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%I M1915 #84 Apr 13 2022 13:25:18
%S 4,2,9,10,42,79,252,582,1645,4106,11070,28459,75348,195898,515073,
%T 1344906,3526786,9223895,24163596,63236638,165595269,433469962,
%U 1134942774,2971150995,7778845732,20364843314,53316562617,139583423242,365436006810,956720876191,2504732642460
%N a(n) = 1 + F(2*n+1) + (-1)^n*L(n).
%C Previous name was: From sum of 1/F(n), where F() = Fibonacci numbers A000045.
%C 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]
%C It would be nice to have a more precise definition. - _N. J. A. Sloane_, May 10 2008
%C From _Russell Jay Hendel_, Feb 24 2015: (Start)
%C 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:
%C (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.
%C (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}.
%C (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.
%C (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.
%C (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.
%C (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).
%C (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)
%C 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
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H W. E. Greig, <a href="http://www.fq.math.ca/Scanned/16-2/greig.pdf">On generalized G_{j,k} numbers</a>, Fib. Quart., 16 (1978), 166-170.
%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3, 1, -7, 5, -1).
%F a(n) = 1 + F(2*n+1) + (-1)^n*L(n). - _Russell Jay Hendel_, Feb 25 2015
%p FL := proc(n) local a,p; a := (1+sqrt(5))/2; p := m -> a^m - (-a)^(-m);
%p 1 + (p(n-1) + p(n+1) + p(2*n-1))/sqrt(5) end: A006172 := n -> FL(-n):
%p seq(round(evalf(A006172(n),32)), n=0..30); # _Peter Luschny_, Mar 09 2015
%t 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 *)
%t 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 *)
%t LinearRecurrence[{3, 1, -7, 5, -1}, {4, 2, 9, 10, 42}, 31] (* or *)
%t 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 *)
%o (PARI)
%o \ps {31};
%o C(x) = 1/(1-x);
%o L(x) = (2+x)/(1+x-x^2);
%o B(x) = (1-x)/(1-3*x+x^2);
%o H(x) = C(x)+L(x)+B(x);
%o Ser(H(x),x)
%o Vec(Ser(H(x),x))
%o /* _Russell Jay Hendel_ Feb 28, 2015 */
%Y Cf. A000032, A000045, A005522, A001519, A254884.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_
%E a(0), a(13)-a(30) from _Russell Jay Hendel_, Feb 25 2015
%E New name (using the formula from _Russell Jay Hendel_) from _Joerg Arndt_, Mar 09 2015
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