontend

A163437 Number of different fixed (possibly) disconnected polyominoes (of any area) bounded tightly by an n X n square.

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A077958 Expansion of 1/(1-2*x^3).

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A163434 Number of different fixed (possibly) disconnected tetrominoes bounded tightly by an n X n square.

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A103115 a(n) = 6*n*(n-1) - 1.

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A162245 Triangle T(n,m) = 6*m*n + 3*m + 3*n + 1 read by rows.

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Magma

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A289181 Number of 6-cycles in the n X n knight graph.

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Formula

A103115 a(n) = 6n(n-1) - 1.

A162245 Triangle T(n,m) = 6mn + 3m + 3n + 1 read by rows.