A248161 -id:A248161 - 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

A121801 Expansion of 2x^2(3-x)/((1+x)(1-3x+x^2)).

Original entry on oeis.org

0, 6, 10, 32, 78, 210, 544, 1430, 3738, 9792, 25630, 67106, 175680, 459942, 1204138, 3152480, 8253294, 21607410, 56568928, 148099382, 387729210, 1015088256, 2657535550, 6957518402, 18215019648, 47687540550, 124847601994

Offset: 1

Roger L. Bagula and Gary W. Adamson, Aug 27 2006

a(n) = the area of an irregular quadrilateral with vertices at the points (L(n),L(n+2)), (F(n+2),F(n+3)), (F(n+3),F(n+2)) and (L(n+2),L(n)), with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, Jun 16 2014

a(n+1) appears also as the fourth 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), A059929(n), a(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - Wolfdieter Lang, Nov 01 2014

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

List([1..30], n-> 2*(Lucas(1,-1,2*n+1)[2] +4*(-1)^n)/5 ); # G. C. Greubel, Jul 22 2019

I:=[0,6,10]; [n le 3 select I[n] else 2*Self(n-1)+2*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Nov 02 2014

[2*(Lucas(2*n+1) +4*(-1)^n)/5: n in [1..30]]; // G. C. Greubel, Jul 22 2019

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]] n = 4 - 1; M = Table[If[Sum[b[i, k]*b[j, k], {k, 0, n}] == 0, 1, 0], {j, 0, n}, {i, 0, n}] v[1] = {0, 1, 2, 3} v[n_] := v[n] = M.v[n - 1] a = Table[Floor[v[n][[1]]], {n, 1, 50}] Det[M - x*IdentityMatrix[4]] Factor[%] aaa = Table[x /. NSolve[Det[M - x*IdentityMatrix[4]] == 0, x][[n]], {n, 1, 4}] Abs[aaa] a1 = Table[N[a[[n]]/a[[n - 1]]], {n, 7, 50}]
CoefficientList[Series[2*x*(3-x)/((1+x)*(1-3*x+x^2)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Jun 16 2014 *)
LinearRecurrence[{2,2,-1},{0,6,10},30] (* Harvey P. Dale, Jan 06 2015 *)
With[{F=Fibonacci}, Table[2*(F[n]*F[n+1] +(-1)^n), {n,30}]] (* G. C. Greubel, Jul 22 2019 *)

concat(0,Vec(2*(3-x)/((1+x)*(1-3*x+x^2))+O(x^30))) \\ Charles R Greathouse IV, Sep 25 2012

vector(30, n, f=fibonacci; 2*(f(n)*f(n+1)+(-1)^n) ) \\ G. C. Greubel, Jul 22 2019

[2*(lucas_number2(2*n+1,1,-1) +4*(-1)^n)/5 for n in (1..30)] # G. C. Greubel, Jul 22 2019

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = -2*A121646(n+1).
G.f.: 2*x^2*(3-x)/((1+x)*(x^2-3*x+1)) (see name).
From Wolfdieter Lang, Nov 01 2014: (Start)
G.f.: (-10 + 8/(1+x) + 2*(1+x)/(1-3*x+x^2))/5 (partial fraction decomposition).
a(n) = (8*(-1)^n + 2*(F(2*(n+1)) + F(2*n)))/5 for n >= 1. a(0) = 0.
(End)
a(n) = 2*(Fibonacci(n)*Fibonacci(n+1) + (-1)^n). - G. C. Greubel, Jul 22 2019

Edited by the Associate Editors of the OEIS, Aug 18 2009

A248156 Inverse Riordan triangle of A106513: Riordan ((1 - 2*x^2 )/(1 + x), x/(1+x)).

Original entry on oeis.org

1, -2, 1, 1, -3, 1, 0, 4, -4, 1, -1, -4, 8, -5, 1, 2, 3, -12, 13, -6, 1, -3, -1, 15, -25, 19, -7, 1, 4, -2, -16, 40, -44, 26, -8, 1, -5, 6, 14, -56, 84, -70, 34, -9, 1, 6, -11, -8, 70, -140, 154, -104, 43, -10, 1, -7, 17, -3, -78, 210, -294, 258, -147, 53, -11, 1, 8, -24, 20, 75, -288, 504, -552, 405, -200, 64, -12, 1

Offset: 0

Wolfdieter Lang, Oct 05 2014

Row sums have o.g.f. (1 - 2*x)/(1 + x): [1, -1, repeat(-1, 1)].

			The triangle T(n,k) begins:
  n\k  0   1   2   3    4    5    6    7    8    9
  0:   1
  1:  -2   1
  2:   1  -3   1
  3:   0   4  -4   1
  4:  -1  -4   8  -5    1
  5:   2   3 -12  13   -6    1
  6:  -3  -1  15 -25   19   -7    1
  7:   4  -2 -16  40  -44   26   -8    1
  8:  -5   6  14 -56   84  -70   34   -9    1
  9:   6 -11  -8  70 -140  154 -104   43  -10    1
  ...
For more rows see the link.
Recurrence from A-sequence: T(5,2) = T(4,1) - T(4,2) = -4 - 8 = -12.
Recurrence from the Z-sequence: T(5,0) = -(2*(-1) + 3*(-4) + 7*8 + 17*(-5) + 41*1) = 2.
Standard recurrence for T(n,0): T(3,0) = -2*T(2,0) - T(1,0) = -2*1 - (-2) = 0.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Wolfdieter Lang, First 13 rows of the triangle

Cf. A006232, A083318, A106513, A248157, A248158, A248159, A248160, A248161.
Columns: A248157 (k=0), A248158 (k=1), A248159 (k=2), A248160 (k=3).
Diagonals: A000012 (k=n), A022958(n+3) (k=n-1), -A034856(n-1) (k=n-2), A000297(n-4) (k=n-3), A014309(n-3) (k=n-4).
Sums: (-1)^n*A001611(n) (diagonal), (-1)^n*A083318(n) (alternating sign row).

function T(n,k) // T = A248156
  if k eq n then return 1;
  elif k eq 0 then return (-1)^n*(3-n);
  else return T(n-1,k-1) - T(n-1,k);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 27 2025

T[n_, k_] := SeriesCoefficient[x^k*(1 - 2*x^2)/(1 + x)^(k + 2), {x, 0, n}]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 09 2014 *)
T[n_, k_]:= T[n, k]= If[k==n,1, If[k==0,(-1)^n*(3-n), T[n-1,k-1]-T[n-1,k]]];
Table[T[n,k], {n,0,25}, {k,0,n}]//Flatten (* G. C. Greubel, May 27 2025 *)

def T(n,k): # T = A248156
    if (k==n): return 1
    elif (k==0): return (-1)^n*(3-n)
    else: return T(n-1,k-1) - T(n-1,k)
print(flatten([[T(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, May 27 2025

O.g.f. row polynomials R(n,x) = Sum_{k=0..n} T(n,k)*x^k = [(-z)^n] (1 - 2*z^2)/( (1 + z)*(1 + (1-x)*z)).
O.g.f. column m: x^m*(1 - 2*x^2)/(1 + x)^(m+2), m >= 0.
The A-sequence is [1, -1], implying the recurrence T(n,k) = T(n-1, k-1) - T(n-1, k), n >= k > = 1.
The Z-sequence is -[2, 3, 7, 17, 41, 99, 239, 577, 1393, ...] = A248161, implying the recurrence T(n, 0) = Sum_{k=0..n-1} T(n-1,k)*Z(k). See the W. Lang link under A006232 for Riordan A- and Z-sequences.
The standard recurrence for the sequence for column k=0 is T(0,0) = 1 and T(n,0) = -2*T(n-1,0) - T(n-2,0), n >= 3, with T(1,0) = -2 and T(2,0) = 1.
From G. C. Greubel, May 27 2025: (Start)
Sum_{k=0..n} T(n, k) = (-1)^(n+1) + 2*([n=0] - [n=1]).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = the repeated pattern of [1, -2, 0, 3, -4, 2]. (End)

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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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A121801 Expansion of 2*x^2*(3-x)/((1+x)*(1-3*x+x^2)).

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A248156 Inverse Riordan triangle of A106513: Riordan ((1 - 2*x^2 )/(1 + x), x/(1+x)).

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A121801 Expansion of 2x^2(3-x)/((1+x)(1-3x+x^2)).