cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A006172 a(n) = 1 + F(2*n+1) + (-1)^n*L(n).

Original entry on oeis.org

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
Offset: 0

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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)/((x-1)*(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 Greig 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 Greig paper: G_k = 1 + L_k + F_{2k-1} (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 Fibonacci-Lucas 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 Greig 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 Greig papers are a bit hard to read since non-standard notation and terminology are used. For example, the generalized Fibonacci and Lucas numbers are indicated by P with subscripts rather than by F. Similarly, Greig refers to F_{2k-1} 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 Greig paper). Let E be the translation operator so that E(f)(x) = f(x+1). Then, for example, (E^2-E-1)(L_n) = L_{n+2} - L_{n+1}-L_n =0 sequence A000032). Thus the operator E^2-E-1 annihilates the Lucas sequence. Similarly, (E^2-3E+1) annihilates the bisected Fibonacci numbers (sequence A001519) and (E-1) 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 (E-1)(E^2-3E+1)(E^2-E-1). 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/(1-x) (sequence A000012). Since the g.f. for Lucas sequence is L(x)=(2-x)/(1-x-x^2)(sequence A000032), it follows that the g.f. for the negative Lucas sequence L(-x), is (2+x)/(1+x-x^2). Similarly, since the g.f. for the bisected Fibonacci numbers F_{2k-1} 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) = (1-x)/(x^2-3x+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/(1-x) + (1-x)/(1-3x+x^2) + (2+x)/(1+x-x^2). This is a partial fraction decomposition of a rational function, and is an equivalent g.f. for the H_k, H(x) = (4-10x-x^2+9x^3-3x^4) / (x^5-5x^4+7x^3-x^2-3x+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(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. 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).

Crossrefs

Programs

  • Maple
    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: 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[n-1]), {n, 0, 30}] (* recall that L_n = F_{n+1}+ F_{n-1}. - Russell Jay Hendel, Feb 25 2015 *)
    Series[1/(1-x)+ (2+x)/(1+x-x^2)+(1-x)/(1-3x+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 *)
  • PARI
    \ps {31};
    C(x) = 1/(1-x);
    L(x) = (2+x)/(1+x-x^2);
    B(x) = (1-x)/(1-3*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 */

Formula

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

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