A103434 -id:A103434 - cp's OEIS frontend

A049684 a(n) = Fibonacci(2n)^2.

0, 1, 9, 64, 441, 3025, 20736, 142129, 974169, 6677056, 45765225, 313679521, 2149991424, 14736260449, 101003831721, 692290561600, 4745030099481, 32522920134769, 222915410843904, 1527884955772561, 10472279279564025, 71778070001175616, 491974210728665289

Offset: 0

Clark Kimberling

This is the r=9 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.

Apparently, this sequence consists of those nonnegative integers k for which x*(x^2-1)*y*(y^2-1) = k*(k^2-1) has a solution in nonnegative integers x, y. If k = a(n), x = A000045(2*n-1) and y = A000045(2*n+1) are a solution. See A374375 for numbers k*(k^2-1) that can be written as a product of two or more factors of the form x*(x^2-1). - Pontus von Brömssen, Jul 14 2024

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 27.
H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
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.
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 1, k=2.
R. Stephan, Boring proof of a nonlinearity
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

First differences give A033890.
First differences of A103434.
Bisection of A007598 and A064841.
a(n) = A064170(n+2) - 1 = (1/5) A081070.
Cf. A000032, A000045, A001622, A001906, A056854, A065563, A092184, A374375.

Join[{a=0, b=1}, Table[c=7*b-1*a+2; a=b; b=c, {n, 60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
Fibonacci[Range[0, 40, 2]]^2 (* Harvey P. Dale, Mar 22 2012 *)
Table[Fibonacci[n - 1] Fibonacci[n + 1] - 1, {n, 0, 40, 2}] (* Bruno Berselli, Feb 12 2015 *)
LinearRecurrence[{8, -8, 1},{0, 1, 9},21] (* Ray Chandler, Sep 23 2015 *)

numlib::fibonacci(2*n)^2 $ n = 0..35; // Zerinvary Lajos, May 13 2008

PARI
```
a(n)=fibonacci(2*n)^2
```

[fibonacci(2*n)^2 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009

G.f.: (x+x^2) / ((1-x)*(1-7*x+x^2)).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) with n>2, a(0)=0, a(1)=1, a(2)=9.
a(n) = 7*a(n-1) - a(n-2) + 2 = A001906(n)^2.
a(n) = (A000032(4*n)-2)/5. [This is in Koshy's book (reference under A065563) on p. 88, attributed to Lucas 1876.] - Wolfdieter Lang, Aug 27 2012
a(n) = 1/5*(-2 + ( (7+sqrt(45))/2 )^n + ( (7-sqrt(45))/2 )^n). - Ralf Stephan, Apr 14 2004
a(n) = 2*(T(n, 7/2)-1)/5 with twice the Chebyshev polynomials of the first kind evaluated at x=7/2: 2*T(n, 7/2)= A056854(n). - Wolfdieter Lang, Oct 18 2004
a(n) = F(2*n-1)*F(2*n+1)-1, see A064170 - Bruno Berselli, Feb 12 2015
a(n) = Sum_{i=1..n} F(4*i-2) for n>0. - Bruno Berselli, Aug 25 2015
From Peter Bala, Nov 20 2019: (Start)
Sum_{n >= 1} 1/(a(n) + 1) = (sqrt(5) - 1)/2.
Sum_{n >= 1} 1/(a(n) + 4) = (3*sqrt(5) - 2)/16. More generally, it appears that
Sum_{n >= 1} 1/(a(n) + F(2*k+1)^2) = ((2*k+1)*F(2*k+1)*sqrt(5) - Lucas(2*k+1))/ (2*F(2*k+1)*F(4*k+2)) for k = 0,1,2,....
Sum_{n >= 2} 1/(a(n) - 1) = (8 - 3*sqrt(5))/9. (End)
E.g.f.: (1/5)*(-2*exp(x) + exp((16*x)/(1 + sqrt(5))^4) + exp((1/2)*(7 + 3*sqrt(5))*x)). - Stefano Spezia, Nov 23 2019
Product_{n>=2} (1 - 1/a(n)) = phi^2/3, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021
a(n) = A092521(n-1)+A092521(n). - R. J. Mathar, Nov 22 2024

Better description and more terms from Michael Somos

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

A203169 Sum of the fourth powers of the first n even-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 82, 4178, 198659, 9349284, 439330980, 20639983621, 969645224182, 45552722051318, 2140008541351943, 100534850436141384, 4722997973709689160, 221880369994471370761, 10423654392318557192602, 489689876072761951752602

Offset: 0

Stuart Clary, Dec 30 2011

Natural bilateral extension (brackets mark index 0): ..., -9349284, -198659, -4178, -82, -1, 0, [0], 1, 82, 4178, 198659, 9349284, ... That is, a(-n) = -a(n-1).

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

Cf. A203170, A203171, A203172.

Cf. A027941, A103434, A163198.

a[n_Integer] := (1/75)(Fibonacci[8n+4] - 12*Fibonacci[4n+2] + 9*(2*n+1)); Table[a[n], {n, 0, 20}]

Let F(n) be the Fibonacci number A000045(n).
a(n) = sum_{k=1..n} F(2k)^4.
Closed form: a(n) = (1/75)(F(8n+4) - 12 F(4n+2) + 9(2 n + 1)).
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 + 26 x^2 + 26 x^3 + x^4)/(1 - 56 x + 440 x^2 - 770 x^3 + 440 x^4 - 56 x^5 + x^6) = x(1 + x)(1 + 25 x + x^2)/((1 - x)^2 (1 - 7 x + x^2)(1 - 47 x + x^2)).

cp's OEIS Frontend

A049684 a(n) = Fibonacci(2n)^2.

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

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

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

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A203169 Sum of the fourth powers of the first n even-indexed Fibonacci numbers.

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