

A006172


a(n) = 1 + F(2*n+1) + (1)^n*L(n).
(Formerly M1915)


3



4, 2, 9, 10, 42, 79, 252, 582, 1645, 4106, 11070, 28459, 75348, 195898, 515073, 1344906, 3526786, 9223895, 24163596, 63236638, 165595269, 433469962, 1134942774, 2971150995, 7778845732, 20364843314, 53316562617, 139583423242, 365436006810, 956720876191, 2504732642460
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,1


COMMENTS

Previous name was: From sum of 1/F(n), where F() = Fibonacci numbers A000045.
The g.f. (2 3*x + 19*x^2  17*x^3 + 4*x^4)/((x1)*(x^2  x  1)*(x^2  3*x + 1)) conjectured by Simon Plouffe in his 1992 dissertation is probably wrong. [The correct g.f. is (4  10*x  x^2 + 9*x^3  3*x^4)/((x  1)*(x^2  x  1)*(x^2  3*x + 1)). From Bruno Berselli, Mar 09 2015]
It would be nice to have a more precise definition.  N. J. A. Sloane, May 10 2008
From Russell Jay Hendel, Feb 24 2015: (Start)
A more precise definition can be found in the Grieg paper cited in the Links section. The sequence 4, 2, 9, 10, 42, ... is given by {H_k}_{k >= 0} = {G_{k}}_{k >= 0}, with G_{k} defined by equation (1) in the Grieg paper: G_k = 1 + L_k + F_{2k1} (sequence A005522). This formula is simpler than the formulas given in the Name and Formula fields of A005522. We make 7 comments:
(Comment I) Using the defining FibonacciLucas recursion to extend the Fibonacci and Lucas numbers to negative subscripts, we have H_k = 1 + F_{2k+1} + (1)^k L_k.
(Comment II) The Name field of this sequence describes it as "From sum of 1/F(n)." This is clarified in references [2,3] of the Grieg paper. It turns out that for positive subscripts G_k = F_{2k} C_k, with C_k = Sum_{m>=0, 1/F_{kb}) with b=2^m. In other words, the G_k = C_k F_{2k} are the numerators of the sequence of C_k = G_k / F_{2k}.
(Comment III) Paradoxically, although the G_k for positive subscripts are numerators of the sequence of infinite reciprocal sums, the G_k for nonpositive subscripts have nothing to do with reciprocal sums.
(Comment IV) The Grieg papers are a bit hard to read since nonstandard notation and terminology are used. For example, the generalized Fibonacci and Lucas numbers are indicated by P with subscripts rather than by F. Similarly, Grieg refers to F_{2k1} as the "bisected" Fibonacci numbers (sequence A001519), a term which rarely appears in modern books on the Fibonacci numbers.
(Comment V) The defining recursion for the G_k 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)(x) = f(x+1). Then, for example, (E^2E1)(L_n) = L_{n+2}  L_{n+1}L_n =0 sequence A000032). Thus the operator E^2E1 annihilates the Lucas sequence. Similarly, (E^23E+1) annihilates the bisected Fibonacci numbers (sequence A001519) and (E1) annihilates the constant sequence (sequence A000012). It follows that the product of these annihilators, annihilates the sum of the sequences, which is G_k. So the defining recursion may be obtained by expanding (E1)(E^23E+1)(E^2E1). Care, however, must be taken in applying the recursion since we are applying it to nonpositive subscripts. The sequence H_k satisfies H_n  5 H_{n+1} + 7 H_{n+2}  H_{n+3}  3 H_{n+4} + H_{n+5} = 0. For example, 2  5*9 + 7*10  1*42  3*79 + 1*252 = 0.
(Comment VI) Using standard techniques, we may obtain the g.f. of the {H_k}_{k>=0} from the defining recursion, of order 5. Since H_k = 1 + F_{2k+1} + (1)^k L_k, it follows that the g.f. for the H_k is the product of the g.f. for the three summands 1, F_{2k+1},(1)^k L_k. The g.f. for the constant sequence, 1, is C(x)=1/(1x) (sequence A000012). Since the g.f. for Lucas sequence is L(x)=(2x)/(1xx^2)(sequence A000032), it follows that the g.f. for the negative Lucas sequence L(x), is (2+x)/(1+xx^2). Similarly, since the g.f. for the bisected Fibonacci numbers F_{2k1} is B(x)=(1  2x)/(1  3x + x^2) (sequence A001519), it follows that the g.f. for B_{2k+1} is 1/x (B(x)1)= B2(x) = (1x)/(x^23x+1). It follows that the g.f. for the H_k is the sum of these g.f.s: C(x) + L(x) + B2(x) = 1/(1x) + (1x)/(13x+x^2) + (2+x)/(1+xx^2). This is a partial fraction decomposition of a rational function, and is an equivalent g.f. for the H_k, H(x) = (410xx^2+9x^33x^4) / (x^55x^4+7x^3x^23x+1).
(Comment VII) We can now comment on the g.f., P(x), given by Simon Plouffe and mentioned at the beginning of this Comment section. Expanding P(x) as a series, we obtain 2 + 9x + 10x^2 + 42x^3 + ..., the g.f. of {H_k}_{k>=1}. In fact, it is easily checked that P(x) = 1/x (H(x)4). This settles the Plouffe conjecture as follows: (i) H(x) is the rational function g.f. of {H_k}_{k>=0}, (ii) P(x) is the rational function g.f. of {H_k}_{k>=1}, and (iii) C(x)+L(x)+B2(x) is the partial fraction g.f. of {H_k}_{k>=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. FL(n)/FL(n) > phi^2 for n > oo and G(n) = FL(n)  FL(n) = A254884(n) has G(n) = G(n) for all n in Z.  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

Table of n, a(n) for n=0..30.
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.
Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
Index entries for linear recurrences with constant coefficients, signature (3, 1, 7, 5, 1).


FORMULA

a(n) = 1 + F(2*n+1) + (1)^n*L(n).  Russell Jay Hendel, Feb 25 2015


MAPLE

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: A006172 := n > FL(n):
seq(round(evalf(A006172(n), 32)), n=0..30); # Peter Luschny, Mar 09 2015


MATHEMATICA

Table[1+Fibonacci[2n+1]+(1)^n*(Fibonacci[n+1]+Fibonacci[n1]), {n, 0, 30}] (* recall that L_n = F_{n+1}+ F_{n1}.  Russell Jay Hendel, Feb 25 2015 *)
Series[1/(1x)+ (2+x)/(1+xx^2)+(1x)/(13x+x^2), {x, 0, 30}] (* Russell Jay Hendel, Feb 25 2015 *)
LinearRecurrence[{3, 1, 7, 5, 1}, {4, 2, 9, 10, 42}, 31] (* or *)
CoefficientList[ Series[(4  10x  x^2 + 9x^3  3x^4)/(1  3x  x^2 + 7x^3  5x^4 + x^5), {x, 0, 30}], x] (* Robert G. Wilson v, Mar 01 2015 *)


PROG

(PARI)
\ps {31};
C(x) = 1/(1x);
L(x) = (2+x)/(1+xx^2);
B(x) = (1x)/(13*x+x^2);
H(x) = C(x)+L(x)+B(x);
Ser(H(x), x)
Vec(Ser(H(x), x))
/* Russell Jay Hendel Feb 28, 2015 */


CROSSREFS

Cf. A000032, A000045, A005522, A001519, A254884.
Sequence in context: A242049 A179398 A233295 * A171631 A052915 A130273
Adjacent sequences: A006169 A006170 A006171 * A006173 A006174 A006175


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


EXTENSIONS

a(0), a(13)a(30) from Russell Jay Hendel, Feb 25 2015
New name (using the formula from Russell Jay Hendel) from Joerg Arndt, Mar 09 2015


STATUS

approved



