A054318 -id:A054318 - cp's OEIS frontend

A233427 Number A(n,k) of tilings of a k X n rectangle using pentominoes of any shape; square array A(n,k), n>=0, k>=0, read by antidiagonals.

1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 5, 0, 0, 5, 0, 1, 1, 0, 0, 56, 0, 56, 0, 0, 1, 1, 0, 0, 0, 501, 501, 0, 0, 0, 1, 1, 0, 0, 0, 0, 4006, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 27950, 27950, 0, 0, 0, 1, 1, 1, 0, 45, 0, 0, 214689, 0, 214689, 0, 0, 45, 0, 1

Offset: 0

Alois P. Heinz, Dec 09 2013

			A(5,2) = A(2,5) = 5:
  ._________. ._________. ._________. ._________. ._________.
  |_________| | ._____| | | |_____. | |   ._|   | |   |_.   |
  |_________| |_|_______| |_______|_| |___|_____| |_____|___|.
Square array A(n,k) begins:
  1, 1,  1,    1,      1,         1,          1, ...
  1, 0,  0,    0,      0,         1,          0, ...
  1, 0,  0,    0,      0,         5,          0, ...
  1, 0,  0,    0,      0,        56,          0, ...
  1, 0,  0,    0,      0,       501,          0, ...
  1, 1,  5,   56,    501,      4006,      27950, ...
  1, 0,  0,    0,      0,     27950,          0, ...
  1, 0,  0,    0,      0,    214689,          0, ...
  1, 0,  0,    0,      0,   1696781,          0, ...
  1, 0,  0,    0,      0,  13205354,          0, ...
  1, 1, 45, 7670, 890989, 101698212, 7845888732, ...
  ...

Liang Kai, Antidiagonals n = 0..26, flattened (Antidiagonals n = 0..17 from Alois P. Heinz)
R. S. Harris, Counting Polyomino Tilings
Liang Kai, Solving tiling enumeration problems by tensor network contractions, arXiv:2503.17698 [math.CO], 2025.
Wikipedia, Pentomino

Columns (or rows) include: A000012, A054318, A233428, A233429, A174249, A233430.
Cf. A099390, A233320, A230031, A246902, A247117, A278657.
Row sums of A247702, A247703, A247704, A247705, A247706, A247707, A247708, A247709, A247710, A247711, A247712, A247713 give A(n,5).

A(n,k) = 0 <=> n*k mod 5 > 0.

A350923 a(0) = 2, a(1) = 2, and a(n) = 10*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

2, 2, 14, 134, 1322, 13082, 129494, 1281854, 12689042, 125608562, 1243396574, 12308357174, 121840175162, 1206093394442, 11939093769254, 118184844298094, 1169909349211682, 11580908647818722, 114639177128975534, 1134810862641936614, 11233469449290390602, 111199883630261969402

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Essentially the same as A157085. - R. J. Mathar, Feb 07 2022

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A031138, A054318, A097784, A253175, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350921, A350922, A350924, A350925, A350926.

LinearRecurrence[{11, -11, 1}, {2, 2, 14}, 25] (* Paolo Xausa, May 30 2025 *)

G.f.: 2*(1 - 10*x + 7*x^2)/((1 - x)*(1 - 10*x + x^2)). - Stefano Spezia, Jan 22 2022
From Hugo Pfoertner, Jan 22 2022: (Start)
a(n) = A031138(n) + 1.
a(n) = 3*A054318(n) - 1.
a(n) = 12*A097784(n-2) + 2 for n >= 2. (End)
a(n) = 2 * A253175(n) for n>=1. - Alois P. Heinz, Jan 22 2022

A211955 Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 6, 10, 4, 1, 10, 30, 28, 8, 1, 15, 70, 112, 72, 16, 1, 21, 140, 336, 360, 176, 32, 1, 28, 252, 840, 1320, 1056, 416, 64, 1, 36, 420, 1848, 3960, 4576, 2912, 960, 128, 1, 45, 660, 3696, 10296, 16016, 14560, 7680, 2176, 256

Offset: 0

Peter Bala, Apr 30 2012

Let b(n,x) = Sum_{k = 0..n} binomial(n+k,2*k)*x^k denote the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients (in ascending powers of x) of the related polynomial sequence R(n,x) := (1/2)*b(n,2*x) + 1/2. Several sequences already in the database are of the form (R(n,x))n>=0 for a fixed value of x. These include A101265 (x = 1), A011900 (x = 2), A182432 (x = 3), A054318 (x = 4) as well as signed versions of A133872 (x = -1), A109613(x = -2), A146983 (x = -3) and A084159 (x = -4).

The polynomials R(n,x) factorize in the ring Z[x] as R(n,x) = P(n,x)*P(n+1,x) for n >= 1: explicitly, P(2*n,x) = 1/2*(b(2*n,2*x) + 1)/b(n,2*x) and P(2*n+1,x) = b(n,2*x). The coefficients of P(n,x) occur in several tables in the database, although without the connection to the Morgan-Voyce polynomials being noted - see A211956 for more details. In terms of T(n,x), the Chebyshev polynomials of the first kind, we have P(2*n,x) = T(2*n,u) and P(2*n+1,x) = 1/u * T(2*n+1,u), where u = sqrt((x+2)/2). Hence R(n,x) = 1/u * T(n,u) * T(n+1,u).

			Triangle begins
.n\k.|..0....1....2....3....4....5....6
= = = = = = = = = = = = = = = = = = = =
..0..|..1
..1..|..1....1
..2..|..1....3....2
..3..|..1....6...10....4
..4..|..1...10...30...28....8
..5..|..1...15...70..112...72...16
..6..|..1...21..140..336..360..176...32

Eric Weisstein's World of Mathematics, Morgan-Voyce polynomials

Cf. A011900, A084159, A085478, A101265 (row sums), A109613, A112373, A123519, A133872 (alt row sums), A146983, A182432, A204021, A208513, A211956, A211957.

T(n,0) = 1; T(n,k) = 2^(k-1)*binomial(n+k,2*k) for k > 0.
O.g.f. for column k (except column 0): 2^(k-1)*x^k/(1-x)^(2*k+1).
O.g.f.: (1-t*(x+2)+t^2)/((1-t)*(1-2*t(x+1)+t^2)) = 1 + (1+x)*t + (1+3*x+2*x^2)*t^2 + ....
Removing the first column from the triangle produces the Riordan array (x/(1-x)^3, 2*x/(1-x)^2).
The row polynomials R(n,x) := 1/2*b(n,2*x) + 1/2 = 1 + x*Sum_{k = 1..n} binomial(n+k,2*k)*(2*x)^(k-1).
Recurrence equation: R(n,x) = 2*(1+x)*R(n-1,x) - R(n-2,x) - x with initial conditions R(0,x) = 1, R(1,x) = 1+x.
Another recurrence is R(n,x)*R(n-2,x) = R(n-1,x)*(R(n-1,x) + x).
With P(n,x) as defined in the Comments section we have (x+2)/x - {Sum_{k = 0..2n} 1/R(k,x)}^2 = 2/(x*P(2*n+1,x)^2); (x+2)/x - {Sum_{k = 0..2n+1} 1/R(k,x)}^2 = (x+2)/(x*P(2*n+2,x)^2); consequently Sum_{k >= 0} 1/R(k,x) = sqrt((x+2)/x) for either x > 0 or x <= -2.
Row sums R(n,1) = A101265(n+1); Alt. row sums R(n,-1) = A133872(n+1);
R(n,2) = A011900(n); R(n,-2) = (-1)^n * A109613(n); R(n,3) = A182432;
R(n,-3) = (-1)^n * A146983(n); R(n,4) = A054318(n+1); R(n,-4) = (-1)^n * A084159(n).

A182432 Recurrence a(n)a(n-2) = a(n-1)(a(n-1) + 3) with a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 28, 217, 1705, 13420, 105652, 831793, 6548689, 51557716, 405913036, 3195746569, 25160059513, 198084729532, 1559517776740, 12278057484385, 96664942098337, 761041479302308, 5991666892320124, 47172293659258681, 371386682381749321

Offset: 0

Peter Bala, Apr 30 2012

The non-linear recurrence equation a(n)*a(n-2) = a(n-1)*(a(n-1) + r) with initial conditions a(0) = 1, a(1) = 1 + r has the solution a(n) = 1/2 + (1/2)*Sum_{k = 0..n} (2*r)^k*binomial(n+k,2*k) = 1/2 + b(n,2*r)/2, where b(n,x) are the Morgan-Voyce polynomials of A085478. The recurrence produces sequences A101265 (r = 1), A011900 (r = 2) and A054318 (r = 4), as well as signed versions of A133872 (r = -1), A109613 (r = -2), A146983 (r = -3) and A084159(r = -4).
Also the indices of centered pentagonal numbers (A005891) which are also centered triangular numbers (A005448). - Colin Barker, Jan 01 2015
Also positive integers y in the solutions to 3*x^2 - 5*y^2 - 3*x + 5*y = 0. - Colin Barker, Jan 01 2015

Seiichi Manyama, Table of n, a(n) for n = 0..1116 (first 201 terms from Vincenzo Librandi)
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Eric Weisstein's World of Mathematics, Morgan-Voyce Polynomials
Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A011900, A054318, A070997, A101265, A109613, A133872, A146983, A211955.

I:=[1, 4, 28]; [n le 3 select I[n] else 9*Self(n-1)-9*Self(n-2)+Self(n-3): n in [1..25]]; // Vincenzo Librandi, May 18 2012

RecurrenceTable[{a[0]==1,a[1]==4,a[n]==(a[-1+n] (3+a[-1+n]))/a [-2+n]}, a[n],{n,30}] (* or *) LinearRecurrence[{9,-9,1},{1,4,28},30] (* Harvey P. Dale, May 14 2012 *)

Vec((1-5*x+x^2)/((1-x)*(1-8*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 01 2015

a(n) = 1/2 + (1/2)*Sum_{k = 0..n} 6^k*binomial(n+k,2*k).
a(n) = R(n,3) where R(n,x) denotes the row polynomials of A211955.
a(n) = (1/u)*T(n,u)*T(n+1,u) with u = sqrt(5/2) and T(n,x) the n-th Chebyshev polynomial of the first kind.
Recurrence equation: a(n) = 8*a(n-1) - a(n-2) - 3 with a(0) = 1 and a(1) = 4.
O.g.f.: (1 - 5*x + x^2)/((1 - x)*(1 - 8*x + x^2)) = 1 + 4*x + 28*x^2 + ....
Sum_{n >= 0} 1/a(n) = sqrt(5/3); 5 - 3*(Sum_{k = 0..2*n} 1/a(k))^2 = 2/A070997(n)^2.
a(0) = 1, a(1) = 4, a(2) = 28, a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3). - Harvey P. Dale, May 14 2012

A253475 Indices of centered square numbers (A001844) which are also centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 6, 55, 540, 5341, 52866, 523315, 5180280, 51279481, 507614526, 5024865775, 49741043220, 492385566421, 4874114620986, 48248760643435, 477613491813360, 4727886157490161, 46801248083088246, 463284594673392295, 4586044698650834700, 45397162391834954701

Offset: 1

Colin Barker, Jan 02 2015

Also positive integers x in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0, the corresponding values of y being A054318.
Also indices of centered hexagonal numbers (A003215) which are also hexagonal numbers (A000384).
Also indices of terms in sequence A193218 which are the square root of a sum of 5th powers (A000539). - Daniel Poveda Parrilla, Jun 10 2017

			6 is in the sequence because the 6th centered square number is 61, which is also the 5th centered hexagonal number.

Colin Barker, Table of n, a(n) for n = 1..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A000384, A000539, A001844, A003215, A054318, A193218, A253175.

LinearRecurrence[{11, -11, 1}, {1, 6, 55}, 25] (* Paolo Xausa, May 30 2025 *)

Vec(x*(5*x-1)/((x-1)*(x^2-10*x+1)) + O(x^100))

a(n) = 11*a(n-1)-11*a(n-2)+a(n-3).
G.f.: x*(5*x-1) / ((x-1)*(x^2-10*x+1)).
a(n) = sqrt((-2-(5-2*sqrt(6))^n-(5+2*sqrt(6))^n)*(2-(5-2*sqrt(6))^(1+n)-(5+2*sqrt(6))^(1+n)))/(4*sqrt(2)). - Gerry Martens, Jun 04 2015
2*a(n) = 1+A054320(n-1). - R. J. Mathar, Feb 07 2022

A087125 Indices k of hex numbers H(k) that are also triangular.

Original entry on oeis.org

0, 5, 54, 539, 5340, 52865, 523314, 5180279, 51279480, 507614525, 5024865774, 49741043219, 492385566420, 4874114620985, 48248760643434, 477613491813359, 4727886157490160, 46801248083088245, 463284594673392294, 4586044698650834699, 45397162391834954700

Offset: 0

Eric W. Weisstein, Aug 14 2003

From the law of cosines, the non-Pythagorean triple {a(n), a(n)+1=A253475(n+1), A072256(n+1)} forms a near-isosceles triangle with the angle bounded by the consecutive sides equal to the regular tetrahedron's central angle (see A156546 and A247412). This implies also that a(n) are those numbers k such that (16/3)*A000217(k)+1 is a perfect square. - Federico Provvedi, Apr 04 2023

Colin Barker, Table of n, a(n) for n = 0..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A006244, A031138, A072256, A054318.

[Round((-4-(5-2*Sqrt(6))^n*(-2+Sqrt(6)) + (2+Sqrt(6))*(5 + 2*Sqrt(6))^n)/8): n in [0..25]]; // G. C. Greubel, Nov 04 2017

CoefficientList[Series[(-x^2+5*x)/((1-x)*(1-10*x+x^2)), {x, 0, 25}], x] (* G. C. Greubel, Nov 04 2017 *)
LinearRecurrence[{11,-11,1},{0,5,54},30] (* Harvey P. Dale, Jun 14 2022 *)
Table[(x Sqrt[z^(2 n + 1) + z^-(2 n + 1) - 2] - 4) / 8 //. {x -> Sqrt[2], y -> Sqrt[3], z -> (5 + 2 x y)}, {n, 0, 100}] // Round (* Federico Provvedi, Apr 16 2023 *)

concat(0, Vec(x*(x-5)/((x-1)*(x^2-10*x+1)) + O(x^50))) \\ Colin Barker, Jun 23 2015

G.f.: (-x^2+5*x)/((1-x)*(1-10*x+x^2)).
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) for n > 2. - Colin Barker, Jun 23 2015
a(n) = (-4 - (5-2*sqrt(6))^n*(-2 + sqrt(6)) + (2+sqrt(6))*(5+2*sqrt(6))^n)/8. - Colin Barker, Mar 05 2016
a(n) = 10*a(n-1) - a(n-2) + 4 for n > 1. - Charlie Marion, Feb 14 2023
a(n) = ((x^(n+1)+1)*(x^n-1))/(2*x^n*(x-1)), with x=5+2*sqrt(6). - Federico Provvedi, Apr 04 2023
a(n) = sqrt(3*A161680(A054318(n+1)) + 1/4) - 1/2 = floor(sqrt(3*A000217(A054318(n+1)-1) + 1/4)). - Federico Provvedi, Apr 16 2023

cp's OEIS Frontend

A233427 Number A(n,k) of tilings of a k X n rectangle using pentominoes of any shape; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A350923 a(0) = 2, a(1) = 2, and a(n) = 10*a(n-1) - a(n-2) - 4 for n >= 2.

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A211955 Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.

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A182432 Recurrence a(n)*a(n-2) = a(n-1)*(a(n-1) + 3) with a(0) = 1, a(1) = 4.

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A253475 Indices of centered square numbers (A001844) which are also centered hexagonal numbers (A003215).

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A087125 Indices k of hex numbers H(k) that are also triangular.

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A182432 Recurrence a(n)a(n-2) = a(n-1)(a(n-1) + 3) with a(0) = 1, a(1) = 4.