A153386 -id:A153386 - cp's OEIS frontend

A001906 F(2n) = bisection of Fibonacci sequence: a(n) = 3*a(n-1) - a(n-2).

0, 1, 3, 8, 21, 55, 144, 377, 987, 2584, 6765, 17711, 46368, 121393, 317811, 832040, 2178309, 5702887, 14930352, 39088169, 102334155, 267914296, 701408733, 1836311903, 4807526976, 12586269025, 32951280099, 86267571272, 225851433717, 591286729879, 1548008755920

Offset: 0

N. J. A. Sloane

Apart from initial term, same as A088305.
Second column of array A102310 and of A028412.
Numbers k such that 5*k^2 + 4 is a square. - Gregory V. Richardson, Oct 13 2002
Apart from initial terms, also Pisot sequences E(3,8), P(3,8), T(3,8). See A008776 for definitions of Pisot sequences.
Binomial transform of A000045. - Paul Barry, Apr 11 2003
Number of walks of length 2n+1 in the path graph P_4 from one end to the other one. Example: a(2)=3 because in the path ABCD we have ABABCD, ABCBCD and ABCDCD. - Emeric Deutsch, Apr 02 2004
Simplest example of a second-order recurrence with the sixth term a square.
Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 3. - Lekraj Beedassy, Jun 11 2004
a(n) (for n > 0) is the smallest positive integer that cannot be created by summing at most n values chosen among the previous terms (with repeats allowed). - Andrew Weimholt, Jul 20 2004
All nonnegative integer solutions of Pell equation b(n)^2 - 5*a(n)^2 = +4 together with b(n) = A005248(n), n >= 0. - Wolfdieter Lang, Aug 31 2004
a(n+1) is a Chebyshev transform of 3^n (A000244), where the sequence with g.f. G(x) is sent to the sequence with g.f. (1/(1+x^2))G(x/(1+x^2)). - Paul Barry, Oct 25 2004
a(n) is the number of distinct products of matrices A, B, C, in (A+B+C)^n where commutator [A,B] = 0 but C does not commute with A or B. - Paul D. Hanna and Max Alekseyev, Feb 01 2006
Number of binary words with exactly k-1 strictly increasing runs. Example: a(3)=F(6)=8 because we have 0|0,1|0,1|1,0|01,01|0,1|01,01|1 and 01|01. Column sums of A119900. - Emeric Deutsch, Jul 23 2006
See Table 1 on page 411 of Lukovits and Janezic paper. - Parthasarathy Nambi, Aug 22 2006
Inverse: With phi = (sqrt(5) + 1)/2, log_phi((sqrt(5) a(n) + sqrt(5 a(n)^2 + 4))/2) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007
[1,3,8,21,55,144,...] is the Hankel transform of [1,1,4,17,75,339,1558,...](see A026378). - Philippe Deléham, Apr 13 2007
The Diophantine equation a(n) = m has a solution (for m >= 1) if and only if floor(arcsinh(sqrt(5)*m/2)/log(phi)) <> floor(arccosh(sqrt(5)*m/2)/log(phi)) where phi is the golden ratio. An equivalent condition is A130259(m) = A130260(m). - Hieronymus Fischer, May 25 2007
a(n+1) = AB^(n)(1), n >= 0, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 1=`1`, 3=`10`, 8=`100`, 21=`1000`, ..., in Wythoff code.
Equals row sums of triangles A140069, A140736 and A140737. - Gary W. Adamson, May 25 2008
a(n) is also the number of idempotent order-preserving partial transformations (of an n-element chain) of width n (width(alpha) = max(Im(alpha))). Equivalently, it is the number of idempotent order-preserving full transformations (of an n-element chain). - Abdullahi Umar, Sep 08 2008
a(n) is the number of ways that a string of 0,1 and 2 of size (n-1) can be arranged with no 12-pairs. - Udita Katugampola, Sep 24 2008
Starting with offset 1 = row sums of triangle A175011. - Gary W. Adamson, Apr 03 2010
As a fraction: 1/71 = 0.01408450... or 1/9701 = 0.0001030821.... - Mark Dols, May 18 2010
Sum of the products of the elements in the compositions of n (example for n=3: the compositions are 1+1+1, 1+2, 2+1, and 3; a(3) = 1*1*1 + 1*2 + 2*1 + 3 = 8). - Dylon Hamilton, Jun 20 2010, Geoffrey Critzer, Joerg Arndt, Dec 06 2010
a(n) relates to regular polygons with even numbers of edges such that Product_{k=1..(n-2)/2} (1 + 4*cos^2 k*Pi/n) = even-indexed Fibonacci numbers with a(n) relating to the 2*n-gons. The constants as products = roots to even-indexed rows of triangle A152063. For example: a(5) = 55 satisfies the product formula relating to the 10-gon. - Gary W. Adamson, Aug 15 2010
Alternatively, product of roots to x^4 - 12x^3 + 51x^2 - 90x + 55, (10th row of triangle A152063) = (4.618...)*(3.618...)*(2.381...)*(1.381...) = 55. - Gary W. Adamson, Aug 15 2010
a(n) is the number of generalized compositions of n when there are i different types of i, (i=1,2,...). - Milan Janjic, Aug 26 2010
Starting with "1" = row sums of triangle A180339, and eigensequence of triangle A137710. - Gary W. Adamson, Aug 28 2010
a(2) = 3 is the only prime.
Number of nonisomorphic graded posets with 0 and uniform hasse graph of rank n > 0, with exactly 2 elements of each rank level above 0. (Uniform used in the sense of Retakh, Serconek, and Wilson.  Graded used in Stanley's sense that every maximal chain has the same length n.) - David Nacin, Feb 13 2012
Pisano period lengths: 1, 3, 4, 3, 10, 12, 8, 6, 12, 30, 5, 12, 14, 24, 20, 12, 18, 12, 9, 30, ... - R. J. Mathar, Aug 10 2012
Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy + 1. - Michel Lagneau, Feb 01 2014
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2}. - Milan Janjic, Jan 25 2015
With a(0) = 0, for n > 1, a(n) is the smallest number not already in the sequence such that a(n)^2 - a(n-1)^2 is a Fibonacci number. - Derek Orr, Jun 08 2015
Let T be the tree generated by these rules: 0 is in T, and if p is in T, then p + 1 is in T and x*p is in T and y*p is in T. The n-th generation of T consists of A001906(n) polynomials, for n >= 0. - Clark Kimberling, Nov 24 2015
For n > 0, a(n) = exactly the maximum area of a quadrilateral with sides in order of lengths F(n), F(n), L(n), and L(n) with L(n)=A000032(n). - J. M. Bergot, Jan 20 2016
a(n) = twice the area of a triangle with vertices at (L(n+1), L(n+2)), (F(n+1), F(n+1)), and (L(n+2), L(n+1)), with L(n)=A000032(n). - J. M. Bergot, Apr 20 2016
Except for the initial 0, this is the p-INVERT of (1,1,1,1,1,...) for p(S) = 1 - S - S^2; see A291000. - Clark Kimberling, Aug 24 2017
a(n+1) is the number of spanning trees of the graph T_n, where T_n is a sequence of n triangles, where adjacent triangles share an edge. - Kevin Long, May 07 2018
a(n) is the number of ways to partition [n] such that each block is a run of consecutive numbers, and each block has a fixed point, e.g., for n=3, 12|3 with 1 and 3 as fixed points is valid, but 13|2 is not valid as 1 and 3 do not form a run. Consequently, a(n) also counts the spanning trees of the graph given by taking a path with n vertices and adding another vertex adjacent to all of them. - Kevin Long, May 11 2018
From Wolfdieter Lang, May 31 2018: (Start)
The preceding comment can be paraphrased as follows. a(n) is the row sum of the array A305309 for n >= 1. The array A305309(n, k) gives the sum of the products of the block lengths of the set partition of [n] := {1, 2, ..., n} with A048996(n, k) blocks of consecutive numbers, corresponding to the compositions obtained from the k-th partition of n in Abramowitz-Stegun order. See the comments and examples at A305309.
{a(n)} also gives the infinite sequence of nonnegative numbers k for which k * ||k*phi|| < 1/sqrt(5), where the irrational number phi = A001622 (golden section), and ||x|| is the absolute value of the difference between x and the nearest integer. See, e.g., the Havil reference, pp. 171-172. (End)
This Chebyshev sequence a(n) = S(n-1, 3) (see a formula below) is related to the bisection of Fibonacci sequences {F(a,b;n)}{n>=0} with input F(a,b;0) = a and F(a,b;1) = b, by F(a,b;2*k) = (a+b)*S(k-1, 3) - a*S(k-2, 3) and F(a,b;2*k+1) = b*S(k, 3) + (a-b)*S(k-1, 3), for k >= 0, and S(-2, 3) = -1. Proof via the o.g.f.s GFeven(a,b,t) = (a - t*(2*a-b))/(1 - 3*t + t^2) and GFodd(a,b,t) = (b + t*(a-b))/(1 - 3*t + t^2). The special case a = 0, b = 1 gives back F(2*k) = S(k-1, 3) = a(k). - _Wolfdieter Lang, Jun 07 2019
a(n) is the number of tilings of two n X 1 rectangles joined orthogonally at a common end-square (so to have 2n-1 squares in a right-angle V shape) with only 1 X 1 and 2 X 1 tiles. This is a consequence of F(2n) = F(n+1)*F(n) + F(n)*F(n-1). - Nathaniel Gregg, Oct 10 2021
These are the denominators of the upper convergents to the golden ratio, tau; they are also the numerators of the lower convergents (viz. 1/1 < 3/2 < 8/5 < 21/13 < ... < tau < ... 13/8 < 5/3 < 2/1). - Clark Kimberling, Jan 02 2022
For n > 1, a(n) is the smallest Fibonacci number of unit equilateral triangle tiles needed to make an isosceles trapezoid of height F(n) triangles. - Kiran Ananthpur Bacche, Sep 01 2024

			G.f. = x + 3*x^2 + 8*x^3 + 21*x^4 + 55*x^5 + 144*x^6 + 377*x^7 + 987*x^8 + ...
a(3) = 8 because there are exactly 8 idempotent order-preserving full transformations on a 3-element chain, namely: (1,2,3)->(1,1,1),(1,2,3)->(2,2,2),(1,2,3)->(3,3,3),(1,2,3)->(1,1,3),(1,2,3)->(2,2,3),(1,2,3)->(1,2,2),(1,2,3)->(1,3,3),(1,2,3)->(1,2,3)-mappings are coordinate-wise. - _Abdullahi Umar_, Sep 08 2008

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Index entries for "core" sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-1).
Index entries for two-way infinite sequences

Fibonacci A000045 = union of this sequence and A001519.
Cf. A000041, A000032, A001622, A048996, A052529, A055991, A078812, A305309, A058038 (pairwise products), A153386.
Inverse sequences A130259 and A130260.
Cf. A033888, A033890, A249450, A344212.

a001906 n = a001906_list !! n
a001906_list =
   0 : 1 : zipWith (-) (map (* 3) $ tail a001906_list) a001906_list
-- Reinhard Zumkeller, Oct 03 2011

[Fibonacci(2*n): n in [0..30]]; // Vincenzo Librandi, Sep 10 2014

with(combstruct): SeqSeqSeqL := [T, {T=Sequence(S, card > 0), S=Sequence(U, card > 1), U=Sequence(Z, card >0)}, unlabeled]: seq(count(SeqSeqSeqL, size=n+1), n=0..28); # Zerinvary Lajos, Apr 04 2009
H := (n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -4):
a := n -> `if`(n = 0, 0, H(2*n, 1, 1/2)):
seq(simplify(a(n)), n=0..30); # Peter Luschny, Sep 03 2019
A001906 := proc(n)
    combinat[fibonacci](2*n) ;
end proc:
seq(A001906(n),n=0..20) ; # R. J. Mathar, Jan 11 2024

f[n_] := Fibonacci[2n]; Array[f, 28, 0] (* or *)
LinearRecurrence[{3, -1}, {0, 1}, 28] (* Robert G. Wilson v, Jul 13 2011 *)
Take[Fibonacci[Range[0,60]],{1,-1,2}] (* Harvey P. Dale, May 23 2012 *)
Table[ ChebyshevU[n-1, 3/2], {n, 0, 30}] (* Jean-François Alcover, Jan 25 2013, after Michael Somos *)
CoefficientList[Series[(x)/(1 - 3x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Sep 10 2014 *)

makelist(fib(2*n),n,0,30); /* Martin Ettl, Oct 21 2012 */

numlib::fibonacci(2*n) $ n = 0..35; // Zerinvary Lajos, May 09 2008

{a(n) = fibonacci(2*n)}; /* Michael Somos, Dec 06 2002 */

{a(n) = subst( poltchebi(n+1)*4 - poltchebi(n)*6, x, 3/2)/5}; /* Michael Somos, Dec 06 2002 */

{a(n) = polchebyshev( n-1, 2, 3/2)}; /* Michael Somos Jun 18 2011 */

Vec(x/(1-3*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Oct 24 2012

def a(n, adict={0:0, 1:1}):
    if n in adict:
        return adict[n]
    adict[n]=3*a(n-1) - a(n-2)
    return adict[n] # David Nacin, Mar 04 2012

[lucas_number1(n,3,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

[fibonacci(2*n) for n in range(0, 28)] # Zerinvary Lajos, May 15 2009

G.f.: x / (1 - 3*x + x^2). -  Simon Plouffe in his 1992 dissertation
a(n) = 3*a(n-1) - a(n-2) = A000045(2*n).
a(n) = -a(-n).
a(n) = A060921(n-1, 0), n >= 1.
a(n) = sqrt((A005248(n)^2 - 4)/5).
a(n) = A007598(n) - A007598(n-2), n > 1.
a(n) = (ap^n - am^n)/(ap-am), with ap := (3+sqrt(5))/2, am := (3-sqrt(5))/2.
Invert transform of natural numbers: a(n) = Sum_{k=1..n} k*a(n-k), a(0) = 1. - Vladeta Jovovic, Apr 27 2001
a(n) = S(n-1, 3) with S(n, x) = U(n, x/2) Chebyshev's polynomials of the 2nd kind, see A049310.
a(n) = Sum_{k=0..n} binomial(n, k)*F(k). - Benoit Cloitre, Sep 03 2002
Limit_{n->infinity} a(n)/a(n-1) = 1 + phi = (3 + sqrt(5))/2. This sequence includes all of the elements of A033888 combined with A033890.
a(0)=0, a(1)=1, a(2)=3, a(n)*a(n-2) + 1 = a(n-1)^2. - Benoit Cloitre, Dec 06 2002
a(n) = n + Sum_{k=0..n-1} Sum_{i=0..k} a(i) = n + A054452(n). - Benoit Cloitre, Jan 26 2003
a(n) = Sum_{k=1..n} binomial(n+k-1, n-k). - Vladeta Jovovic, Mar 23 2003
E.g.f.: (2/sqrt(5))*exp(3*x/2)*sinh(sqrt(5)*x/2). - Paul Barry, Apr 11 2003
Second diagonal of array defined by T(i, 1) = T(1, j) = 1, T(i, j) = Max(T(i-1, j) + T(i-1, j-1); T(i-1, j-1) + T(i, j-1)). - Benoit Cloitre, Aug 05 2003
a(n) = F(n)*L(n) = A000045(n)*A000032(n). - Lekraj Beedassy, Nov 17 2003
F(2n+2) = 1, 3, 8, ... is the binomial transform of F(n+2). - Paul Barry, Apr 24 2004
Partial sums of A001519(n). - Lekraj Beedassy, Jun 11 2004
a(n) = Sum_{i=0..n-1} binomial(2*n-1-i, i)*5^(n-i-1)*(-1)^i. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
a(n) = Sum_{k=0..n} binomial(n+k, n-k-1) = Sum_{k=0..n} binomial(n+k, 2k+1).
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*3^(n-2*k). - Paul Barry, Oct 25 2004
a(n) = (n*L(n) - F(n))/5 = Sum_{k=0..n-1} (-1)^n*L(2*n-2*k-1).
The i-th term of the sequence is the entry (1, 2) in the i-th power of the 2 X 2 matrix M = ((1, 1), (1, 2)). - Simone Severini, Oct 15 2005
Computation suggests that this sequence is the Hankel transform of A005807. The Hankel transform of {a(n)} is Det[{{a(1), ..., a(n)}, {a(2), ..., a(n+1)}, ..., {a(n), ..., a(2n-1)}}]. - John W. Layman, Jul 21 2000
a(n+1) = (A005248(n+1) - A001519(n))/2. - Creighton Dement, Aug 15 2004
a(n+1) = Sum_{i=0..n} Sum_{j=0..n} binomial(n-i, j)*binomial(n-j, i). - N. J. A. Sloane, Feb 20 2005
a(n) = (2/sqrt(5))*sinh(2*n*psi), where psi:=log(phi) and phi=(1+sqrt(5))/2. - Hieronymus Fischer, Apr 24 2007
a(n) = ((phi+1)^n - A001519(n))/phi with phi=(1+sqrt(5))/2. - Reinhard Zumkeller, Nov 22 2007
Row sums of triangle A135871. - Gary W. Adamson, Dec 02 2007
a(n)^2 = Sum_{k=1..n} a(2*k-1). This is a property of any sequence S(n) such that S(n) = B*S(n-1) - S(n-2) with S(0) = 0 and S(1) = 1 including {0,1,2,3,...} where B = 2. - Kenneth J Ramsey, Mar 23 2008
a(n) = 1/sqrt(5)*(phi^(2*n+2) - phi^(-2*n-2)), where phi = (1+sqrt(5))/2, the golden ratio. - Udita Katugampola (SIU), Sep 24 2008
If p[i] = i and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i<=j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n) = det(A). - Milan Janjic, May 02 2010
If p[i] = Stirling2(i,2) and if A is the Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i<=j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, May 08 2010
a(n) = F(2*n+10) mod F(2*n+5).
a(n) = 1 + a(n-1) + Sum_{i=1..n-1} a(i), with a(0)=0. - Gary W. Adamson, Feb 19 2011
a(n) is equal to the permanent of the (n-1) X (n-1) Hessenberg matrix with 3's along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
a(n), n > 1 is equal to the determinant of an (n-x) X (n-1) tridiagonal matrix with 3's in the main diagonal, 1's in the super and subdiagonals, and the rest 0's. - Gary W. Adamson, Jun 27 2011
a(n) = b such that Integral_{x=0..Pi/2} sin(n*x)/(3/2-cos(x)) dx = c + b*log(3). - Francesco Daddi, Aug 01 2011
a(n+1) = Sum_{k=0..n} A101950(n,k)*2^k. - Philippe Deléham, Feb 10 2012
G.f.: A(x) = x/(1-3*x+x^2) = G(0)/sqrt(5); where G(k)= 1 -(a^k)/(1 - b*x/(b*x - 2*(a^k)/G(k+1))), a = (7-3*sqrt(5))/2, b = 3+sqrt(5), if |x|<(3-sqrt(5))/2 = 0.3819660...; (continued fraction 3 kind, 3-step ). - Sergei N. Gladkovskii, Jun 25 2012
a(n) = 2^n*b(n;1/2) = -b(n;-1), where b(n;d), n=0,1,...,d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also Witula's et al. papers). - Roman Witula, Jul 12 2012
Product_{n>=1} (1 + 1/a(n)) = 1 + sqrt(5). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = (1/6)*(1 + sqrt(5)). - Peter Bala, Dec 23 2012
G.f.: x/(1-2*x) + x^2/(1-2*x)/(Q(0)-x) where Q(k) = 1 - x/(x*k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Feb 23 2013
G.f.: G(0)/2 - 1, where G(k) = 1 + 1/( 1 - x/(x + (1-x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013
G.f.: x*G(0)/(2-3*x), where G(k) = 1 + 1/( 1 - x*(5*k-9)/(x*(5*k-4) - 6/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 17 2013
Sum_{n>=1} 1/(a(n) + 1/a(n)) = 1. Compare with A001519, A049660 and A049670. - Peter Bala, Nov 29 2013
a(n) = U(n-1,3/2) where U(n-1,x) is Chebyshev polynomial of the second kind. - Milan Janjic, Jan 25 2015
The o.g.f. A(x) satisfies A(x) + A(-x) + 6*A(x)*A(-x) = 0. The o.g.f. for A004187 equals -A(sqrt(x))*A(-sqrt(x)). - Peter Bala, Apr 02 2015
For n > 1, a(n) = (3*F(n+1)^2 + 2*F(n-2)*F(n+1) - F(n-2)^2)/4. - J. M. Bergot, Feb 16 2016
For n > 3, a(n) = floor(MA) - 4 for n even and floor(MA) + 5 for n odd.  MA is the maximum area of a quadrilateral with lengths of sides in order L(n), L(n), F(n-3), F(n+3), with L(n)=A000032(n). The ratio of the longer diagonal to the shorter approaches 5/3. - J. M. Bergot, Feb 16 2016
a(n+1) = Sum_{j=0..n} Sum_{k=0..j} binomial(n-j,k)*binomial(j,k)*2^(j-k). - Tony Foster III, Sep 18 2017
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} C(k+i,k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = Sum_{k=1..A000041(n)} A305309(n, k), n >= 1. Also row sums of triangle A078812.- Wolfdieter Lang, May 31 2018
a(n) = H(2*n, 1, 1/2) for n > 0 where H(n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -4). - Peter Luschny, Sep 03 2019
Sum_{n>=1} 1/a(n) = A153386. - Amiram Eldar, Oct 04 2020
a(n) = A249450(n) + 2. - Leo Tavares, Oct 10 2021
a(n) = -2/(sqrt(5)*tan(2*arctan(phi^(2*n)))), where phi = A001622 is the golden ratio. - Diego Rattaggi, Nov 21 2021
a(n) = sinh(2*n*arcsinh(1/2))/sqrt(5/4). - Peter Luschny, May 21 2022
From Amiram Eldar, Dec 02 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 1 + 1/sqrt(5) (A344212).
Product_{n>=2} (1 + (-1)^n/a(n)) = (5/6) * (1 + 1/sqrt(5)). (End)
a(n) = Sum_{k>=0} Fibonacci(2*n*k)/(Lucas(2*n)^(k+1)). - Diego Rattaggi, Jan 12 2025
Sum_{n>=0} a(n)/3^n = 3. - Diego Rattaggi, Jan 20 2025

A079586 Decimal expansion of Sum_{k>=1} 1/F(k) where F(k) is the k-th Fibonacci number A000045(k).

Original entry on oeis.org

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

Offset: 1

Benoit Cloitre, Jan 26 2003

André-Jeannin proved that this constant is irrational.

This constant does not belong to the quadratic number field Q(sqrt(5)) (Bundschuh and Väänänen, 1994). - Amiram Eldar, Oct 30 2020

			3.35988566624317755317201130291892717968890513373...

Daniel Duverney, Number Theory, World Scientific, 2010, 5.22, pp.75-76.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 358.

Kenny Lau, Table of n, a(n) for n = 1..10000 (First 1000 terms computed by Joerg Arndt)
Richard André-Jeannin, Irrationalité de la somme des inverses de certaines suites récurrentes, Comptes Rendus de l'Académie des Sciences - Series I - Mathematics 308:19 (1989), pp. 539-541.
Richard André-Jeannin, Problem H-450, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 29, No. 1 (1991), p. 89; Comparable, Solution to Problem H-450 by Paul S. Bruckman, ibid., Vol. 30, No. 2 (1992), p. 191-192.
Richard André-Jeannin, Sequences of Integers Satisfying Recurrence Relations, The Fibonacci Quarterly, Vol. 29, No. 3 (1991), pp. 205-208;
Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
Paul S. Bruckman, Problem B-602, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 25, No. 3 (1987), p. 279; Fibonacci Infinite Series, Solution to Problem B-602 by C. Georghiou, ibid., Vol. 26, No. 3 (1988), pp. 281-282.
Peter Bundschuh and Keijo Väänänen, Arithmetical investigations of a certain infinite product, Compositio Mathematica, Vol. 91, No. 2 (1994), pp. 175-199.
Daniel Duverney, Irrationalité de la somme des inverses de la suite Fibonacci, Elemente der Mathematik, Vol. 52, No. 1 (1997), pp. 31-36.
William Gosper, Acceleration of Series, Artificial Intelligence Memo #304 (1974).
W. E. Greig, Sums of Fibonacci reciprocals, The Fibonacci Quarterly, Vol. 15, No. 1 (1977), pp. 46-48.
Peter Griffin, Acceleration of the Sum of Fibonacci Reciprocals, The Fibonacci Quarterly, Vol. 30, No. 2 (1992), pp. 179-181.
Sarah H. Holliday and Takao Komatsu, On the sum of reciprocal generalized Fibonacci numbers, Integers 11A (2011), Article 11. Alternate link.
A. F. Horadam, Elliptic functions and Lambert series in the summation of reciprocals in certain recurrence-generated sequences, The Fibonacci Quarterly, Vol. 26, No.2 (May-1988), pp. 98-114.
Fredrik Johansson, The reciprocal Fibonacci constant.
Paul Kinlaw, Michael Morris, and Samanthak Thiagarajan, Sums related to the Fibonacci sequence, Husson University (2021).
Tapani Matala-Aho and Marc Prévost, Quantitative irrationality for sums of reciprocals of Fibonacci and Lucas numbers, Ramanujan J., Vol. 11 (2006), pp. 249-261.
Z.W. M. Trzaska, Fibonacci Polynomials their Properties and Applications, Z. Anal. Anwend. 15 (1996), no. 3, pp. 729-746.
Eric Weisstein's World of Mathematics, Reciprocal Fibonacci Constant.
Wikipedia, Reciprocal Fibonacci constant.

Cf. A000045, A000796, A001622, A002163, A002390, A084119, A093540, A016628, A105199, A153386, A153387.

Cf. A350901, A350902, A350903, A350904.

Digits := 120: c := Pi/2 + I*arccsch(2):
Jeannin := n -> sqrt(5/4)*add(I^(1-j)/sin(j*c), j = 1..n):
evalf(Jeannin(1000)); # Peter Luschny, Nov 15 2023

digits = 105; Sqrt[5]*NSum[(-1)^n/(GoldenRatio^(2*n + 1) - (-1)^n), {n, 0, Infinity}, WorkingPrecision -> digits, NSumTerms -> digits] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Apr 09 2013 *)
First@RealDigits[Sqrt[5]/4 ((Log[5] + 2 QPolyGamma[1, 1/GoldenRatio^4] - 4 QPolyGamma[1, 1/GoldenRatio^2])/(2 Log[GoldenRatio]) + EllipticTheta[2, 0, 1/GoldenRatio^2]^2), 10, 105] (* Vladimir Reshetnikov, Nov 18 2015 *)

/* Fast computation without splitting into even and odd indices, see the Arndt reference */
lambert2(x, a, S)=
{
/* Return G(x,a) = Sum_{n>=1} a*x^n/(1-a*x^n) (generalized Lambert series)
   computed as Sum_{n=1..S} x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) )
   As series in x correct up to order S^2.
   We also have G(x,a) = Sum_{n>=1} a^n*x^n/(1-x^n) */
    return( sum(n=1,S, x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) ) ) );
}
inv_fib_sum(p=1, q=1, S)=
{
/* Return Sum_{n>=1} 1/f(n) where f(0)=0, f(1)=1, f(n) = p*f(n-1) + q*f(n-1)
   computed using generalized Lambert series.
   Must have p^2+4*q > 0 */
    my(al,be);
    \\ Note: the q here is -q in the Horadam paper.
    \\ The following numerical examples are for p=q=1:
    al=1/2*(p+sqrt(p^2+4*q));  \\ == +1.6180339887498...
    be=1/2*(p-sqrt(p^2+4*q));  \\ == -0.6180339887498...
    return( (al-be)*( 1/(al-1) + lambert2(be/al, 1/al, S) ) ); \\ == 3.3598856...
}
default(realprecision,100);
S = 1000; /* (be/al)^S == -0.381966^S == -1.05856*10^418 << 10^-100 */
inv_fib_sum(1,1,S) /* 3.3598856... */ /* Joerg Arndt, Jan 30 2011 */

suminf(k=1, 1/(fibonacci(k))) \\ Michel Marcus, Feb 19 2019

m=120; numerical_approx(sum(1/fibonacci(k) for k in (1..10*m)), digits=m) # G. C. Greubel, Feb 20 2019

Alternating series representation: 3 + Sum_{k >= 1} (-1)^(k+1)/(F(k)*F(k+1)*F(k+2)). - Peter Bala, Nov 30 2013
From Amiram Eldar, Oct 04 2020: (Start)
Equals sqrt(5) * Sum_{k>=0} (1/(phi^(2*k+1) - 1) - 2*phi^(2*k+1)/(phi^(4*(2*k+1)) - 1)), where phi is the golden ratio (A001622) (Greig, 1977).
Equals sqrt(5) * Sum_{k>=0} (-1)^k/(phi^(2*k+1) - (-1)^k) (Griffin, 1992).
Equals A153386 + A153387. (End)
From Gleb Koloskov, Sep 14 2021: (Start)
Equals 1 + c1*(c2 + 32*Integral_{x=0..infinity} f(x) dx),
where c1 = sqrt(5)/(8*log(phi)) = A002163/(8*A002390),
      c2 = 2*arctan(2)+log(5) = 2*A105199+A016628,
      phi = (1+sqrt(5))/2 = A001622,
f(x) = sin(x)*(4+cos(2*x))/((exp(Pi*x/log(phi))-1)*(2*cos(2*x)+3)*(7-2*cos(2*x))) (End)
From Amiram Eldar, Jan 27 2022: (Start)
Equals 3 + 2 * Sum_{k>=1} 1/(F(2*k-1)*F(2*k+1)*F(2*k+2)) (Bruckman, 1987).
Equals 2 + Sum_{k>=1} 1/A350901(k) (André-Jeannin, Problem H-450, 1991).
Equals lim_{n->oo} A350903(n)/(A350904(n)*A350902(n)) (André-Jeannin, 1991). (End)
Equals sqrt(5/4)*Sum_{j>=1} i^(1-j)/sin(j*c) where c = Pi/2 + i*arccsch(2). - Peter Luschny, Nov 15 2023
Equals lim_{n->oo} A203006(n)/A003266(n) (Z.W. M. Trzaska, 1996). - Raul Prisacariu, Sep 04 2024

A265288 Decimal expansion of Sum_{n >= 1} (phi - c(2*n-1)), where phi is the golden ratio (A001622), and c(n) is the n-th convergent to the continued fraction expansion of phi.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Dec 06 2015

Define the lower deviance of x > 0 by dL(x) = Sum_{n>=1} (x - c(2*n-1,x)), where c(k,x) = k-th convergent to x. The greatest lower deviance occurs when x = golden ratio, so that this constant is the absolute maximal lower deviance.
Guide to related constants (as sequences):
   x          Sum{x-c(2*n-1)}  Sum{c(2*n)-x}  Sum|c(2*n)-c(2*n-1)|
(1+sqrt(5))/2   A265288          A265289         A265290
sqrt(2)         A265291          A265292         A265293
sqrt(3)         A265294          A265295         A265296
sqrt(5)         A265297          A265298         A265299
sqrt(6)         A265300          A265301         A265302
sqrt(8)         A265303          A265304         A265305
   e            A265306          A265307         A265308

			0.75720437504600733864782526067377483...
The convergents to x are c(1) = 1, c(2) = 2, c(3) = 3/2, c(4) = 5/3, ..., so that
A265288 = (x - 1) + (x - 3/2) + (x - 8/5) + ... ;
A265289 = (2 - x) + (5/3 - x) + (13/8 - x ) + ... ;
A265290 = (2 - 1) + (5/3 - 3/2) + (13/8 - 8/5) + ...

Cf. A000045, A001227, A001622, A153386, A265289, A265290.

x := -(3 - sqrt(5))/2:
evalf(sqrt(5)*add(x^(n*(n+1)/2)/(x^n - 1), n = 1..24), 100); # Peter Bala, Aug 21 2022

x = GoldenRatio; z = 600; c = Convergents[x, z];
s1 = Sum[x - c[[2 k - 1]], {k, 1, z/2}]; N[s1, 200]
s2 = Sum[c[[2 k]] - x, {k, 1, z/2}]; N[s2, 200]
N[s1 + s2, 200]
RealDigits[s1, 10, 120][[1]]  (* A265288 *)
RealDigits[s2, 10, 120][[1]]  (* A265289 *)
RealDigits[s1 + s2, 10, 120][[1]] (* A265290 *)

Equals Sum_{k >= 1} 1/(phi^(2*k-1) * F(2*k-1)), where F(k) is the k-th Fibonacci number (A000045). - Amiram Eldar, Oct 05 2020
From Peter Bala, Aug 19 2022: (Start)
The constant equals Sum_{k >= 1} (-1)^(k+1)/F(2*k). The constant also equals (3/5)*Sum_{k >= 1} (-1)^(k+1)/(F(2*k)*F(2*k+2)*F(2*k+4)) + 11/15.
A rapidly converging series for the constant is sqrt(5) * Sum_{k >= 1} x^(k*(k+1)/2)/ (x^k - 1) at x = phi - 2 = -(3 - sqrt(5))/2. (End)

A153387 Decimal expansion of Sum_{n>=1} 1/Fibonacci(2*n-1).

Original entry on oeis.org

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

Offset: 1

Eric W. Weisstein, Dec 25 2008

Borwein et al. express the sum in terms of theta functions. - N. J. A. Sloane, May 16 2011

Duverney et al. (1997) proved that this constant is transcendental. - Amiram Eldar, Oct 30 2020

			1.8245151574069245681...

J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See pp. 202-203.

Joerg Arndt, Table of n, a(n) for n = 1..1000
Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
L. Carlitz, Problem B-110, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; An Infinite Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 469-470.
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.
Index entries for transcendental numbers

Cf. A000032, A002878, A079586, A153386, A190649.

sumpos(n=1, 1/fibonacci(2*n-1)) \\ Michel Marcus, Feb 05 2022

Equals sqrt(5)/4 * (T(b^2)^2 - T(b)^2) where T(q) = 1 + 2*Sum_{n>=1} q^(n^2) and b = 1/2*(1-sqrt(5)); see the Arndt reference and the references cited there. - Joerg Arndt, Feb 01 2014

Equals sqrt(5) * Sum_{n>=0} (-1)^n/Lucas(2*n+1) (Carlitz, 1967). - Amiram Eldar, Feb 05 2022

Definition reconciled to sequence and example by Clark Kimberling, Aug 06 2013

A158933 Decimal expansion of Sum_{n>=1} ((-1)^(n+1))/F(n) where F(n) is the n-th Fibonacci number A000045(n).

Original entry on oeis.org

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

Offset: 0

Michel Lagneau, Mar 26 2011

André-Jeannin (1989) proved that this constant is irrational, and Tachiya (2004) proved that it does not belong to the quadratic number field Q(sqrt(5)). - Amiram Eldar, Oct 30 2020

			0.2891446485706715831123055096157291669...

Richard André-Jeannin, Irrationalité de la somme des inverses de certaines suites récurrentes, Comptes Rendus de l'Académie des Sciences - Series I - Mathematics, Vol. 308, No. 19 (1989), pp. 539-541.
Yohei Tachiya, Irrationality of certain Lambert series, Tokyo Journal of Mathematics, Vol. 27, No. 1 (2004), pp. 75-85.
Eric Weisstein's World of Mathematics, Reciprocal Fibonacci Constant.

Cf. A000045, A001622, A079586.

Cf. A153386, A153387, A324007.

with(combinat, fibonacci):Digits:=100:s:=0:for n from 1 to 2000 do: a1:=fibonacci(n):s:=s+evalf(1/a1)*(-1)^(n+1):od:print(s):

digits = 95; NSum[(-1)^(n+1)*(1/Fibonacci[n]), {n, 1, Infinity}, WorkingPrecision -> digits+1, NSumTerms -> digits] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Jan 28 2014 *)

-sumalt(n=1,(-1)^n/fibonacci(n)) \\ Charles R Greathouse IV, Oct 03 2016

Equals sqrt(5) * Sum_{k>=0} (-1)^k/(phi^(2*k+1) + (-1)^k), where phi is the golden ratio (A001622). - Amiram Eldar, Oct 04 2020
Equals A153387 - A153386. - Joerg Arndt, Oct 04 2020
Equals 1 - A324007. - Amiram Eldar, Feb 09 2023

Offset corrected by Arkadiusz Wesolowski, Jun 28 2011

A256178 Expansion of exp( Sum_{n >= 1} L(2n)L(4n)x^n/n ), where L(n) = A000032(n) is a Lucas number.

Original entry on oeis.org

1, 21, 385, 6930, 124410, 2232594, 40062659, 718896255, 12900072515, 231482415780, 4153783429236, 74536619356836, 1337505365115205, 24000559953034665, 430672573790340805, 7728105768275278134, 138675231255170368494

Offset: 0

Peter Bala, Mar 18 2015

Let L(n) = A000032(n) denote the n-th Lucas number.
For a fixed positive integer k, the power series expansion of exp( Sum_{n >= 1} L(k*n)x^n/n ) has integer coefficients given by the formula F(k*n)/F(k), where F(n) = A000045(n) [Johnson, 2.22].
The power series expansion of exp( Sum_{n >= 1} L(k*n)*L(2*k*n) *x^n/n ) has integer coefficients given by ( F(k*(n + 1))*F(k*(n + 2))*F(k*(n + 3)) )/( F(k)*F(2*k)*F(3*k) )
The present sequence is the particular case k = 2. See A001655 for the case k = 1.

B. Johnson, Fibonacci Identities by Matrix Methods and Generalisation to Related Sequences
Eric Weisstein's World of Mathematics, Fibonacci Number
Index entries for linear recurrences with constant coefficients, signature (21,-56,21,-1).

Cf. A000032, A000045, A001655, A010048, A153386, A265288.

seq((1/24)*fibonacci(2*n+2)*fibonacci(2*n+4)*fibonacci(2*n+6), n = 0 .. 16);

Table[1/8 * Sum[Fibonacci[2*k + 2]*Fibonacci[6*n - 6*k + 6], {k, 0, n}], {n, 0, 17}] (* or *) RecurrenceTable[{a[n] == 21*a[n - 1] - 56*a[n - 2] + 21*a[n - 3] - a[n - 4], a[1] == 1, a[2] == 21, a[3] == 385, a[4] == 6930}, a, {n, 17}] (* Michael De Vlieger, Mar 18 2015 *)

a(n) = ( F(2*n + 2)*F(2*n + 4)*F(2*n + 6) )/( F(2)*F(4)*F(6) ).
a(n) = (1/8) * Sum_{k = 0..n} F(2*k + 2)*F(6*n - 6*k + 6).
O.g.f.: 1/( (1 - 3*x + x^2)*(1 - 18*x + x^2) ) = 1/8 * Sum_{n >= 0} F(2*n + 2)*x^n * Sum_{n >= 0} F(6*n + 6)*x^n.
O.g.f. also equals exp( Sum_{n >= 1} trace( M^(2*n) + M^(6*n) )*x^n/n ), where M is the 2X2 matrix [ 1, 1; 1, 0 ].
Recurrences: a(n) = 21*a(n-1) - 56*a(n-2) + 21*a(n-3) - a(n-4).
Also a(0) = 1 and for n >= 1, a(n) = (1/n)*Sum_{k = 1..n} L(2*k)*L(4*k)*a(n-k).
From Peter Bala, Aug 19 2022: (Start)
Sum_{n >= 0} 1/a(n) = 40/3 - 8*Sum_{n >= 1} 1/F(2*n) = 40/3 - 8*A153386.
Sum_{n >= 0} (-1)^n/a(n) = - 88/3 + 40*Sum_{n >= 1} (-1)^(n+1)/F(2*n). Cf. A265288. (End)

A360928 Decimal expansion of Sum_{i>=0} 1/(phi^(4*i+2) - 1) where phi = (1+sqrt(5))/2 is the golden ratio.

Original entry on oeis.org

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

Offset: 0

Kevin Ryde, Feb 25 2023

			0.68663856568126063939655676567056596...

Kevin Ryde, Table of n, a(n) for n = 0..10000
W. E. Greig, Sums of Fibonacci Reciprocals, The Fibonacci Quarterly, Vol. 15, No. 1, February 1977, pp. 46-48 (see equation (15)).

Cf. A001622 (phi), A153386.

RealDigits[Sum[1/(GoldenRatio^(4*i + 2) - 1), {i, 0, Infinity}], 10, 105][[1]] (* Amiram Eldar, Feb 26 2023 *)
RealDigits[(Log[(5 + 3*Sqrt[5])/2] + QPolyGamma[0, 1/2, (7 + 3*Sqrt[5])/2]) / Log[(7 - 3*Sqrt[5])/2], 10, 105][[1]] (* Vaclav Kotesovec, Feb 26 2023 *)

sumpos(i=0, 1/(((1+sqrt(5))/2)^(4*i+2) - 1)) \\ Michel Marcus, Feb 26 2023

Equals Sum_{j>=1} phi^(2*j)/(phi^(4*j) - 1) [Greig, equation (15)].

Equals A153386 / sqrt(5) [Greig, equations (13) and (14)].

A360958 Decimal expansion of Sum_{i>=1} 1/Fibonacci(3*i).

Original entry on oeis.org

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

Offset: 0

Kevin Ryde, Feb 28 2023

Sum of reciprocals of the even Fibonacci numbers, so Sum_{i>=1} 1/A014445(i)

			.66350213893302819713588109594999329...

Kevin Ryde, Table of n, a(n) for n = 0..10000

Cf. A014445, A079586, A153386, A360957.

Equals A079586 - A360957.

A371647 Decimal expansion of Sum_{k>=0} 1/Fibonacci(5^k).

Original entry on oeis.org

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

Offset: 1

Amiram Eldar, Mar 31 2024

This constant is a transcendental number (Nyblom, 2001).

			1.20001332889036987670776409546835505643055506881380...

M. A. Nyblom, A Theorem on Transcendence of Infinite Series II, Journal of Number Theory, Vol. 91, No. 1 (2001), pp. 71-80.
Index entries for transcendental numbers.

Cf. A000045, A145232, A371648.

Similar constants: A079585, A079586, A153386, A153387, A371649.

RealDigits[Sum[1/Fibonacci[5^k], {k, 0, 10}], 10, 120][[1]]

PARI
```
suminf(k = 0, 1/fibonacci(5^k))
```

Equals Sum_{k>=0} 1/A145232(k).

A079587 Continued fraction expansion of Sum_{k>=1} 1/F(k) where F(k) is the k-th Fibonacci number A000045(k).

Original entry on oeis.org

3, 2, 1, 3, 1, 1, 13, 2, 3, 3, 2, 1, 1, 6, 3, 2, 4, 362, 2, 4, 8, 6, 30, 50, 1, 6, 3, 3, 2, 7, 2, 3, 1, 3, 2, 1, 1, 1, 1, 3, 4, 1, 1, 1, 172, 3, 26, 1, 18, 8, 2, 1, 1, 2, 8, 2, 72, 2, 3, 1, 1, 1, 1, 5, 2, 33, 1, 2, 52, 1, 3, 5, 1, 11, 1, 1, 3, 13, 2, 2, 4, 2, 14, 4, 1, 1, 10, 4, 4, 3, 5, 17, 3, 4, 2, 4, 1

Offset: 0

Benoit Cloitre, Jan 26 2003

			3.35988566624....

Eric Weisstein's World of Mathematics, Reciprocal Fibonacci Constant.
Wikipedia, Reciprocal Fibonacci constant

Cf. A079586 (decimal expansion), A153386, A153387.

Offset changed by Andrew Howroyd, Jul 06 2024

cp's OEIS Frontend

A001906 F(2n) = bisection of Fibonacci sequence: a(n) = 3*a(n-1) - a(n-2).

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A079586 Decimal expansion of Sum_{k>=1} 1/F(k) where F(k) is the k-th Fibonacci number A000045(k).

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A265288 Decimal expansion of Sum_{n >= 1} (phi - c(2*n-1)), where phi is the golden ratio (A001622), and c(n) is the n-th convergent to the continued fraction expansion of phi.

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A153387 Decimal expansion of Sum_{n>=1} 1/Fibonacci(2*n-1).

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A158933 Decimal expansion of Sum_{n>=1} ((-1)^(n+1))/F(n) where F(n) is the n-th Fibonacci number A000045(n).

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A256178 Expansion of exp( Sum_{n >= 1} L(2*n)*L(4*n)*x^n/n ), where L(n) = A000032(n) is a Lucas number.

A256178 Expansion of exp( Sum_{n >= 1} L(2n)L(4n)x^n/n ), where L(n) = A000032(n) is a Lucas number.