A105394 -id:A105394 - cp's OEIS frontend

A001254 Squares of Lucas numbers.

4, 1, 9, 16, 49, 121, 324, 841, 2209, 5776, 15129, 39601, 103684, 271441, 710649, 1860496, 4870849, 12752041, 33385284, 87403801, 228826129, 599074576, 1568397609, 4106118241, 10749957124, 28143753121, 73681302249, 192900153616, 505019158609, 1322157322201, 3461452808004, 9062201101801, 23725150497409

Offset: 0

N. J. A. Sloane

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 36, 60.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 97.
Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Wiley, New York, 2001. [Note that Identity 34.7 on page 404 is wrong. - Alonso del Arte, Sep 07 2010]

G. C. Greubel, Table of n, a(n) for n = 0..2375 (terms 0..200 from T. D. Noe)
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Tanya Khovanova, Recursive Sequences
T. Mansour, A note on sum of k-th power of Horadam's sequence, arXiv:math/0302015 [math.CO], 2003.
P. Stanica, Generating functions, weighted and non-weighted sums for powers of second-order recurrence sequences, arXiv:math/0010149 [math.CO], 2000.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000204.
Cf. A007598, A079291.
With alternating signs, cf. A075150.
Bisection of A001638 and A006499. First differences of A005970.
Second row of array A103324.

[ Lucas(n)^2 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011

with(combinat):seq(5*fibonacci(n)^2+4*(-1)^n, n=0..26)

Table[LucasL[n]^2, {n, 0, 29}] (* Alonso del Arte, Apr 11 2011 *)
LinearRecurrence[{2, 2, -1}, {4, 1, 9}, 33] (* Jean-François Alcover, Jan 07 2019 *)

a(n)=5*fibonacci(n)^2 + 4*(-1)^n \\ Charles R Greathouse IV, Sep 24 2015

from sympy import lucas
def a(n):  return lucas(n)**2
print([a(n) for n in range(33)]) # Michael S. Branicky, Apr 01 2021

a(n) = A000032(n)^2.
G.f.: ( 4-7*x-x^2 ) / ( (1+x)*(x^2-3*x+1) ). - Len Smiley, Nov 30 2001
From Ralf Stephan, Feb 08 2003: (Start)
a(n) = r^n + (1/r)^n + 2*(-1)^n, with r=(3+sqrt(5))/2.
a(n+3) = 2*a(n+2) + 2*a(n+1) - a(n). (End)
a(n) = L(2*n) + 2*(-1)^n = L(n-1)*L(n+1) + 5(-1)^n.
a(n) = 5*Fibonacci(n)^2 + 4*(-1)^n.
a(n) + a(n+1) = A106729(n). - R. J. Mathar, Nov 17 2011
E.g.f.: 2*exp(-x)*(exp(5*x/2)*cosh(sqrt(5)*x/2)+1). - Wolfdieter Lang, Jan 14 2012
a(n) = 1/4*( a(n-2) - a(n-1) - a(n+1) + a(n+2) ). The same recurrence holds for A007598. - Peter Bala, Aug 18 2015
For n>1, a(n)=(10*F(2*n-1) + 2*L(n-2)*L(n+1))/4 where F(n)=A000045(n), L(n)=A000204(n). - J. M. Bergot, Nov 25 2015
a(n) = (L(n-2)*L(n+2) + L(n-1)*L(n+1))/2 with L(k)=A000032(k). - J. M. Bergot, May 25 2017
From Peter Bala, Nov 13 2019: (Start)
Sum_{n >= 1} 1/a(n) = (1/8)*( theta_3(beta)^4 - 1 ) = A105394, where beta = (3 - sqrt(5))/2 and theta_3(q) = 1 + 2*Sum_{n >= 1} q^(n^2) is a theta function. See Borwein and Borwein, Exercise 7(f), p. 97.
Sum_{n >= 1} 1/(a(n) - 5) = (3 - sqrt(5))/6; Sum_{n >= 1} (-1)^n/(a(n) - 5) = (15 - sqrt(5))/30; Sum_{n >= 1} 1/(a(2*n) - 5) = (5 - sqrt(5))/10.
Sum_{n >= 1} 1/(a(n) - 25/a(n)) = 2/9.
Conjecture: Sum_{n >= 1} 1/(a(n) - 5*(-1)^n*F(2*k+1)^2) = 1/(2*a(2*k+1)) for k = 0,1,2,.... (End)
a(n) = 3*a(n-1) - a(n-2) + 10*(-1)^n. - Greg Dresden, May 18 2020

A105393 Decimal expansion of sum of reciprocals of squares of Fibonacci numbers.

Original entry on oeis.org

2, 4, 2, 6, 3, 2, 0, 7, 5, 1, 1, 6, 7, 2, 4, 1, 1, 8, 7, 7, 4, 1, 5, 6, 9, 4, 1, 2, 9, 2, 6, 6, 2, 0, 3, 7, 4, 3, 2, 0, 2, 5, 9, 7, 7, 4, 5, 1, 3, 8, 3, 0, 9, 0, 5, 1, 1, 0, 1, 0, 2, 8, 3, 4, 5, 4, 6, 6, 1, 1, 9, 3, 7, 5, 1, 1, 1, 9, 7, 8, 6, 3, 6, 8, 7, 7, 5, 3, 8, 9, 8, 1, 5, 2, 1, 5, 3, 6, 3, 6, 3, 7, 9, 2, 1

Offset: 1

Jonathan Vos Post, Apr 04 2005

Known to be transcendental. - Benoit Cloitre, Jan 07 2006
Compare with Sum_{n >= 1} 1/(F(n)^2 + 1) = (5*sqrt(5) - 3)/6 and Sum_{n >= 3} 1/(F(n)^2 - 1) = (43 - 15*sqrt(5))/18. - Peter Bala, Nov 19 2019
Duverney et al. (1997) proved that this constant is transcendental. - Amiram Eldar, Oct 30 2020

			2.426320751167241187741569...

Richard André-Jeannin, Irrationalité de la somme des inverses de certaines suites récurrentes, C. R. Acad. Sci. Paris Ser. I Math., Vol. 308, No. 19 (1989), pp. 539-541.
Daniel Duverney, Keiji Nishioka, Kumiko Nishioka and Iekata Shiokawa, Transcendence of Rogers-Ramanujan continued fraction and reciprocal sums of Fibonacci numbers, Proceedings of the Japan Academy, Series A, Mathematical Sciences, Vol. 73, No. 7 (1997), pp. 140-142.
Michel Waldschmidt, Elliptic functions and transcendence, in: Krishnaswami Alladi (ed.), Surveys in number theory, Springer, New York, NY, 2008, pp. 143-188, alternative link. See Corollary 51.
Eric Weisstein's World of Mathematics, Fibonacci Number.
Eric Weisstein's World of Mathematics, Lucas Number.
Eric Weisstein's World of Mathematics, Reciprocal Fibonacci Constant.
Index entries for transcendental numbers

Cf. A000045, A007598 (squares of Fibonacci numbers).

Cf. A079586, A093540, A105394.

RealDigits[Total[1/Fibonacci[Range[500]]^2],10,120][[1]] (* Harvey P. Dale, May 31 2016 *)

sum(k=1,500,1./fibonacci(k)^2) \\ Benoit Cloitre, Jan 07 2006

Equals Sum_{k>=1} 1/F(k)^2 = 2.4263207511672411877... - Benoit Cloitre, Jan 07 2006

More terms from Benoit Cloitre, Jan 07 2006

A333209 a(n) is the denominator of Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)), with Lucas(i) as defined in A000032.

Original entry on oeis.org

2, 36, 7392, 1688148, 197412831, 21085413226416, 101768454084335346, 60343478516053297339236, 73240105330540144095414793632, 1956470757376233684880813258936380492, 32802418997525523144166495047229414174839, 202042966989952174292936124782341088713724476716231

Offset: 1

A.H.M. Smeets, Mar 11 2020

The numerators are given in A333208.
See A333088 and A333089 for similar fractions for infinite sums of Fibonacci numbers.
Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)) is a fraction for n > 0.
Sum_{i >= 0} 1/Lucas(i)^2 = 1/4 + A105394, i.e., the n = 0 case, is believed to be transcendental.

			These infinite sums begin: 1/2, 7/36, 551/7392, ...

A.H.M. Smeets, Table of n, a(n) for n = 1..54

Cf. A000032, A105394, A132318, A333088, A333089, A333208.

a[n_] := Denominator[Sum[1/(LucasL[2 i - 2]*LucasL[2 i - 1]), {i, 1, n}]/Fibonacci[2 n]]; Array[a, 12] (* Amiram Eldar, Mar 11 2020 *)

from math import gcd
f0, f1, g0, g1, snum, sden, n = 1, 1, 1, 2, 0, 1, 0
while n < 12:
    n = n+1
    snum, sden = g0*g1*snum+sden, sden*g0*g1
    d = gcd(snum,sden*f0)
    print(n,sden*f0//d)
f0, f1, g0, g1 = 2*f0+f1, f0+f1, 2*g0+g1, g0+g1 # A.H.M. Smeets, Nov 30 2020

from math import gcd
f0, f1, g0, g1, snum, sden, n = 1, 1, 1, 2, 0, 1, 0
while n < 12:
    n = n+1
    snum, sden = g0*g1*snum+sden, sden*g0*g1
    d = gcd(snum,sden*f0)
    print(n,sden*f0//d)
    f0, f1, g0, g1 = 2*f0+f1, f0+f1, 2*g0+g1, g0+g1 # A.H.M. Smeets, Nov 30 2020

a(n) = denominator of (1/Fibonacci(2n)) * Sum_{i = 1..n} 1/(Lucas(2i-2)*Lucas(2i-1)).
Lim_{n -> inf} (A333208(n)/a(n)) / (A333208(n-1)/a(n-1)) = 1 - 1/phi = 1/phi^2 = A132318.
The following generalization holds: (Start)
Let H_(a,b) (n) be defined by H_(a,b) (0) = a, H_(a,b) (1) = b and H_(a,b) (n) = H_(a,b) (n-1) + H_(a,b) (n-2) for n > 1, then
Sum_{i >= 0} 1/(H_(a,b) (i)*H_(a,b) (i+2n)) = (1/Fibonacci(2n)) * Sum_{i=1..n} 1/(H_(a,b) (2i-2)*H_(a,b) (2i-1)) for n > 0, and are thus all fractions. Specially, H_(0,1) are the Fibonacci numbers A000045, H_(2,1) as here, are the Lucas numbers A000032, and H_(3,1) are the Pibonacci numbers A104449. (End)

A333208 a(n) is the numerator of Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)), with Lucas(i) as defined in A000032.

Original entry on oeis.org

1, 7, 551, 48091, 2148268, 87644575267, 161577754532123, 36595152483523582367, 16965509829762630129638831, 173107561150078104051618631740949, 1108595900580419409151086339986148307, 2608169750203411467722731179728125652086612772

Offset: 1

A.H.M. Smeets, Mar 11 2020

nonn

The denominators are given in A333209.
See A333088 and A333089 for similar fractions for infinite sums of Fibonacci numbers.
Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)) is a fraction for n > 0.
Sum_{i >= 0} 1/Lucas(i)^2 = 1/4 + A105394, i.e., the n = 0 case, is believed to be transcendental.

			These infinite sums begin: 1/2, 7/36, 551/7392, ...

Cf. A000032, A105394, A132318, A333088, A333089, A333209.

a[n_] := Numerator[Sum[1/(LucasL[2 i - 2]*LucasL[2 i - 1]), {i, 1, n}]/Fibonacci[2 n]]; Array[a, 12] (* Amiram Eldar, Mar 11 2020 *)

a(n) = numerator of (1/Fibonacci(2n)) * Sum_{i=1..n} 1/(Lucas(2i-2)*Lucas(2i-1)).
Lim_{n -> inf} (a(n)/A333209(n)) / (a(n-1)/A333209(n-1)) = 1 - 1/phi = 1/phi^2 = A132318.
The following generalization holds: (Start)
Let H_(a,b) (n) be defined by H_(a,b) (0) = a, H_(a,b) (1) = b and H_(a,b) (n) = H_(a,b) (n-1) + H_(a,b) (n-2) for n > 1, then
Sum_{i >= 0} 1/(H_(a,b) (i)*H_(a,b) (i+2n)) = (1/Fibonacci(2n)) * Sum_{i=1..n} 1/(H_(a,b) (2i-2)*H_(a,b) (2i-1)) for n > 0, and are thus all fractions. Specially, H_(0,1) are the Fibonacci numbers A000045, H_(2,1) as here, are the Lucas numbers A000032, and H_(3,1) are the Pibonacci numbers A104449. (End)

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A001254 Squares of Lucas numbers.

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A105393 Decimal expansion of sum of reciprocals of squares of Fibonacci numbers.

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A333209 a(n) is the denominator of Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)), with Lucas(i) as defined in A000032.

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A333208 a(n) is the numerator of Sum_{i >= 0} 1/(Lucas(i)*Lucas(i+2n)), with Lucas(i) as defined in A000032.

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