A001519 -id:A001519 - cp's OEIS frontend

A093129 Binomial transform of Fibonacci(2n-1) (A001519).

1, 2, 5, 15, 50, 175, 625, 2250, 8125, 29375, 106250, 384375, 1390625, 5031250, 18203125, 65859375, 238281250, 862109375, 3119140625, 11285156250, 40830078125, 147724609375, 534472656250, 1933740234375, 6996337890625

Offset: 0

Paul Barry, Mar 23 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-5).

Cf. A000032, A000045, A001519, A020876, A030191, A093123.

a:=[1,2];; for n in [3..30] do a[n]:=5*(a[n-1]-a[n-2]); od; a; # G. C. Greubel, Dec 27 2019

I:=[1,2]; [n le 2 select I[n] else 5*(Self(n-1) - Self(n-2)): n in [1..30]]; // G. C. Greubel, Dec 27 2019

a:= n-> (<<0|1>, <-5|5>>^n. <<1,2>>)[1,1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Aug 29 2015

LinearRecurrence[{5, -5}, {1, 2}, 25] (* Jean-François Alcover, May 11 2019 *)
Table[If[EvenQ[n], 5^(n/2)*Fibonacci[n-1], 5^((n-1)/2)*LucasL[n-1]], {n,0,30}] (* G. C. Greubel, Dec 27 2019 *)

my(x='x+O('x^30)); Vec((1-3*x)/(1-5*x+5*x^2)) \\ G. C. Greubel, Dec 27 2019

[lucas_number2(n,5,5) for n in range(-1,25)] # Zerinvary Lajos, Jul 08 2008

G.f.: (1-3*x)/(1-5*x+5*x^2).
a(n) = (5-sqrt(5))*((5+sqrt(5))/2)^n/10 + (5+sqrt(5))*((5-sqrt(5))/2)^n/10.
a(n) = A093123(n)/2^n.
a(n) = A020876(n-1). - R. J. Mathar, Sep 05 2008
a(n) = A030191(n) - 3*A030191(n-1). - R. J. Mathar, Jun 29 2012
a(2*n) = 5^n*Fibonacci(2*n-1), a(2*n+1) = 5^n*Lucas(2*n). - G. C. Greubel, Dec 27 2019
E.g.f.: (1/10)*exp((1/2)*(5-sqrt(5))*x)*(5 + sqrt(5) + (5 - sqrt(5))*exp(sqrt(5)*x)). - Stefano Spezia, Dec 28 2019

A130255 Maximal index k of an odd Fibonacci number (A001519) such that A001519(k) = Fibonacci(2k-1) <= n (the 'lower' odd Fibonacci Inverse).

Original entry on oeis.org

1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

Hieronymus Fischer, May 24 2007, Jul 02 2007

nonn

Inverse of the odd Fibonacci sequence (A001519), nearly, since a(A001519(n))=n except for n=0 (see A130256 for another version). a(n)+1 is the number of odd Fibonacci numbers (A001519) <= n (for n >= 1).

			a(10)=3 because A001519(3) = 5 <= 10, but A001519(4) = 13 > 10.

G. C. Greubel, Table of n, a(n) for n = 1..10000

Cf. partial sums A130257. Other related sequences: A000045, A130233, A130237, A130239, A130256, A130259, A104160. Lucas inverse: A130241 - A130248.

phi:=(1+Sqrt(5))/2; [Floor((1 +Argsinh(Sqrt(5)*n/2)/Log(phi))/2): n in [1..100]]; // G. C. Greubel, Sep 09 2018

Table[Floor[(1 +ArcSinh[Sqrt[5]*n/2]/Log[GoldenRatio])/2], {n, 1, 100}] (* G. C. Greubel, Sep 09 2018 *)

phi=(1+sqrt(5))/2; vector(100, n, floor((1 +asinh(sqrt(5)*n/2)/log(phi))/2)) \\ G. C. Greubel, Sep 09 2018

a(n) = floor((1 + arcsinh(sqrt(5)*n/2)/log(phi))/2).
a(n) = floor((1 + arccosh(sqrt(5)*n/2)/log(phi))/2).
a(n) = floor((1 + log_phi(sqrt(5)*n))/2) for n >= 1, where phi = (1 + sqrt(5))/2.
G.f.: g(x) = 1/(1-x)*Sum_{k>=1} x^Fibonacci(2k-1).
a(n) = floor((1/2)*(1 + log_phi(sqrt(5)*n + 1))) for n >= 1.

A130256 Minimal index k of an odd Fibonacci number A001519 such that A001519(k) = Fibonacci(2*k-1) >= n (the 'upper' odd Fibonacci Inverse).

Original entry on oeis.org

0, 0, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset: 0

Hieronymus Fischer, May 24 2007, Jul 02 2007

nonn

Inverse of the odd Fibonacci sequence (A001519), nearly, since a(A001519(n))=n except for n=1 (see A130255 for another version).

a(n+1) is the number of odd Fibonacci numbers (A001519) <= n (for n >= 0).

			a(10)=4 because A001519(4) = 13 >= 10, but A001519(3) = 5 < 10.

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. partial sums A130258.
Other related sequences: A000045, A001906, A130234, A130237, A130239, A130255, A130260.
Lucas inverse: A130241 - A130248.

[0,0] cat [Ceiling((1/2)*(1 + Log(Sqrt(5)*n-1)/(Log((1+Sqrt(5))/2)))): n in [2..100]]; // G. C. Greubel, Sep 12 2018

Join[{0, 0}, Table[Ceiling[1/2*(1 + Log[GoldenRatio, (Sqrt[5]*n - 1)])], {n, 2, 100}]] (* G. C. Greubel, Sep 12 2018 *)

for(n=0,100, print1(if(n==0, 0, if(n==1, 0, ceil((1/2)*(1 + log(sqrt(5)*n-1)/(log((1+sqrt(5))/2)))))), ", ")) \\ G. C. Greubel, Sep 12 2018

a(n) = ceiling((1+arccosh(sqrt(5)*n/2)/log(phi))/2), where phi=(1+sqrt(5))/2.
G.f.: (x/(1-x))*Sum_{k>=0} x^Fibonacci(2*k-1).
a(n) = ceiling((1/2)*(1+log_phi(sqrt(5)*n-1))) for n >= 2, where phi=(1+sqrt(5))/2.

A154626 a(n) = 2^n*A001519(n).

Original entry on oeis.org

1, 2, 8, 40, 208, 1088, 5696, 29824, 156160, 817664, 4281344, 22417408, 117379072, 614604800, 3218112512, 16850255872, 88229085184, 461973487616, 2418924584960, 12665653559296, 66318223015936, 347246723858432, 1818207451086848, 9520257811087360

Offset: 0

Paul Barry, Jan 13 2009

Hankel transform of 1,1,3,11,45,... (see A026375). Binomial transform of A015448.
From Gary W. Adamson, Jul 22 2016: (Start)
A production matrix for the sequence is M =
  1, 1, 0, 0, 0, ...
  1, 0, 5, 0, 0, ...
  1, 0, 0, 5, 0, ...
  1, 0, 0, 0, 5, ...
  ...
Take powers of M, extracting the upper left terms; getting
the sequence starting (1, 1, 2, 8, 40, 208, ...). (End)
The sequence is N=5 in an infinite set of INVERT transforms of powers of N prefaced with a "1". (1, 2, 8, 40, ...) is the INVERT transform of (1, 1, 5, 25, 125, ...). The first six of such sequences are shown in A006012 (N=3). - Gary W. Adamson, Jul 24 2016
From Gary W. Adamson, Jul 27 2016: (Start)
The sequence is the first in an infinite set in which we perform the operation for matrix M (Cf. Jul 22 2016), but change the left border successively from (1, 1, 1, 1, ...) then to (1, 2, 2, 2, ...), then (1, 3, 3, 3, ...) ...; generally (1, N, N, N, ...). Extracting the upper left terms of each matrix operation, we obtain the infinite set beginning:
  N=1  (A154626): 1,  2,  8,  40,  208,  1088, ...
  N=2  (A084120): 1,  3, 15,  81,  441,  1403, ...
  N=3  (A180034): 1,  4, 22, 124,  700,  3952, ...
  N=4  (A001653): 1,  5, 29, 169,  985,  5741, ...
  N=5  (A000400): 1,  6, 36, 216, 1296,  7776, ...
  N=6  (A015451): 1,  7, 43, 265, 1633, 10063, ...
  N=7  (A180029): 1,  8, 50, 316, 1996, 12608, ...
  N=8  (A180028): 1,  9, 57, 369, 1285, 15417, ...
  N=9  (.......): 1, 10, 64, 424, 2800, 18496, ...
  N=10 (A123361): 1, 11, 71, 481, 3241, 21851, ...
  N=11 (.......): 1, 12, 78, 540, 3708, 25488, ...
... Each of the sequences begins (1, (N+1), (7*N + 1),
  (40*N + (N-1)^2), ... (End)
The set of infinite sequences shown (Cf. comment of Jul 27 2016), can be generated from the matrices P = [(1,N; 1,5]^n, (N=1,2,3,...) by extracting the upper left terms. Example: N=6 sequence (A015451): (1, 7, 43, 265, ...) can be generated from the matrix P = [(1,6); (1,5)]^n. - Gary W. Adamson, Jul 28 2016

Karl V. Keller, Jr., Table of n, a(n) for n = 0..1000
Bishal Deb, Cyclic sieving phenomena via combinatorics of continued fractions, arXiv:2508.13709 [math.CO], 2025. See p. 42.
Index entries for linear recurrences with constant coefficients, signature (6,-4).

Cf. A006012, A084120, A180034, A001653, A000400, A015451, A180029, A180028, A123362.

[n le 2 select (n) else 6*Self(n-1)-4*Self(n-2): n in [1..25]]; // Vincenzo Librandi, May 15 2015

LinearRecurrence[{6, -4}, {1, 2}, 30] (* Vincenzo Librandi, May 15 2015 *)

Vec((1-4*x) / (1-6*x+4*x^2) + O(x^30)) \\ Colin Barker, Sep 22 2017

G.f.: (1 - 4*x) / (1 - 6*x + 4*x^2).
a(n) = A084326(n+1) - 4*A084326(n). - R. J. Mathar, Jul 19 2012
From Colin Barker, Sep 22 2017: (Start)
a(n) = (((3-sqrt(5))^n*(1+sqrt(5)) + (-1+sqrt(5))*(3+sqrt(5))^n)) / (2*sqrt(5)).
a(n) = 6*a(n-1) - 4*a(n-2) for n>1. (End)
E.g.f.: exp(3*x)*(5*cosh(sqrt(5)*x) - sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, Aug 26 2025

A246638 Sequence a(n) = 2 + 3*A001519(n+1) appearing in a certain four circle touching problem together with A246639.

Original entry on oeis.org

5, 8, 17, 41, 104, 269, 701, 1832, 4793, 12545, 32840, 85973, 225077, 589256, 1542689, 4038809, 10573736, 27682397, 72473453, 189737960, 496740425, 1300483313, 3404709512, 8913645221, 23336226149, 61095033224, 159948873521, 418751587337, 1096305888488, 2870166078125, 7514192345885

Offset: 0

Wolfdieter Lang, Aug 31 2014

This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032.
a(n), together with b(n) = A246639(n) appears as the curvature c(n) =  a(n) + (4*b(n)/5)*phi (phi = (1+sqrt(5))/2, golden section) of the circle which touches i) a circle of radius 5/4 (in some length units) divided by a chord of length 2 into two unequal parts, and ii) the two touching circles in the smaller part which have curvatures A240926(n) and  A240926(n+1), both also touching the circle with radius 5/4. See the illustration of Kival Ngaokrajang's link given in A240926, where the first circles in the smaller (upper) part are shown. The present circles will lie in the region between the large circle and two of these circles in the upper part.
Descartes' theorem on touching circles (see the links) is applied here as  c(n) = -4/5 + A(n) + A(n+1) + 2*sqrt((-4/5 )*(A(n) + A(n+1)) + A(n)*A(n+1)), with A(n) = A240926(n), n >= 0.
For the proof for the first formula for a(n) given below use the formula for the curvature A240926(n) = 2 + 2*S(n, 3) - 3* S(n-1, 3) (see the W. Lang link in A240926, part II) in c(n) and compare with a(n) from c(n) =  a(n) + (4*b(n)/5)*phi. This is done by using standard S-polynomial identities like the three term recurrence and the Cassini-Simson type identity
  S(n, x)*S(n-2, x) = -1 + S(n-1, x)^2 (here for x=3). This implies S(n, 3)*S(n-1, 3) = (-1 + S(n, 3)^2 + S(n-1, 3)^2)/3. See also the mentioned link, part III a).
a(n) appears also in the curvature for the touching circles and chord problem in the smaller part of a circle with radius 5/4 dissected by a chord of length 2, together with A246640, where details are given.

			a(1) = 8 because c(n) = -4/5 + 5 + 9 + 2*sqrt((-4/5 )*(5 + 9) + 5*9) = 4*(2+(13/5)*phi). This is also 8 + (4*13/5)*phi with A246639(1) = 13.

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Descartes' Circle Theorem.
Wikipedia, Descartes' Theorem.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A246639, A049310, A001519, A115032.

I:=[5,8,17]; [n le 3 select I[n] else 4*Self(n-1) - 4*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{4, -4, 1}, {5, 8, 17}, 30] (* or *)  CoefficientList[ Series[(5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

Vec((5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016

a(n) = 2 + 3*(S(n,3) - S(n-1,3)) = 2 + 3*A001519(n+1), n >= 0, with Chebyshev's S-polynomials (see A049310).
O.g.f.: (5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n >=1, a(-2) = 8, a(-1) = 5, a(0) = 5.
a(n) = (2^(-1-n)*(5*2^(2+n)-3*(3-sqrt(5))^n*(-5+sqrt(5))+3*(3+sqrt(5))^n*(5+sqrt(5))))/5. - Colin Barker, Nov 02 2016

A169691 Let T be the sequence Fibonacci(2n+1), n>=0 (cf. A001519); sequence lists the differences T(j)-T(i) for i

Original entry on oeis.org

1, 3, 4, 8, 11, 12, 21, 29, 32, 33, 55, 76, 84, 87, 88, 144, 199, 220, 228, 231, 232, 377, 521, 576, 597, 605, 608, 609, 987, 1364, 1508, 1563, 1584, 1592, 1595, 1596, 2584, 3571, 3948, 4092, 4147, 4168, 4176, 4179, 4180, 6765, 9349, 10336, 10713, 10857, 10912, 10933

Offset: 1

N. J. A. Sloane, Apr 14 2010

nonn

See Comments in A169691.

Cf. A000045, A001519, A169690, A169692, A169693, A169677-A169680.

Module[{nn=31,fbs},fbs=Fibonacci[Range[1,nn,2]];Sort[Flatten[Table[ fbs[[n]]- Take[fbs,n-1],{n,nn/2}]]]] (* Harvey P. Dale, Aug 30 2015 *)

A005522 a(n) = 1 + L(n) + F(2n-1) with {L(n)}A000032)%20and%20F(2n-1)">{n>=0} the Lucas numbers (A000032) and F(2*n-1){n>=0} the bisected Fibonacci numbers (A001519).

Original entry on oeis.org

4, 3, 6, 10, 21, 46, 108, 263, 658, 1674, 4305, 11146, 28980, 75547, 197262, 515594, 1348477, 3528150, 9233244, 24167167, 63261114, 165604618, 433534041, 1134967250, 2971318756, 7778909811, 20365282518, 53316730378, 139584573093, 365436446014, 956723886540

Offset: 0

N. J. A. Sloane

From Russell Jay Hendel, Mar 02 2015: (Start)
The Name field was changed. The former Name-field entry was:
a(n) = Fibonacci(2n)*(1 + Fibonacci(n-1))/Fibonacci(n) for n even; a(n) = 2 + Fibonacci(2n)*(1 + Fibonacci(n-1))/Fibonacci(n) for n odd.
The new entry in the Name field is based on Equation (1) of the Grieg paper cited in the Links section. This formula is simpler than the former formulas given in the Name and Formula fields of this sequence. We make 6 comments:
(Comment I) The former formula in the Name section is consistent with a(0)=4 since 0 is a removable singularity of the former formula in the Name section. To see this we use the continuous Binet form, F(x) = (alpha^x - beta^x)/sqrt(5) with alpha = (sqrt(5)+1)/2 and beta = 1-alpha. Hence, for all nonzero x, F(2x)/F(x) = (alpha^(2x) - beta^(2x))/(alpha^x - beta^x) = alpha^x + beta^x. It immediately follows that lim_{x->0} F(2x)/F(x) = alpha^0 + beta^0 = 2. Consequently, lim_{x->0} Fibonacci(2x)*(1 + Fibonacci(x-1))/Fibonacci(x) = 2(1+Fibonacci(-1)) = 4 as required.
(Comment II) As pointed out in references [2,3] of the Grieg paper, a(n) = F(2n) C(n), with C(n) = Sum_{m>=0} 1/F(n*b), b=2^m, and {F(n)}_{n>=0}, the Fibonacci numbers (A000045). In other words, the a(n) = C(n) F(2n) are the numerators of the sequence of C(n) = a(n) / F(2n), which are the values of infinite sums of reciprocals of subsets of the Fibonacci numbers.
(Comment III) The probable reason that the elegant formulas in the Grieg papers were overlooked is that the Grieg papers are difficult to read because of their nonstandard notation and terminology. For example, the generalized Fibonacci and Lucas numbers are indicated by P with subscripts rather than by F. Similarly, Grieg refers to the sequence of odd-indexed Fibonacci numbers as the "bisected" Fibonacci numbers (A001519), a term which rarely appears in modern books on the Fibonacci numbers.
(Comment IV) The defining recursion for a(n) is best given in factored form using annihilating operators (Equation (12) in the Grieg paper). Let E be the translation operator so that E(f)(n) = f(n+1). Then, for example, (E^2-E-1)(L(n)) = L(n+2) - L(n+1) - L(n) = 0 (A000032). Thus the operator E^2-E-1 annihilates the Lucas sequence. Similarly, (E^2-3E+1) annihilates the bisected Fibonacci numbers (A001519) and (E-1) annihilates the constant sequence (A000012). It follows that the product of these annihilators annihilates the sum of the sequences, which is a(n). So the defining recursion may be obtained by expanding (E-1)(E^2-3E+1)(E^2-E-1). It follows that the sequence a(n) satisfies a(n) - 5 a(n-1) + 7 a(n-2) - a(n-3) - 3 a(n-4) + a(n-5) = 0. For example, 1*108 - 5*46 + 7*21 - 1*10 - 3*6 + 1*3 = 0. Note that the derivation of this recursion, based on annihilators, is more straightforward than the derivation in the Grieg paper based on central difference operators.
(Comment V) Using standard techniques, we may obtain the g.f. of the a(n) from the defining recursion, of order 5. Since a(n) = 1 + F(2n-1) + L(n), it follows that the g.f. for the a(n) is the product of the g.f. for the three summands 1, F(2n-1), L(n). The g.f. for the constant sequence, 1, is C(x)=1/(1-x) (A000012); the g.f. for the Lucas sequence is L(x)=(2-x)/(1-x-x^2)(A000032); and the g.f. for the bisected Fibonacci numbers F(2n-1) is B(x)=(1 - 2x)/(1 - 3x + x^2) (sequence A001519). It follows that the g.f. for the a(n) is the sum of these g.f.s: C(x) + L(x) + B(x) = 1/(1-x) + (1-2x)/(1-3x+x^2) + (2-x)/(1-x-x^2). This is a partial fraction decomposition of a rational function, that is an equivalent g.f. for the a(n), A(x) = (4 - 17x + 19x^2 - 3x^3 - 2x^4)/(1 - 5x + 7x^2 - x^3 - 3x^4 + x^5).
(Comment VI) We can now comment on the g.f., P(x), conjectured by Simon Plouffe and mentioned in the MAPLE section below. Expanding P(x) as a series, we obtain the following g.f. of the a(n): 3 + 6x + 10x^2 + 21x^3 ... . In fact, it is easily checked that P(x) = 1/x (A(x)-4). This settles the Plouffe conjecture as follows: (i) A(x) is the rational function g.f. of a(n){n>=0}, (ii) P(x) is the rational function g.f. of a(n){n>=1}, and (iii) C(x)+L(x)+B(x) is the partial fraction g.f. of a(n)_{n>=0}. (End)
Let phi = (1+sqrt(5))/2, p(n) = phi^n - (-phi)^(-n) and FL(n) = 1 + (p(n-1) + p(n+1) + p(2*n-1))/sqrt(5). FL is a doubly infinite sequence with FL(n) = A005522(n) and FL(-n) = A006172(n) for n >= 0. - Peter Luschny, Mar 09 2015

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000
W. E. Greig, Sums of Fibonacci Reciprocals, The Fibonacci Quarterly, Vol. 15, No. 1 (1977), pp. 46-48.
W. E. Greig, On generalized G_{j,k} numbers, Fib. Quart., 16 (1978), 166-170.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Index entries for linear recurrences with constant coefficients, signature (5,-7,1,3,-1).

Cf. A000032, A000045, A001519, A001622, A006172, A079586, A254884.

[1 + Lucas(n) + Fibonacci(2*n - 1): n in [0..30]]; // Vincenzo Librandi, Mar 08 2015

with(combinat): a:=proc(n) if n mod 2 = 0 then fibonacci(2*n)*(1+fibonacci(n-1))/fibonacci(n) else 2+fibonacci(2*n)*(1+fibonacci(n-1))/fibonacci(n) fi end: seq(a(n),n=1..32); # Emeric Deutsch, Apr 01 2005
A005522:=-(-3+9*z-z**2-10*z**3+4*z**4)/(z-1)/(z**2-3*z+1)/(z**2+z-1); # conjectured by Simon Plouffe in his 1992 dissertation
FL := proc(n) local a,p; a := (1+sqrt(5))/2; p := m -> a^m - (-a)^(-m);
1 + (p(n-1) + p(n+1) + p(2*n-1))/sqrt(5) end: A005522 := n -> FL(n):
seq(round(evalf(A005522(n),32)), n=0..30); # Peter Luschny, Mar 09 2015

Table[1+Fibonacci[2n-1]+LucasL[n], {n, 0, 30}] (* Russell Jay Hendel, Mar 02 2015 *) (* modified by G. C. Greubel, Jan 01 2018 *)
Series[1/(1-x)+ (2-x)/(1-x-x^2)+(1-2*x)/(1-3*x+x^2), {x, 0, 30}] (* Russell Jay Hendel, Mar 02 2015 *)
LinearRecurrence[{5,-7,1,3,-1}, {4, 3, 6, 10, 21}, 31] (* Russell Jay Hendel, Mar 02 2015 *)
CoefficientList[ Series[(4 - 17x + 19x^2 - 3x^3 - 2x^4)/(1 - 5x + 7x^2 - x^3 - 3x^4 + x^5), {x, 0, 30}], x] (* Russell Jay Hendel, Mar 02 2015 *)

\ps {31};
C(x) = 1/(1-x);
L(x) = (2-x)/(1-x-x^2);
B(x) = (1-2*x)/(1-3*x+x^2);
A(x) = C(x)+L(x)+B(x);
Ser(A(x),x)
Vec(Ser(A(x),x))
\\ Russell Jay Hendel, Mar 02 2015

{lucas(n) = fibonacci(n+1) + fibonacci(n-1)};
for(n=0,30, print1(1 + fibonacci(2*n-1) + lucas(n), ", ")) \\ G. C. Greubel, Jan 01 2018

a(n) = ((sqrt(5)+5)/10)*(((3+sqrt(5))/2)^(n-1)) + ((5-sqrt(5))/10)*(((3-sqrt(5))/2)^(n-1)) + ((1+sqrt(5))/2)^n + ((1-sqrt(5))/2)^n + 1. - Tim Monahan, Jul 23 2011
a(n) = 1 + L(n) + F(2*n-1), with L(n) the Lucas numbers, sequence A000032 and F(2n-1) the bisected Fibonacci numbers, sequence A001519. - Russell Jay Hendel, Mar 02 2015
a(n) = 5*a(n-1) - 7*a(n-2) + a(n-3) + 3*a(n-4) - a(n-5). - Russell Jay Hendel, Mar 02 2015
Sum_{n>=1} (a(2*n-1)/Fibonacci(4*n-2) - 1/phi) = A079586, where phi is the golden ratio (A001622) (Greig, 1977). - Amiram Eldar, Jan 29 2022

Edited by Emeric Deutsch, Apr 01 2005
a(0), a(29), a(30) added by Russell Jay Hendel, Mar 02 2015
Name changed by Russell Jay Hendel, Mar 02 2015

A246639 Sequence a(n) = 3 + 5*A001519(n+1) appearing in a certain three circle touching problem, together with A246638.

Original entry on oeis.org

8, 13, 28, 68, 173, 448, 1168, 3053, 7988, 20908, 54733, 143288, 375128, 982093, 2571148, 6731348, 17622893, 46137328, 120789088, 316229933, 827900708, 2167472188, 5674515853, 14856075368, 38893710248, 101825055373, 266581455868, 697919312228, 1827176480813, 4783610130208, 12523653909808

Offset: 0

Wolfdieter Lang, Sep 01 2014

See the comments under A246638. The curvature c(n) defined there is c(n) = A246638(n) + (4*a(n)/5)*phi with phi = (1+sqrt(5))/2, the golden section. It lives in the quadratic number field Q(sqrt(5)). Descartes' theorem on touching circles gives c(n) = -4/5 + A(n) + A(n+1) + 2*sqrt((-4/5 )*(A(n) + A(n+1)) + A(n)*A(n+1)), with A(n) = A240926(n), n >= 0. For the proof of the first formula given below one compares this a(n) with the a(n) in c(n) given above. This uses standard Chebyshev S-polynomial identities with x = 3, like the three term recurrence and the Cassini-Simson type identity S(n, x)*S(n-2, x) = -1 + S(n-1, x)^2 (here for x=3). This implies S(n, 3)*S(n-1, 3) = (-1 + S(n, 3)^2 + S(n-1, 3)^2)/3. See also the W. Lang link in A240926, part III a).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A246638, A049310, A001519, A115032.

I:=[8, 13, 28]; [n le 3 select I[n] else 4*Self(n-1) -4*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

CoefficientList[Series[(8-19*x+8*x^2)/((1-x)*(1-3*x+x^2)), {x, 0, 50}], x] (* or *) LinearRecurrence[{4,-4,1}, {8, 13, 28}, 30] (* G. C. Greubel, Dec 20 2017 *)

Vec((8-19*x+8*x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016

a(n) = 3 + 5*(S(n, 3) - S(n-1, 3)) =  3 + 5*A001519(n+1), n >= 0, with Chebyshev S-polynomials (see A049310) with S(-1, x) = 0.
O.g.f.: (8-19*x+8*x^2)/((1-x)*(1-3*x+x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n >=1, a(-2) = 13, a(-1) = 8, a(0) = 8.
a(n) = 2^(-1-n)*(3*2^(1+n)-(3-sqrt(5))^n*(-5+sqrt(5))+(3+sqrt(5))^n*(5+sqrt(5))). - Colin Barker, Nov 02 2016

A246640 Sequence a(n) = 1 + A001519(n+1) appearing in a certain touching problem for three circles and a chord, together with A246638.

Original entry on oeis.org

2, 3, 6, 14, 35, 90, 234, 611, 1598, 4182, 10947, 28658, 75026, 196419, 514230, 1346270, 3524579, 9227466, 24157818, 63245987, 165580142, 433494438, 1134903171, 2971215074, 7778742050, 20365011075, 53316291174, 139583862446, 365435296163, 956722026042, 2504730781962

Offset: 0

Wolfdieter Lang, Sep 03 2014

Essentially the same as A093467 and A032908.
This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032.
a(n), together with b(n) = A246638(n), appears in a curvature c(n) =  b(n) + 4*a(n)*phi, with phi = (1+sqrt(5))/2, the golden section. This is an integer in the real quadratic field Q(sqrt(5)). c(n) is the curvature of the circle which touches i) a chord of length 2 (in some length units) of a circle of radius 5/4 which is divided by this chord in two unequal parts, and ii) the two touching circles in the smaller part which have curvatures A240926(n) and A240926(n+1). These two touching circles touch also the circle with radius 5/4 and the chord. See the illustration of Kival Ngaokrajang's link given in A240926, where the first circles in the smaller (upper) part are shown. c(n) is an integer in the real quadratic field  Q(sqrt(5)).
From Descartes' theorem on touching circles (see the links) one has here: c(n) = A(n) + A(n+1) + 2*sqrt(A(n)*A(n+1)),
  with A(n) = A240926(n), n >= 0. In this application the chord has curvature 0.
For the proof for the first formula for a(n) given below use the formula for the curvature A240926(n) = 2 + 2*S(n, 3) - 3* S(n-1, 3) (see the W. Lang link found in A240926, part II) in c(n) from Descartes' formula and compare it with a(n) from c(n) = A246638(n) + 4*a(n)*(1+sqrt(5))/2. This can be done by using standard S-polynomial identities like the three term recurrence and the Cassini-Simson type identity (see a comment on A246638) which implies the formula S(n, 3)*S(n-1, 3) = (-1 + S(n, 3)^2 + S(n-1, 3)^2)/3. See also the above mentioned W. Lang link part III b).

			a(1) = 3 because c(1) = 0 +  5 + 9 + 2*sqrt(5*9) = 8 + 12*phi which is indeed 8 + 4*3*phi, with 8 = A246638(1).

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Descartes' Circle Theorem.
Wikipedia, Descartes' Theorem.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A246638, A240926, A049310, A001519.

I:=[2, 3, 6]; [n le 3 select I[n] else 4*Self(n-1) - 4*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{4,-4,1},{2, 3, 6}, 30] (* or *) CoefficientList[ Series[ (2-5*x+ 2*x^2)/ ((1-x)*(1-3*x+x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

Vec((2-5*x+2*x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016

a(n) = 1 + S(n, 3) - S(n-1, 3) = 1 + A001519(n+1), n>=0, with Chebyshev's S-polynomials (see A049310), and S(-1, x) = 0.
O.g.f.: (2-5*x+2*x^2)/((1-x)*(1-3*x+x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n >= 1, with a(-2) = 3, a(-1) = 2 and a(0) = 2.
a(n) = 1+(2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/sqrt(5). - Colin Barker, Nov 02 2016

cp's OEIS Frontend

A093129 Binomial transform of Fibonacci(2n-1) (A001519).

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A130255 Maximal index k of an odd Fibonacci number (A001519) such that A001519(k) = Fibonacci(2k-1) <= n (the 'lower' odd Fibonacci Inverse).

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A130256 Minimal index k of an odd Fibonacci number A001519 such that A001519(k) = Fibonacci(2*k-1) >= n (the 'upper' odd Fibonacci Inverse).

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A154626 a(n) = 2^n*A001519(n).

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A246638 Sequence a(n) = 2 + 3*A001519(n+1) appearing in a certain four circle touching problem together with A246639.

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A169691 Let T be the sequence Fibonacci(2n+1), n>=0 (cf. A001519); sequence lists the differences T(j)-T(i) for i

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A005522 a(n) = 1 + L(n) + F(2*n-1) with {L(n)}A000032)%20and%20F(2*n-1)">{n>=0} the Lucas numbers (A000032) and F(2*n-1){n>=0} the bisected Fibonacci numbers (A001519).

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A005522 a(n) = 1 + L(n) + F(2n-1) with {L(n)}A000032)%20and%20F(2n-1)">{n>=0} the Lucas numbers (A000032) and F(2*n-1){n>=0} the bisected Fibonacci numbers (A001519).