

A005522


a(n) = 1 + L(n) + F(2*n1) with {L(n)}_{n>=0} the Lucas numbers (A000032) and F(2*n1)_{n>=0} the bisected Fibonacci numbers (A001519).
(Formerly M2537)


3



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
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OFFSET

0,1


COMMENTS

From Russell Jay Hendel, Mar 02 2015: (Start)
The Name field was changed. The former Namefield entry was:
a(n) = Fibonacci(2n)*(1 + Fibonacci(n1))/Fibonacci(n) for n even; a(n) = 2 + Fibonacci(2n)*(1 + Fibonacci(n1))/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 = 1alpha. 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(x1))/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 oddindexed 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^2E1)(L(n)) = L(n+2)  L(n+1)  L(n) = 0 (A000032). Thus the operator E^2E1 annihilates the Lucas sequence. Similarly, (E^23E+1) annihilates the bisected Fibonacci numbers (A001519) and (E1) 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 (E1)(E^23E+1)(E^2E1). It follows that the sequence a(n) satisfies a(n)  5 a(n1) + 7 a(n2)  a(n3)  3 a(n4) + a(n5) = 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(2n1) + 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(2n1), L(n). The g.f. for the constant sequence, 1, is C(x)=1/(1x) (A000012); the g.f. for the Lucas sequence is L(x)=(2x)/(1xx^2)(A000032); and the g.f. for the bisected Fibonacci numbers F(2n1) 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/(1x) + (12x)/(13x+x^2) + (2x)/(1xx^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(n1) + p(n+1) + p(2*n1))/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


REFERENCES

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


LINKS

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. 4648.
W. E. Greig, On generalized G_{j,k} numbers, Fib. Quart., 16 (1978), 166170.
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).


FORMULA

a(n) = ((sqrt(5)+5)/10)*(((3+sqrt(5))/2)^(n1)) + ((5sqrt(5))/10)*(((3sqrt(5))/2)^(n1)) + ((1+sqrt(5))/2)^n + ((1sqrt(5))/2)^n + 1.  Tim Monahan, Jul 23 2011
a(n) = 1 + L(n) + F(2*n1), with L(n) the Lucas numbers, sequence A000032 and F(2n1) the bisected Fibonacci numbers, sequence A001519.  Russell Jay Hendel, Mar 02 2015
a(n) = 5*a(n1)  7*a(n2) + a(n3) + 3*a(n4)  a(n5).  Russell Jay Hendel, Mar 02 2015
Sum_{n>=1} (a(2*n1)/Fibonacci(4*n2)  1/phi) = A079586, where phi is the golden ratio (A001622) (Greig, 1977).  Amiram Eldar, Jan 29 2022


MAPLE

with(combinat): a:=proc(n) if n mod 2 = 0 then fibonacci(2*n)*(1+fibonacci(n1))/fibonacci(n) else 2+fibonacci(2*n)*(1+fibonacci(n1))/fibonacci(n) fi end: seq(a(n), n=1..32); # Emeric Deutsch, Apr 01 2005
A005522:=(3+9*zz**210*z**3+4*z**4)/(z1)/(z**23*z+1)/(z**2+z1); # 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(n1) + p(n+1) + p(2*n1))/sqrt(5) end: A005522 := n > FL(n):
seq(round(evalf(A005522(n), 32)), n=0..30); # Peter Luschny, Mar 09 2015


MATHEMATICA

Table[1+Fibonacci[2n1]+LucasL[n]), {n, 0, 30}] (* Russell Jay Hendel, Mar 02 2015 *)(* modified by G. C. Greubel, Jan 01 2018 *)
Series[1/(1x)+ (2x)/(1xx^2)+(12*x)/(13*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 *)


PROG

(PARI)
\ps {31};
C(x) = 1/(1x);
L(x) = (2x)/(1xx^2);
B(x) = (12*x)/(13*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
(Magma) [1 + Lucas(n) + Fibonacci(2*n  1): n in [0..30]]; // Vincenzo Librandi, Mar 08 2015
(PARI) {lucas(n) = fibonacci(n+1) + fibonacci(n1)};
for(n=0, 30, print1(1 + fibonacci(2*n1) + lucas(n), ", ")) \\ G. C. Greubel, Jan 01 2018


CROSSREFS

Cf. A000032, A000045, A001519, A001622, A006172, A079586, A254884.
Sequence in context: A128754 A353401 A196889 * A276202 A215336 A328650
Adjacent sequences: A005519 A005520 A005521 * A005523 A005524 A005525


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


EXTENSIONS

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


STATUS

approved



