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A006172
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a(n) = 1 + F(2*n+1) + (-1)^n*L(n).
(Formerly M1915)
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3
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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
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OFFSET
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0,1
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COMMENTS
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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
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_{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 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 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, Grieg 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 Grieg 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
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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MAPLE
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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):
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MATHEMATICA
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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 *)
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PROG
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(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))
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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