A103433 -id:A103433 - cp's OEIS frontend

A081068 a(n) = (Lucas(4n+2) + 2)/5, or Fibonacci(2n+1)^2, or A081067(n)/5.

1, 4, 25, 169, 1156, 7921, 54289, 372100, 2550409, 17480761, 119814916, 821223649, 5628750625, 38580030724, 264431464441, 1812440220361, 12422650078084, 85146110326225, 583600122205489, 4000054745112196, 27416783093579881

Offset: 0

R. K. Guy, Mar 04 2003

Indices of 12-gonal numbers which are also squares (A342709). - Bernard Schott, Mar 19 2021
Values of y in solutions of x^2 = 5*y^2 - 4*y in positive integers. See A360467 for how this relates to a problem regarding the subdivision of a square into four triangles of integer area. - Alexander M. Domashenko, Feb 26 2023
And the corresponding x values of x^2 = 5*y^2 - 4*y are in A033890. - Bernard Schott, Feb 26 2023

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 19.
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (See Corollary 1 (vii)).
Pridon Davlianidze, Problem B-1264, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 1 (2020), p. 82; It's All About Catalan, Solution to Problem B-1264, ibid., Vol. 59, No. 1 (2021), pp. 87-88.
Derek Jennings, On Sums of Reciprocals of Fibonacci and Lucas Numbers, The Fibonacci Quarterly, Vol. 32, No. 1 (1994), pp. 18-21.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081067.
Cf. A001622, A051324, A239798, A342709.
Cf. A033890, A360467.

I:=[1, 4, 25]; [n le 3 select I[n] else 8*Self(n-1)-8*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012

[(Lucas(4*n+2) + 2)/5: n in [0..30]]; // G. C. Greubel, Dec 17 2017

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d,`,(luc(4*n+2)+2)/5) od: # James Sellers, Mar 05 2003

CoefficientList[Series[-(1-4*x+x^2)/((x-1)*(x^2-7*x+1)),{x,0,40}],x] (* or *) LinearRecurrence[{8,-8,1},{1,4,25},50] (* Vincenzo Librandi, Jun 26 2012 *)
Table[(LucasL[4*n+2] + 2)/5, {n,0,30}] (* G. C. Greubel, Dec 17 2017 *)

main(size)={ return(concat([1],vector(size,n,fibonacci(2*n+1)^2))) } /* Anders Hellström, Jul 11 2015 */

for(n=0,30, print1(fibonacci(2*n+1)^2, ", ")) \\ G. C. Greubel, Dec 17 2017

a(n) = A001519(n+1)^2 = A122367(n)^2 = A058038(n) + 1.
a(n) = A103433(n+1) - A103433(n).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = Fibonacci(2*n)*Fibonacci(2*n+2) +1. - Gary Detlefs, Apr 01 2012
G.f.: (1-4*x+x^2)/((1-x)*(x^2-7*x+1)). - Colin Barker, Jun 26 2012
Sum_{n>=0} 1/(a(n) + 1) = 1/3*sqrt(5). - Peter Bala, Nov 30 2013
Sum_{n>=0} 1/a(n) = sqrt(5) * Sum_{n>=1} (-1)^(n+1)*n/Fibonacci(2*n) (Jennings, 1994). - Amiram Eldar, Oct 30 2020
Product_{n>=1} (1 + 1/a(n)) = phi^2/2 (A239798), where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021

More terms from James Sellers, Mar 05 2003

A077916 Expansion of (1-x)^(-1)/(1 + 2x - 2x^2 - x^3).

Original entry on oeis.org

1, -1, 5, -10, 30, -74, 199, -515, 1355, -3540, 9276, -24276, 63565, -166405, 435665, -1140574, 2986074, -7817630, 20466835, -53582855, 140281751, -367262376, 961505400, -2517253800, 6590256025, -17253514249, 45170286749, -118257345970, 309601751190, -810547907570

Offset: 0

N. J. A. Sloane, Nov 17 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,4,-1,-1).

Cf. A002571.
Bisections are A103433 and A103434.
Cf. A064831, A000045. A203803, A207969, A207970.

a[0] = 1; a[1] = -1; a[2] = 5; a[3] = -10; a[n_] := a[n] = -a[n-1] + 4 a[n-2] - a[n-3] - a[n-4]; Table[a[n], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)
CoefficientList[Series[(1 - x)^(-1)/(1 + 2*x - 2*x^2 - x^3), {x, 0, 50}], x] (* G. C. Greubel, Dec 25 2017 *)
Table[If[OddQ[n], (Fibonacci[2n+2]+n+1)/5, -(Fibonacci[2n+2]-n-1)/5], {n,1,20}] (* Rigoberto Florez, May 09 2019 *)

Vec((1-x)^(-1)/(1+2*x-2*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

Vec(1/((1-x)^2*(1+3*x+x^2)) + O(x^100)) \\ Altug Alkan, Oct 28 2015

a(n-1) = Sum_{i=1..n} (-1)^(i+1)*Fibonacci(i)*Fibonacci(i+1), n >= 1. - Alexander Adamchuk, Jun 16 2006
From R. J. Mathar, Mar 14 2011: (Start)
G.f.: 1/((1-x)^2*(1+3*x+x^2)).
a(n) = ((-1)^n*A001906(n+2)+n+2)/5. (End)
O.g.f.: exp( Sum_{n >= 1} Lucas(n)^2*(-x)^n/n ) = 1 - x + 5*x^2 - 10*x^3 + .... Cf. A203803. See also A207969 and A207970. - Peter Bala, Apr 03 2014
From Vladimir Reshetnikov, Oct 28 2015: (Start)
Recurrence (5-term): a(0) = 1, a(1) = -1, a(2) = 5, a(3) = -10, a(n) = -a(n-1) + 4*a(n-2) - a(n-3) - a(n-4).
Recurrence (4-term): a(0) = 1, a(1) = -1, a(2) = 5, n*a(n) = (1-2*n)*a(n-1) + (3*n+3)*a(n-2) + (n+1)*a(n-3).
(End)
a(n) = (F(2n+2)+n+1)/5 if n is odd and a(n)= -(F(2n+2)-n-1)/5 if n is even, where F(n) = Fibonacci numbers (A000045). - Rigoberto Florez, May 09 2019

A156088 Alternating sum of the squares of the first n even-indexed Fibonacci numbers.

Original entry on oeis.org

0, -1, 8, -56, 385, -2640, 18096, -124033, 850136, -5826920, 39938305, -273741216, 1876250208, -12860010241, 88143821480, -604146740120, 4140883359361, -28382036775408, 194533374068496, -1333351581704065, 9138927697859960

Offset: 0

Stuart Clary, Feb 04 2009

Apart from signs, same as A092521.

Natural bilateral extension (brackets mark index 0): ..., 2640, -385, 56, -8, 1, 0, [0], -1, 8, -56, 385, -2640, 18096, ... This is (-a(n))-reversed followed by a(n). That is, a(-n) = -a(n-1).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-8,-8,-1).

Cf. A000032, A000045, A081079, A092521, A103434, A103433, A156089.

A156088:= func< n | (-1)^n*(Lucas(4*n+2)-3)/15 >; // G. C. Greubel, Jun 12 2025

a[n_]:= If[n >= 0, Sum[(-1)^k Fibonacci[2k]^2, {k,n}], Sum[ -(-1)^k Fibonacci[-2k]^2, {k,-n-1}]];
LinearRecurrence[{-8,-8,-1}, {0,-1,8}, 41] (* G. C. Greubel, Jun 12 2025 *)

def A156088(n): return (-1)^n*(lucas_number2(4*n+2,1,-1) -3)//15 # G. C. Greubel, Jun 12 2025

Let F(n) be the n-th Fibonacci number, A000045(n), and L(n) be the n-th Lucas number, A000032(n), then: (Start)
a(n) = Sum_{k=1..n} (-1)^k F(2*k)^2.
Closed form: a(n) = (-1)^n * (L(4*n+2) - 3)/15.
Factored closed form: a(n) = (1/3) * (-1)^n * F(n)*L(n)*F(n+1)*L(n+1) = (1/3)*(-1)^n * F(2*n)*F(2*n+2).
Recurrence: a(n) + 8*a(n-1) + 8*a(n-2) + a(n-3) = 0.
G.f.: -x/(1 + 8*x + 8*x^2 + x^3) = -x/((1 + x)(1 + 7*x + x^2)). (End)
From G. C. Greubel, Jun 12 2025: (Start)
a(n) = (-1)^n*A081079(n)/15.
E.g.f.: (1/15)*( exp(-7*x/2)*( 3*cosh(p*x) - sqrt(5)*sinh(p*x) ) - 3*exp(-x) ), where p = 3*sqrt(5)/2. (End)

A156089 Alternating sum of the squares of the first n odd-indexed Fibonacci numbers.

Original entry on oeis.org

0, -1, 3, -22, 147, -1009, 6912, -47377, 324723, -2225686, 15255075, -104559841, 716663808, -4912086817, 33667943907, -230763520534, 1581676699827, -10840973378257, 74305136947968, -509294985257521, 3490759759854675

Offset: 0

Stuart Clary, Feb 04 2009

Natural bilateral extension (brackets mark index 0): ..., 6912, -1009, 147, -22, 3, -1, [0], -1, 3, -22, 147, -1009, 6912, ... This is a(n)-reversed followed by a(n), without repeating the 0. That is, a(-n) = a(n).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-7,0,7,1).

Cf. A000032, A000045, A103434, A103433, A156088.

A156089:= func< n | ((-1)^n*(Lucas(4*n)+3) - 5)/15 >; // G. C. Greubel, Jun 12 2025

a[n_]:= If[n >= 0, Sum[(-1)^k Fibonacci[2k-1]^2, {k,n} ], -Sum[-(-1)^k Fibonacci[-2k+1]^2, {k,-n}]];
Join[{0},Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[1,43,2]]^2, {-1,1}],2]]] (* Harvey P. Dale, Aug 18 2011 *)
LinearRecurrence[{-7,0,7,1}, {0,-1,3,-22}, 41] (* G. C. Greubel, Jun 12 2025 *)

def A156089(n): return ((-1)^n*(lucas_number2(4*n,1,-1)+3)-5)//15 # G. C. Greubel, Jun 12 2025

Let F(n) be the n-th Fibonacci number, A000045(n), and L(n) be the n-th Lucas number, A000032(n).
a(n) = Sum_{k=1..n} (-1)^k F(2*k-1)^2.
Closed form: a(n) = (-1)^n * (L(4*n) + 3)/15 - 1/3.
Factored closed form: a(n) = (1/3)*F(2*n)^2 if n is even.
Not-so-factored closed form: a(n) = -(F(2*n)^2 + 2)/3 if n is odd.
Recurrence: a(n) + 7*a(n-1) - 7*a(n-3) - a(n-4) = 0.
G.f.: -x*(1 + 4*x + x^2)/(1 + 7*x - 7*x^3 - x^4) = -x*(1 + 4*x + x^2)/((1 - x)*(1 + x)*(1 + 7*x + x^2)).
E.g.f.: (2/15)*( exp(-7*x/2)*cosh(3*sqrt(5)*x/2) - cosh(x) - 4*sinh(x) ). - G. C. Greubel, Jun 12 2025

A203170 Sum of the fourth powers of the first n odd-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 17, 642, 29203, 1365539, 64107780, 3011403301, 141469813301, 6646055880582, 312223061019703, 14667837157106759, 689076118833981960, 32371909717271872585, 1520790680382055836761, 71444790066793903279242

Offset: 0

Stuart Clary, Dec 30 2011

Natural bilateral extension (brackets mark index 0): ..., -1365539, -29203, -642, -17, -1, [0], 1, 17, 642, 29203, 1365539, ... That is, a(-n) = -a(n).

Index entries for linear recurrences with constant coefficients, signature (56,-440,770,-440,56,-1).

Cf. A203169, A203171, A203172.

Cf. A001906, A103433, A163200.

a[n_Integer] := (1/75)(Fibonacci[8n] + 12*Fibonacci[4n] + 18 n); Table[a[n], {n, 0, 20}]

Let F(n) be the Fibonacci number A000045(n).
a(n) = Sum_{k=1..n} F(2k-1)^4.
Closed form: a(n) = (1/75)(F(8n) + 12 F(4n) + 18 n).
Recurrence: a(n) - 56 a(n-1) + 440 a(n-2) - 770 a(n-3) + 440 a(n-4) - 56 a(n-5) + a(n-6) = 0.
G.f.: A(x) = (x - 39 x^2 + 130 x^3 - 39 x^4 + x^5)/(1 - 56 x + 440 x^2 - 770 x^3 + 440 x^4 - 56 x^5 + x^6) = x(1 - 39 x + 130 x^2 - 39 x^3 + x^4)/((1 - x)^2 (1 - 7 x + x^2)(1 - 47 x + x^2)).

A280470 Triangle A106534 with reversed rows.

Original entry on oeis.org

1, 1, 2, 2, 3, 5, 5, 7, 10, 15, 14, 19, 26, 36, 51, 42, 56, 75, 101, 137, 188, 132, 174, 230, 305, 406, 543, 731, 429, 561, 735, 965, 1270, 1676, 2219, 2950, 1430, 1859, 2420, 3155, 4120, 5390, 7066, 9285, 12235, 4862, 6292, 8151, 10571, 13726, 17846, 23236, 30302, 39587, 51822, 16796, 21658, 27950, 36101, 46672

Offset: 0

Tony Foster III, Jan 03 2017

			Fibonacci Determinant Triangle:
    1;
    1,    2;
    2,    3,    5;
    5,    7,   10,   15;
   14,   19,   26,   36,   51;
   42,   56,   75,  101,  137,  188;
  132,  174,  230,  305,  406,  543,  731;
  429,  561,  735,  965, 1270, 1676, 2219, 2950;
  ...

P. Barry, A. Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010) # 10.8.2, page 5
A. Cvetkovi, Predrag Rajkovic, and Milos IvkoviCatalan Numbers, the Hankel Transform, and Fibonacci Numbers, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.3.

Cf. A000108, A001519, A007317, A011971, A103433, A106534, A197649.

&cat [[&+[Binomial(k,j)*Catalan(n-j): j in [0..k]]: k in [0..n]]: n in [0..10]]; // Bruno Berselli, Mar 07 2017

Table[Sum[Binomial[k, j] CatalanNumber[n - j], {j, 0, k}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Mar 08 2017 *)

C(n)=binomial(2*n,n)/(n+1);
T(n,k)=sum(j=0,k,binomial(k,j)*C(n-j));
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()); \\ Joerg Arndt, Jan 15 2017

T(n,k) = Sum_{j=0..k} binomial(k,j) * A000108(n-j). - Joerg Arndt, Jan 15 2017

cp's OEIS Frontend

A081068 a(n) = (Lucas(4*n+2) + 2)/5, or Fibonacci(2*n+1)^2, or A081067(n)/5.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Magma

Maple

Mathematica

PARI

PARI

Formula

Extensions

A077916 Expansion of (1-x)^(-1)/(1 + 2*x - 2*x^2 - x^3).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A156088 Alternating sum of the squares of the first n even-indexed Fibonacci numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

A156089 Alternating sum of the squares of the first n odd-indexed Fibonacci numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

A203170 Sum of the fourth powers of the first n odd-indexed Fibonacci numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A280470 Triangle A106534 with reversed rows.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A081068 a(n) = (Lucas(4n+2) + 2)/5, or Fibonacci(2n+1)^2, or A081067(n)/5.

A077916 Expansion of (1-x)^(-1)/(1 + 2x - 2x^2 - x^3).