A180033 -id:A180033 - cp's OEIS frontend

A180032 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1+x)/(1-5x-7x^2).

1, 6, 37, 227, 1394, 8559, 52553, 322678, 1981261, 12165051, 74694082, 458625767, 2815987409, 17290317414, 106163498933, 651849716563, 4002393075346, 24574913392671, 150891318490777, 926480986202582, 5688644160448349

Offset: 0

Johannes W. Meijer, Aug 09 2010

The a(n) represent the number of n-move routes of a fairy chess piece starting in a given corner or side square (m = 1, 3, 7, 9; 2, 4, 6, 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a white chess queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen.
On a 3 X 3 chessboard there are 2^9 = 512 ways to explode with fury on the central square (we assume here that a red queen might behave like a white queen). The red queen is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program. For the corner and side squares the 512 red queens lead to 17 red queen sequences, see the cross-references for the complete set.
The sequence above corresponds to 8 red queen vectors, i.e., A[5] vectors, with decimal values 239, 367, 431, 463, 487, 491, 493 and 494. The central square leads for these vectors to A152240.
This sequence belongs to a family of sequences with g.f. (1+x)/(1 - 5*x - k*x^2). The members of this family that are red queen sequences are A180030 (k=8), A180032 (k=7; this sequence), A000400 (k=6), A180033 (k=5), A126501 (k=4), A180035 (k=3), A180037 (k=2) A015449 (k=1) and A003948 (k=0). Other members of this family are A030221 (k=-1), A109114 (k=-3), A020989 (k=-4), A166060 (k=-6).
Inverse binomial transform of A054413.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Wikipedia, Alice in Wonderland (2010 film)
Index entries for linear recurrences with constant coefficients, signature (5, 7).

Cf. A180028 (Central square).

Cf. Red queen sequences corner and side squares [decimal value A[5]]: A090018 [511], A135030 [255], A180030 [495], A005668 [127], A180032 [239], A000400 [63], A180033 [47], A001109 [31], A126501 [15], A154244 [23], A180035 [7], A138395 [19], A180037 [3], A084326 [17], A015449 [1], A003463 [16], A003948 [0].

I:=[1,6]; [n le 2 select I[n] else 5*Self(n-1)+7*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

with(LinearAlgebra): nmax:=20; m:=1; A[5]:= [1,1,1,1,0,1,1,1,0]: A:=Matrix([[0,1,1,1,1,0,1,0,1], [1,0,1,1,1,1,0,1,0], [1,1,0,0,1,1,1,0,1], [1,1,0,0,1,1,1,1,0], A[5], [0,1,1,1,1,0,0,1,1], [1,0,1,1,1,0,0,1,1], [0,1,0,1,1,1,1,0,1], [1,0,1,0,1,1,1,1,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);

LinearRecurrence[{5,7},{1,6},40] (* Vincenzo Librandi, Nov 15 2011 *)
CoefficientList[Series[(1+x)/(1-5x-7x^2),{x,0,30}],x] (* Harvey P. Dale, Apr 04 2024 *)

G.f.: (1+x)/(1 - 5*x - 7*x^2).
a(n) = 5*a(n-1) + 7*a(n-2) with a(0) = 1 and a(1) = 6.
a(n) = ((7+9*A)*A^(-n-1) + (7+9*B)*B^(-n-1))/53 with A = (-5+sqrt(53))/14 and B = (-5-sqrt(53))/14.

A180165 Triangle read by rows, derived from an array of sequences generated from (1 + x)/ (1 - rx - rx^2).

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 8, 5, 1, 5, 15, 22, 8, 1, 6, 24, 57, 60, 13, 1, 7, 35, 116, 216, 164, 21, 1, 8, 48, 205, 560, 819, 448, 34, 1, 9, 63, 330, 1200, 2704, 3105, 1224, 55, 1, 10, 80, 497, 2268, 7025, 13056, 11772, 3344, 89, 1, 11, 99, 712, 3920, 15588, 41125, 63040, 44631, 9136, 144

Offset: 1

Gary W. Adamson, Aug 14 2010

Row sums = A180166: (1, 3, 7, 18, 51, 161, 560, 2163, ...).

Rows of the array, with other offsets: (row 1 = A000045 starting with offset 2: (1, 2, 3, 5, 8, 13, ...); and for rows > 1, the entries: A028859, A125145, A086347, and A180033 start with offset 0; with the offset in the present array = 1.

			First few rows of the triangle:
  1;
  1, 2;
  1, 3, 3;
  1, 4, 8, 5;
  1, 5, 15, 22, 8;
  1, 6, 24, 57, 60, 13;
  1, 7, 35, 116, 216, 164, 21;
  1, 8, 48, 205, 560, 819, 448, 34;
  1, 9, 63, 330, 1200, 2704, 3105, 1224, 55;
  1, 10, 80, 497, 2268, 7025, 13056, 11772, 3344, 89;
  1, 11, 99, 712, 3920, 15588, 41125, 63040, 44631, 9136, 144;
  1, 12, 120, 981, 6336, 30919, 107136, 240750, 304384, 169209, 24960, 233;
  ...
As an array A(r,k) by upwards antidiagonals:
        k=1  k=2  k=3   k=4    k=5
  r=1:   1,   2,    3,    5,     8, ...
  r=2:   1,   3,    8,   22,    60, ...
  r=3:   1,   4,   15,   57,   216, ...
  r=4:   1,   5,   24,  116,   560, ...
  r=5:   1,   6,   35,  205,  1200, ...
Row r=5 = A180033 = (1, 6, 35, 205,...) and is generated from (1+x)/(1-5*x-5*x^2); is the INVERT transform of row r=4; and the array term A(5,4) = 205 = 5*35 + 5*6.
Terms A(2,4) and A(2,5) = [22,60] = [0,1; 2,2]^3 * [1,3].

Robert P. P. McKone, Antidiagonals n = 0..199, flattened

Cf. A180166, A000045, A028859, A125145, A086347, A180033.

Cf. A340156.

A180165[a_] := Reverse[Table[Table[CoefficientList[Series[(1 + x)/(1 - r*x - r*x^2), {x, 0, a - 2}], x], {r, 1, a + 1}][[k, n - k]], {n, 1, a}, {k, 1, n - 1}], 2] // Flatten;
A180165[12] (* Robert P. P. McKone, Jan 19 2021 *)

Triangle read by rows, generated from an array of sequences generated from (1 + x)/(1 - r*x - r*x^2); r > 0.
Alternatively, given the array with offset 1, the sequence r-th sequence is generated from a(k) = r*a(k-1) + r*(k-2); a(1) = 1, a(2) = r+1.
With a matrix method, the array can be generated from a 2 X 2 matrix of the form [0,1; r,r] = M, such that M^n * [1,r+1] = [r,n+1; r,n+2].
Also, for r > 1, the (r+1)-th row of the array is the INVERT transform of the r-th row.

a(35) corrected by Robert P. P. McKone, Dec 31 2020

A340156 Square array read by upward antidiagonals: T(n, k) is the number of n-ary strings of length k containing 00.

Original entry on oeis.org

1, 1, 3, 1, 5, 8, 1, 7, 21, 19, 1, 9, 40, 79, 43, 1, 11, 65, 205, 281, 94, 1, 13, 96, 421, 991, 963, 201, 1, 15, 133, 751, 2569, 4612, 3217, 423, 1, 17, 176, 1219, 5531, 15085, 20905, 10547, 880, 1, 19, 225, 1849, 10513, 39186, 86241, 92935, 34089, 1815

Offset: 2

Robert P. P. McKone, Dec 29 2020

			For n = 3 and k = 4, there are 21 strings: {0000, 0001, 0002, 0010, 0011, 0012, 0020, 0021, 0022, 0100, 0200, 1000, 1001, 1002, 1100, 1200, 2000, 2001, 2002, 2100, 2200}.
Square table T(n,k):
     k=2:  k=3:  k=4:   k=5:    k=6:     k=7:
n=2:   1     3     8     19      43       94
n=3:   1     5    21     79     281      963
n=4:   1     7    40    205     991     4612
n=5:   1     9    65    421    2569    15085
n=6:   1    11    96    751    5531    39186
n=7:   1    13   133   1219   10513    87199
n=8:   1    15   176   1849   18271   173608
n=9:   1    17   225   2665   29681   317817

Robert P. P. McKone, Antidiagonals n = 2..100, flattened

Cf. A008466 (row 2), A186244 (row 3), A000567 (column 4).
Cf. A180165 (not containing 00), A340242 (containing 000).
Cf. A000045, A028859, A125145, A086347, A180033, A180167, A322054.
Cf. A005563, A033445.

m[r_] := Normal[With[{p = 1/n}, SparseArray[{Band[{1, 2}] -> p, {i_, 1} /; i <= r -> 1 - p, {r + 1, r + 1} -> 1}]]];
T[n_, k_, r_] := MatrixPower[m[r], k][[1, r + 1]]*n^k;
Reverse[Table[T[n, k - n + 2, 2], {k, 2, 11}, {n, 2, k}], 2] // Flatten (* Robert P. P. McKone, Jan 26 2021 *)

T(n, k) = n^k - A180165(n+1,k-1), where A180165 in the number of strings not containing 00.

m(2) = [1 - 1/n, 1/n, 0; 1 - 1/n, 0, 1/n; 0, 0, 1], is the probability/transition matrix for two consecutive "0" -> "containing 00".

A180167 a(0) = 1, a(1) = 7; a(n)= 6a(n-1) + 6a(n-2) for n>1.

Original entry on oeis.org

1, 7, 48, 330, 2268, 15588, 107136, 736344, 5060880, 34783344, 239065344, 1643092128, 11292944832, 77616221760, 533454999552, 3666427327872, 25199293964544, 173194327754496, 1190361730314240, 8181336348412416, 56230188472359936, 386469148924634112

Offset: 0

Gary W. Adamson, Aug 14 2010

			a(4) = 2268 = 6*a(3) + 6*a(2) = 6*330 + 6*48.
Using the INVERT transform operation, a(3) = 330 = (205, 35, 6, 1) dot
(1, 1, 7, 48) = (205 + 35 + 42 + 48), where (1, 6, 35, 205, 1200, ...) = A180033.
G.f. = 1 + 7*x + 48*x^2 + 330*x^3 + 2268*x^4 + 15588*x^5 + 107136*x^6 + ...

Colin Barker, Table of n, a(n) for n = 0..1000
Jean-Paul Allouche, Jeffrey Shallit, and Manon Stipulanti, Combinatorics on words and generating Dirichlet series of automatic sequences, arXiv:2401.13524 [math.CO], 2025. See p. 14.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015) # 15.4.7.
Index entries for linear recurrences with constant coefficients, signature (6,6).

Cf. A057089, A180033.

CoefficientList[Series[(1 + x)/(1 - 6 x - 6 x^2), {x, 0, 21}], x] (* Michael De Vlieger, Dec 16 2021 *)

Vec((1 + x)/(1 - 6*x - 6*x^2) + O(x^50)) \\ Colin Barker, May 13 2016

G.f.: (1 + x)/(1 - 6*x - 6*x^2); = INVERT transform of A180033
a(n) = ((3-sqrt(15))^n*(-4+sqrt(15))+(3+sqrt(15))^n*(4+sqrt(15)))/(2*sqrt(15)). - Alexander R. Povolotsky, Aug 22 2010, corrected by Colin Barker, May 13 2016
a(n) = A057089(n) + A057089(n-1). - R. J. Mathar, Apr 04 2012
E.g.f.: (4*sqrt(15)*sinh(sqrt(15)*x) + 15*cosh(sqrt(15)*x))*exp(3*x)/15. - Ilya Gutkovskiy, May 13 2016

cp's OEIS Frontend

A180032 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1+x)/(1-5*x-7*x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A180165 Triangle read by rows, derived from an array of sequences generated from (1 + x)/ (1 - r*x - r*x^2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A340156 Square array read by upward antidiagonals: T(n, k) is the number of n-ary strings of length k containing 00.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A180167 a(0) = 1, a(1) = 7; a(n)= 6*a(n-1) + 6*a(n-2) for n>1.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A180032 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1+x)/(1-5x-7x^2).

A180165 Triangle read by rows, derived from an array of sequences generated from (1 + x)/ (1 - rx - rx^2).

A180167 a(0) = 1, a(1) = 7; a(n)= 6a(n-1) + 6a(n-2) for n>1.