A121801 -id:A121801 - cp's OEIS frontend

A059929 a(n) = Fibonacci(n)*Fibonacci(n+2).

0, 2, 3, 10, 24, 65, 168, 442, 1155, 3026, 7920, 20737, 54288, 142130, 372099, 974170, 2550408, 6677057, 17480760, 45765226, 119814915, 313679522, 821223648, 2149991425, 5628750624, 14736260450, 38580030723, 101003831722, 264431464440, 692290561601, 1812440220360

Offset: 0

Henry Bottomley, Feb 09 2001

Expansion of golden ratio (1+sqrt(5))/2 as an infinite product: phi = Product_{i>=0} (1+1/(Fibonacci(2*i+1) * Fibonacci(2*i+3)-1)) * (1-1/(Fibonacci(2*i+2) * Fibonacci(2i+4)+1)). - Thomas Baruchel, Nov 11 2003
Each of these is one short of or one over the square of a Fibonacci number (A007598). This means that a rectangle sized F(n) by F(n + 2) units can't be converted into a square with sides of length F(n + 1) units unless one square unit of material is added or removed. - Alonso del Arte, May 03 2011
These are the integer parts of the numerators of the numbers with continued fraction representations [1, 2, 2, 2, ...], [1, 1, 2, 2, 2, ...], [1, 1, 1, 2, 2, 2, ...], etc., that is, sqrt(2), (2+sqrt(2))/2, 3-sqrt(2), (10+sqrt(2))/7, (24-sqrt(2))/14, etc. - Geoffrey Caveney, May 03 2014
a(n) appears also as the third component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, where F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-A248161(n), A079472(n+1), a(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - Wolfdieter Lang, Nov 01 2014
Numbers with a continued fraction expansion with the repeating sequence of length n [1, 1, ..., 1, 2], n-1 ones followed by a single two, for n > = 1, appear to be equal to (F(n) + sqrt(a(n)))/F(n+1), where F(n) = A000045(n). - R. James Evans, Nov 21 2018
The preceding conjecture is true. Proof: For n >= 1 let c(n) := confrac(repeat(1^{n-1}, 2)) where 1^{k} denotes 1 taken k times. This can be computed, e.g. from [Perron, third and fourth eq. on p. 62], as c(n) = (F(n) + sqrt(F(n+1)^2 - (-1)^n)) / F(n+1), which is the conjectured formula because F(n+1)^2 - (-1)^n = a(n). - Wolfdieter Lang, Jan 05 2019

			G.f. = 2*x + 3*x^2 + 10*x^3 + 24*x^4 + 65*x^5 + 168*x^6 + ... - _Michael Somos_, Mar 18 2022

Oskar Perron, Die Lehre von den Kettenbrüchen, Band I, 3. Auflage, B. G. Teubner, Stuttgart, 1954, pp. 61-61.

Muniru A Asiru, Table of n, a(n) for n = 0..2374 (first 501 terms from Harry J. Smith)
Tim Jones (producer), Engineering Bits & Bytes: The Fibonacci Puzzle, Wayne State University College of Engineering (2011).
Eric H. Kuo, Applications of graphical condensation for enumerating matchings and tilings, arXiv:math/0304090 [math.CO], 2003.
Marc Renault, Properties of the Fibonacci Sequence Under Various Moduli, Master's Thesis, Wake Forest University, 1996.
Michel Waldschmidt, Open Diophantine problems, arXiv:math/0312440 [math.NT], 2003-2004.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Bisection of A070550.
First differences of A059840.
Cf. A000045, A001519, A001654, A007598, A033999, A035513, A079472, A121801, A126358, A147973, A248161.

a:=List([0..30],n->Fibonacci(n)*Fibonacci(n+2));; Print(a); # Muniru A Asiru, Jan 05 2019

[Fibonacci(n)*Fibonacci(n+2): n in [0..30]]; // Vincenzo Librandi, Jul 02 2014

with(combinat): a:=n->fibonacci(n)*fibonacci(n+2): seq(a(n), n=0..26); # Zerinvary Lajos, Oct 07 2007

Table[Fibonacci[n]*Fibonacci[n+2],{n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Nov 17 2009 *)

a(n) = fibonacci(n)*fibonacci(n + 2) \\ Harry J. Smith, Jun 30 2009

from sympy import fibonacci
[fibonacci(n)*fibonacci(n+2) for n in range(30)] # Stefano Spezia, Jan 05 2019

[fibonacci(n)*fibonacci(n+2) for n in range(30)] # G. C. Greubel, Nov 21 2018

a(n) = Fibonacci(n+1)^2 - (-1)^n = A007598(n+1) + A033999(n+1) = A000045(n+1)^2 - A033999(n).
G.f.: (2*x-x^2) / ((1+x)*(1-3*x+x^2)).
Sum_{n>=1} 1/a(n) = 1.
Sum_{n>=1} (-1)^n/a(n) = 2 - sqrt(5).
Sum_{n>=1} 1/a(2n-1) = 1/phi = (sqrt(5) - 1)/2. - Franz Vrabec, Sep 15 2005
Sum_{n>=1} 1/a(2n) = (3 - sqrt(5))/2. - Franz Vrabec, Nov 30 2009
a(n) = ((7+3*sqrt(5))/10)*((3+sqrt(5))/2)^(n-1) + ((7-3*sqrt(5))/10)*((3-sqrt(5))/2)^(n-1) + (3/5)*(-1)^(n-1). - Tim Monahan, Aug 09 2011
a(n) = (Lucas(n+1)^2 - Fibonacci(n+1)^2)/4. - Vincenzo Librandi, Aug 02 2014
a(n) = F(n-2)*F(n) + F(n-1)*F(n) + F(n-2)*F(n+1) + F(n-1)*F(n+1), where F=A000045, F(-2)=-1, F(-1)=1. - Bruno Berselli, Nov 03 2015
a(n) = A035513(1,n-1)*A035513(3,n-1)/2 = A035513(1,n-1)*A035513(4,n-1)/3. - R. J. Mathar, Sep 04 2016
a(n)+a(n+1) = A001519(n+2). - R. J. Mathar, Oct 19 2021
a(n) = 2*A001654(n) - A001654(n-1). - R. J. Mathar, Oct 19 2021
a(n)+a(n+3) = 2*F(2n+5) = A126358(n+2). - Andrés Ventas, Oct 25 2021
Sum_{n>=1} Fibonacci(n+1)/a(n) = 2. - Amiram Eldar, Jan 11 2022
a(n) = a(-2-n) and a(n) + a(n+3) = 2*(a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Mar 18 2022

A079472 Number of perfect matchings on an n X n L-shaped graph.

Original entry on oeis.org

0, 2, 4, 12, 30, 80, 208, 546, 1428, 3740, 9790, 25632, 67104, 175682, 459940, 1204140, 3152478, 8253296, 21607408, 56568930, 148099380, 387729212, 1015088254, 2657535552, 6957518400, 18215019650, 47687540548, 124847601996, 326855265438, 855718194320

Offset: 1

Helen King (h.king(AT)uea.ac.uk), Jan 15 2003

a(n+1) = 2*F(n)*F(n+1) appears as the second component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, with F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-A248161(n), a(n+1), A059929(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - Wolfdieter Lang, Nov 01 2014
a(n+1) is the numerator of the continued fraction [1,...,1,2,1,...,1] with n 1's to the left of the central 2, and n 1's to the right of the central 2. For the denominators, see A061646. - Greg Dresden and Max Liu, Jun 25 2023
For n >= 3, a(n) equals the sum of the sides of the right triangle with side lengths [F(n)*F(n-3), 2*F(n-1)*F(n-2), F(2*n-3)] (n = 4 corresponds to the 3-4-5 right triangle). - Peter Bala, Nov 03 2023

			a(7) = 2*13*8 = 208 = number of matchings. F(7) = 13 F(6) = 8
a(3) = 4 because in the graph with vertex set {(0,0), (1,0), (2,0), (0,1), (1,1), (2,1), (0,2), (1,2)} and edge set {h(0,0), h(1,0), h(0,1), h(1,1), h(0,2), v(0,0), v(0,1), v(1,0), v(1,1), v(2,0)}, where h(i,j) (v(i,j)) is a horizontal (vertical) edge of unit length starting from vertex (i,j), we have the following four perfect matchings: {h(0,0), h(0,1), h(0,2), v(2,0)}, {h(0,0), v(0,1), v(1,1), v(2,0)}, {v(0,0), v(1,0), v(2,0), h(0,2)} and {v(0,0), h(1,0), h(1,1), h(0,2)}. - _Emeric Deutsch_, Dec 30 2004
G.f. = 2*x^2 + 4*x^3 + 12*x^4 + 30*x^5 + 80*x^6 + 208*x^7 + 546*x^8 + ...

Daniele Corradetti, La Metafisica del Numero, 2008.
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. pp. 178, 255.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
P. F. F. Espinosa, J. F. González, J. P. Herrán, A. M. Cañadas, and J. L. Ramírez, On some relationships between snake graphs and Brauer configuration algebras, Algebra Disc. Math. (2022) Vol. 33, No. 2, 29-59.
I. Gutman and S. J. Cyvin, A result on 1-factors related to Fibonacci numbers, The Fibonacci Quarterly, 28 (1990), pp. 81-84.
C.-A. Laisant, Observations sur les triangles rectangles en nombres entiers et les suites de Fibonacci, Nouvelles Annales de Math. (1919, in French) Series 4, Vol. 19, 391-397.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001654, A121646.

List([1..30], n -> 2*Fibonacci(n-1)*Fibonacci(n)); # G. C. Greubel, Jan 07 2019

[2*Fibonacci(n)*Fibonacci(n-1): n in [1..30]]; // Vincenzo Librandi, Jun 29 2014

with(combinat,fibonacci):seq(2*fibonacci(n)*fibonacci(n-1),n=1..30);

LinearRecurrence[{2,2,-1}, {0,2,4}, 30] (* Arkadiusz Wesolowski, Sep 15 2012 *)
Table[(2*Fibonacci[n]*Fibonacci[n-1]), {n,30}] (* Vincenzo Librandi, Jun 29 2014 *)

{a(n) = 2 * fibonacci(n) * fibonacci(n-1)}; \\ Michael Somos, Jun 28 2014

concat(0, Vec(2*x^2/((x+1)*(x^2-3*x+1)) + O(x^40))) \\ Colin Barker, Sep 27 2016

[2*fibonacci(n-1)*fibonacci(n) for n in (1..30)] # G. C. Greubel, Jan 07 2019

a(n) = 2*F(n)*F(n-1) where F(n) are the Fibonacci numbers (A000045).
From Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jan 18 2003: (Start)
a(n) = 2*A001654(n) = F(2*n) - F(n)^2 = A001906(n) - A007598(n).
a(n) = (F(n+1)^2 - F(n-2)^2)/2 = (A007598(n+1) - A007598(n-2))/2.
a(n) = 2*(L(2*n-1) + (-1)^n)/5 = (2/5)*(A002878(n-1) + A033999(n)), where L(n) = A000032(n).
a(n+1) = a(n) + 2*F(n)^2.
G.f.: 2*x^2/((1+x)*(1-3*x+x^2)). (End)
a(n) = Im( (F(n) + i*F(n+1))^2 ) (cf. A121646). - Daniele Corradetti (d.corradetti(AT)gmail.com), May 02 2008
From Michael Somos, Jun 28 2014: (Start)
a(n) = F(n+1)^2 - F(n)^2 - F(n-1)^2.
a(1 - n) = -a(n). (End)
a(n) = ( 2*(-1)^n - (1+sqrt(5))*((3-sqrt(5))/2)^n - (1-sqrt(5))*((3+sqrt(5))/2)^n )/5. - Colin Barker, Sep 27 2016
From Rigoberto Florez, May 06 2020: (Start)
a(n) = F(2n-2) + F(n-1)^2, where F(n) is the n-th Fibonacci number.
a(n) = M^(n+1)[2,1], for n>0 where M=[0,0,1;0,1,2;1,1,1]. (End)
a(n) = F(n)^2 + F(n-1)^2 - F(n-2)^2. - Michael Somos, Mar 02 2023

More terms from Benoit Cloitre and Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jan 18 2003

A226205 a(n) = F(n)^2 - F(n-1)^2 or F(n+1) * F(n-2) where F(n) = A000045(n), the Fibonacci numbers.

Original entry on oeis.org

1, 0, 3, 5, 16, 39, 105, 272, 715, 1869, 4896, 12815, 33553, 87840, 229971, 602069, 1576240, 4126647, 10803705, 28284464, 74049691, 193864605, 507544128, 1328767775, 3478759201, 9107509824, 23843770275, 62423800997, 163427632720, 427859097159, 1120149658761

Offset: 1

Michael Somos, Jun 06 2013

A001519(n)^2 = A079472(n)^2 + a(n)^2 and (A001519(n), A079472(n), a(n)) is a Pythagorean triple.
INVERT transform is A052156. PSUM transform is A007598. SUMADJ transform is A088305. BINOMIAL transform is A039717. BINOMIAL transform with 0 prepended is A112091 with 0 prepended. BINOMIAL transform inverse is A084179(n+1).
In general, the difference between squares of two consecutive terms of a second order linear recurrence having a signature of (c,d) will be a third order recurrence with signature (c^2+d,(c^2+d)*d,-d^3). - Gary Detlefs, Mar 13 2025

			G.f. = x + 3*x^3 + 5*x^4 + 16*x^5 + 39*x^6 + 105*x^7 + 272*x^8 + 715*x^9 + ...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
John P. Bonomo and Montana Ferita, A Small Fib, College Math. J., 2023.
Nurettin Irmak, Product of arbitrary Fibonacci numbers with distance 1 to Fibonomial coefficient, Turk J Math, (2017) 41: 825-828. See p. 828.
C.-A. Laisant, Observations sur les triangles rectangles en nombres entiers et les suites de Fibonacci, Nouvelles Annales de Math. (1919, in French) Série 4, Vol. 19, 391-397.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001519, A001622, A001654, A007598, A033999, A039717, A052156, A079472, A084179, A088305, A112091, A121646, A290565.
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.
Cf. A260259: numbers of the form F(n)*F(n+1)-(-1)^n. - Bruno Berselli, Nov 02 2015

[Fibonacci(n)^2-Fibonacci(n-1)^2: n in [1..40]]; // Vincenzo Librandi, Jun 18 2014

a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<1,0,3>>)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 28 2016

a[ n_] := Fibonacci[n + 1] Fibonacci[n - 2]; (* Michael Somos, Jun 17 2014 *)
CoefficientList[Series[(1 - x)^2/((1 + x) (1 - 3 x + x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 17 2014 *)

{a(n) = fibonacci( n + 1) * fibonacci( n - 2)};

a(n) = round(2^(-1-n)*(-(-1)^n*2^(3+n)-(3-sqrt(5))^n*(1+sqrt(5))+(-1+sqrt(5))*(3+sqrt(5))^n)/5) \\ Colin Barker, Sep 28 2016

lista(nn) = {my(p = (3*x-1)/(x^3-2*x^2-2*x+1)); for (n=1, nn, p = deriv(p, x); print1(subst(p, x, 0)/n!, ", "); ); } \\ Michel Marcus, May 22 2018

G.f.: x * (1 - x)^2 / ((1 + x) * (1 -3*x + x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = - A121646(n).
a(n) = -a(1-n) for all n in Z.
a(n) = A121801(n+1) / 2. - Michael Somos, Jun 17 2014
a(n) = a(n-1) + A000045(n-1)^2 - 2*(-1)^n, for n>1. - Alexander Samokrutov, Sep 07 2015
a(n) = F(n-1)*F(n) - (-1)^n. - Bruno Berselli, Oct 30 2015
a(n) = 2^(-1-n)*(-(-1)^n*2^(3+n)-(3-sqrt(5))^n*(1+sqrt(5))+(-1+sqrt(5))*(3+sqrt(5))^n)/5. - Colin Barker, Sep 28 2016
From Amiram Eldar, Oct 06 2020: (Start)
Sum_{n>=3} 1/a(n) = (1/2) * A290565 - 1/4.
Sum_{n>=3} (-1)^(n+1)/a(n) = (3/2) * (1/phi - 1/2), where phi is the golden ratio (A001622). (End)

A122944 A nonsense sequence.

Original entry on oeis.org

1, 1, -1, -1, -1, 1, 0, 2, 1, -1, 1, -1, -4, -1, 1, 0, -2, 2, 6, 1, -1, 0, 0, 4, -2, -7, -1, 1, 0, 2, -1, -9, 3, 9, 1, -1, 1, 1, -13, 8, 20, -8, -13, -1, 1, 0, -2, -2, 24, -15, -31, 13, 17, 1, -1, 0, 0, 4, 4, -40, 20, 44, -14, -19, -1, 1, 0, 0, 0, -8, -4, 56, -24, -54, 14, 20, 1, -1, 0, 0, 0, 0, 16, 8, -88, 30, 71, -15, -22, -1, 1, 0, 0, 0, 16, 8

Offset: 1

Roger L. Bagula and Gary W. Adamson, Oct 24 2006

			Triangle:
  {1},
  {1, -1},
  {-1, -1, 1},
  {0, 2, 1, -1},
  {1, -1, -4, -1, 1},
  {0, -2, 2,6, 1, -1},
  {0, 0, 4, -2, -7, -1, 1},
  {0, 2, -1, -9, 3, 9,1, -1},
  {1, 1, -13, 8, 20, -8, -13, -1, 1},
  {0, -2, -2, 24, -15, -31, 13,17, 1, -1},
  {0, 0, 4, 4, -40, 20, 44, -14, -19, -1, 1},
  {0, 0, 0, -8, -4, 56, -24, -54, 14, 20, 1, -1}
Polynomials:
  1,
  1 - x,
  -1 - x + x^2,
  2 x + x^2 - x^3,
  1 - x - 4 x^2 - x^3 + x^4,
  -2 x + 2 x^2 + 6 x^3 +x^4 - x^5,
  4 x^2 - 2 x^3 - 7 x^4 - x^5 +x^6,
  2 x - x^2 - 9 x^3 + 3 x^4 + 9 x^5 + x^6 - x^7

Roger Bagula and Gary Adamson, Pascal's Triangle in Gray Code: its Hadamard and IFS

Cf. A121801.

c[i_, k_] := Floor[Mod[i/2^k, 2]]; b[i_, k_] = If[c[i, k] == 0 && c[i, k + 1] == 0, 0, If[c[i, k] == 1 && c[i, k + 1] == 1, 0, 1]];
An[d_] := Table[If[Sum[b[n, k]*b[m, k], {k, 0, d - 1}] == 0, 1, 0], {n, 0, d - 1}, {m, 0, d - 1}]: a=Join[{{1}}, Table[CoefficientList[CharacteristicPolynomial[An[d], x], x], {d, 1, 20}]];
Flatten[a]
RowSum=Table[Apply[Plus, Abs[a[[n]]]], {n, 1, Length[a]}]

A248161 Expansion of (2-x+x^2)/((1+x)(1-3x+x^2)).

Original entry on oeis.org

2, 3, 11, 26, 71, 183, 482, 1259, 3299, 8634, 22607, 59183, 154946, 405651, 1062011, 2780378, 7279127, 19056999, 49891874, 130618619, 341963987, 895273338, 2343856031, 6136294751, 16065028226, 42058789923, 110111341547

Offset: 0

Wolfdieter Lang, Nov 01 2014

The negative of this sequence provides the first component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, where F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-a(n), A079472(n+1), A059929(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given.

G. C. Greubel, Table of n, a(n) for n = 0..1000
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.

Cf. A000045, A059929, A079472, A121801, A248156.

[-(Fibonacci(n)^2 +Fibonacci(n+1)^2 + Fibonacci(n+2)^2 - Fibonacci(n+3)^2): n in [0..30]]; // Vincenzo Librandi, Nov 01 2014

CoefficientList[Series[(2 - x + x^2)/((1 + x) (1 - 3 x + x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Nov 01 2014 *)
With[{F=Fibonacci}, Table[F[2*n+2] +F[n]*F[n+1] +(-1)^n, {n,0,40}]] (* G. C. Greubel, May 30 2025 *)

def A248161(n): return fibonacci(2*n+2) +fibonacci(n)*fibonacci(n+1) +(-1)^n
print([A248161(n) for n in range(41)]) # G. C. Greubel, May 30 2025

a(n) = F(n+3)^2 - (F(n)^2 + F(n+1)^2 + F(n+2)^2), F(n) = A000045(n).
a(n) = (6*F(2*n+2) + F(2*n) + 4*(-1)^n)/5, with the Fibonacci numbers F = A000045.
O.g.f.: (2-x+x^2)/((1+x)*(1-3*x+x^2)) = (4/(1+x) + (x+6)/(1-3*x+x^2))/5.
From G. C. Greubel, May 30 2025: (Start)
a(n) = Fibonacci(2*n+2) + Fibonacci(n)*Fibonacci(n+1) + (-1)^n.
E.g.f.: (1/5)*(exp(3*x/2)*(6*cosh(sqrt(5)*x/2) + 4*sqrt(5)*sinh(sqrt(5)*x/2)) + 4*exp(-x)). (End)

Typo in formula fixed by Vincenzo Librandi, Nov 01 2014

A131218 Gray code/ Silvester-Hadamard binary triangular sequence from 16 X 16 self-similar matrix.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1

Offset: 1

Roger L. Bagula, Sep 27 2007

See A140820 for another version.

			{1},
{1, 1},
{1, 0, 1},
{1, 0, 0, 1},
{1, 1, 0, 1, 1},
{1, 1, 0, 0, 1, 1},
{1, 0, 0, 0, 0, 0, 1},
{1, 0, 0, 0, 0, 0, 0, 1},
{1, 1, 0, 0, 0, 0, 0, 1, 1},
{1, 1, 1, 1, 0, 0, 1, 1, 1, 1},
{1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1},
{1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1},
{1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1},
{1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}

Cf. A121801, A122944, A123949, A140820.

c[i_,k_]:=Floor[Mod[i/2^k,2]];
b[i_,k_]=If[c[i,k]==0&&c[i,k+1]\[Equal]0,0,If[c[ i,k]==1&&c[i,k+1]\[Equal]1,0,1]];
a0=Table[If[Sum[b[i,k]*b[j,k],{k,0,n}]\[Equal]0,1,0],{j,0,n},{i,0,n}];
ListDensityPlot[a0,Mesh\[Rule]False];
c=Delete[Table[Reverse[Table[a0[[n,l-n]],{n,1,l-1}]],{l,1,Dimensions[a0][[1]]+1}],1];
Flatten[c]

a(n,m) = Antidiagonal[HadamardMatrix[n,m]]

This looks interesting, but I do not understand the definition. - N. J. A. Sloane, Oct 16 2008

A123219 Expansion of -x(x^4 + 52x^3 - 122x^2 - 28x + 1) / ((x-1)(x^2 - 34x + 1)(x^2 + 6x + 1)).

Original entry on oeis.org

1, 1, 81, 2401, 83521, 2825761, 96059601, 3262808641, 110841719041, 3765342321601, 127910874833361, 4345203949621921, 147609026049038401, 5014361666349715681, 170340687719412376401, 5786569020271612560001

Offset: 1

Roger L. Bagula and Gary W. Adamson, Oct 05 2006

G. C. Greubel, Table of n, a(n) for n = 1..653
Index entries for linear recurrences with constant coefficients, signature (29,174,-174,-29,1).

Cf. A001333, A090390, A120717, A121801.

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(-x*(x^4+52*x^3-122*x^2-28*x+1)/((x-1)*(x^2-34*x+1)*(x^2+6*x+1)))); // G. C. Greubel, Oct 12 2018

seq(coeff(series(-x*(x^4+52*x^3-122*x^2-28*x+1)/((x-1)*(x^2-34*x+1)*(x^2+6*x+1)),x,n+1), x, n), n = 1 .. 20); # Muniru A Asiru, Oct 13 2018

LinearRecurrence[{29,174,-174,-29,1},{1,1,81,2401,83521},20] (* Harvey P. Dale, Jun 01 2018 *)

x='x+O('x^30); Vec(-x*(x^4+52*x^3-122*x^2-28*x+1)/((x-1)*(x^2-34*x+1)*(x^2+6*x+1))) \\ G. C. Greubel, Oct 12 2018

G.f.: -x*(x^4 + 52*x^3 - 122*x^2 - 28*x + 1) / ((x-1)*(x^2 - 34*x + 1)*(x^2 + 6*x + 1)). - Colin Barker, Jan 04 2013

New name from Colin Barker, Jan 04 2013

Edited by Joerg Arndt, Oct 13 2018

A123949 This sequence needs a definition.

Original entry on oeis.org

1, 1, -1, 1, -2, 1, 0, -2, 3, -1, 1, 0, -2, 0, 1, 0, -2, -1, 3, 1, -1, 0, 0, -3, 6, -2, -2, 1, 0, 2, -9, 15, -11, 3, 1, -1, 1, -4, 2, 6, -1, -6, -1, 2, 1, 0, -2, 7, -1, -11, -3, 8, 4, -1, -1, 0, 0, -3, -6, 4, 18, -9, -2, -3, 0, 1, 0, 0, 0, -4, 3, 19, -29, 11, -2, 2, 1, -1, 0, 0, 0, 0, 4, 0, -25, 16, 26, -20, -4, 2, 1, 0, 0, 0, -4, 11, 7, -63, 63, 8, -34

Offset: 1

Gary W. Adamson and Roger L. Bagula, Oct 26 2006

n-th level Hadamard matrices for Pascal type binary are inverted and multiplied by the corresponding n th level Gray code Hadamard (some of which matrices have determinant zero): the resulting matrix is processed to give a triangular sequence.
Matrices:
1 X 1 {{1}}
2 X 2 {{1, 0}, {0, 1}}
3 X 3 {{1, -1, -1}, {0, 1, 1}, {0, 1, 1}}
4 X 4 {{1, 1, 0, 0}, {0, -1, 0, 0}, {0, -1, 0, 1}, {0, 2, 1, 0}}
5 X 5 {{1, 1, -1, -1, -1}, {0, -1, 0, 0, 0}, {0, -1, 0, 1, 1}, {0, 2, 1, 0, 0}, {0, 0, 1, 1, 1}}
6 X 6 {{1, 1, 0, -1, -1, 0}, {0, -1, -1, 0, 0, -1}, {0, -1, 0, 1, 1, 0}, {0, 2, 1, 0, 0, 1}, {0, 0, 0, 1, 1, 0}. {0, 0, 1, 0, 0, 1}}.
They don't get interesting until 4 X 4!

			Triangle begins
  1;
  1, -1;
  1, -2, 1;
  0, -2, 3, -1;
  1, 0, -2, 0, 1;
  0, -2, -1, 3,1, -1;
  0, 0, -3, 6, -2, -2, 1;
  0, 2, -9, 15, -11,3, 1, -1;
  1, -4, 2, 6, -1, -6, -1, 2, 1;
  0, -2,7, -1, -11, -3, 8, 4, -1, -1;
  0, 0, -3, -6, 4, 18, -9, -2, -3, 0, 1;
Polynomials:
  1;
  1 -  x;
  1 - 2x +  x^2;
  0 - 2x + 3x^2 -   x^3;
  1 + 0x - 2x^2         +   x^4;
  0 - 2x -  x^2 +  3x^3 +   x^4 -  x^5;
  0 + 0x + 3x^2 +  6x^3 -  2x^4 - 2x^5 + x^6;
  0 + 2x - 9x^2 + 15x^3 - 11x^4 + 3x^5 + x^6 -  x^7;
  1 - 4x + 2x^2 +  6x^3 -   x^4 - 6x^5 - x^6 + 2x^7 + x^8;

Cf. A122944, A121801, A122947.

c[i_, k_] := Floor[Mod[i/2^k, 2]];
b[i_, k_] := If[c[i, k] == 0 && c[ i, k + 1] == 0, 0, If[c[i, k] == 1 && c[i, k + 1] == 1, 0, 1]];
An[d_] := Table[If[Sum[b[n, k]*b[m, k], {k, 0, d - 1}] == 0, 1, 0], {n, 0, d - 1}, {m, 0, d - 1}];
Bn[d_] := Table[If[Sum[c[n, k]*c[ m, k], {k, 0, d - 1}] == 0, 1, 0], {n, 0, d - 1}, {m, 0, d - 1}];
Xn[d_] := MatrixPower[Bn[d], -1].An[d];
a = Join[{{1}}, Table[CoefficientList[CharacteristicPolynomial[Xn[d], x], x], {d, 1, 20}]];
Flatten[%]

cp's OEIS Frontend

A059929 a(n) = Fibonacci(n)*Fibonacci(n+2).

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A079472 Number of perfect matchings on an n X n L-shaped graph.

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A226205 a(n) = F(n)^2 - F(n-1)^2 or F(n+1) * F(n-2) where F(n) = A000045(n), the Fibonacci numbers.

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A122944 A nonsense sequence.

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A248161 Expansion of (2-x+x^2)/((1+x)*(1-3*x+x^2)).

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A131218 Gray code/ Silvester-Hadamard binary triangular sequence from 16 X 16 self-similar matrix.

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A248161 Expansion of (2-x+x^2)/((1+x)(1-3x+x^2)).

A123219 Expansion of -x(x^4 + 52x^3 - 122x^2 - 28x + 1) / ((x-1)(x^2 - 34x + 1)(x^2 + 6x + 1)).