|
%I #34 Apr 02 2023 20:28:58
%S 1,2,6,1,24,6,120,36,1,720,240,12,5040,1800,120,1,40320,15120,1200,20,
%T 362880,141120,12600,300,1,3628800,1451520,141120,4200,30,39916800,
%U 16329600,1693440,58800,630,1,479001600,199584000,21772800,846720,11760,42
%N Array of coefficients of numerator polynomials (divided by x) of the n-th approximation of the continued fraction x/(1+x/(2+x/(3+...
%C The row length sequence of this array is 1 + floor((n-1)/2) = A008619(n-1), n >= 1.
%C The array of denominators is found under A084950.
%C The continued fraction 0 + K_{k=1..infinity}(x/k) = x/(1+x/(2+x/(3+... has n-th approximation P(n,x)/Q(n,x). These polynomials satisfy the recurrence q(n,x) = n*q(n-1,x) + x*q(n-2,x), for q replaced by P or Q with inputs P(-1,x) = 1, P(0,x) = 0 and Q(-1,x) = 0 and Q(0,1) = 1. The present array provides the coefficients for Phat(n,x) := P(n,x)/x = sum(a(n,m)*x^m,m=0..floor((n-1)/2)), n >= 1. The recurrence is that of q(n,x) and the inputs are Phat(-1,x) = 1/x and Phat(0,x) =0. For the Q(n,x) coefficients see the companion array A084950. The solution with input q(-1,x) = a and q(0,x) = b is then, due to linearity, q(a,b;n,x) = a*x*Phat(n,x) + b*Q(n,x). The motivation to consider the q(n,x) recurrence stems from e-mails from _Gary Detlefs_, who considered integer x and various inputs and gave explicit formulas.
%C This array coincides with the SW-NE diagonals of the coefficient array |A066667| or A105278 (taken with offset [0,0]) of the generalized Laguerre polynomials n!*L(1,n,x) (parameter alpha = 1).
%C The entries a(n,m) have a combinatorial interpretation in terms of certain so-called labeled Morse code polynomials using dots (length 1) and dashes (of length 2). a(n,m) is the number of possibilities to decorate the n-1 positions 2,...,n with m dashes, m from {0,1,...,floor((n-1)/2)}, and n-1-2*m dots. A dot at position k has a label k and each dash between two neighboring positions has a label x. a(n,m) is the sum of these labeled Morse codes with m dashes after the label x^m has been divided out. E.g., a(6,2) = 6 + 4 + 2 = 12 from the 3 codes: dash dash dot, dash dot dash,and dot dash dash, or (23)(45)6, (23)4(56) and 2(34)(56), and labels (which are in general multiplicative) 6*x^2, 4*x^2 and 2*x^2, respectively.
%C For general Morse code polynomials (Euler's continuants) see the Graham et al. reference given in A221915, p. 302. - _Wolfdieter Lang_, Feb 28 2013
%C Row sums Phat(n,1) = A001053(n+1), n >= 1. Alternating row sums Phat(n,-1) = A058798(n), n >= 1.
%C From _Wolfdieter Lang_, Mar 06 2013 (Start)
%C The recurrence for q(n,x) given above, can be transformed to the one of Bessel functions given in Abramowitz-Stegun (see A103921 for the reference) in the first line of eq. 9.1.27 on p. 361 via i^n*q(n,x)/sqrt(x)^n = C(n+1,-i*2*sqrt(x)) with the imaginary unit i, where C can stand for BesselJ or BesselY. In order to fix the two inputs for the Q or Phat polynomials (given above) one uses a linear combination of these two independent solutions. The Wronskian eq. 9.1.16, p. 360, is used to simplify the coefficients. One can also use an alternative version based on eqs. 9.6.3 and 9.6.5, p. 375, to trade the J and Y polynomials for I and K.
%C This produces the two explicit formulas given below, and also the two versions given for Q in A084950.
%C (End)
%C For large order n the behavior of the row polynomials Phat(n,x) (see above) is known from the one of Bessel functions. See a comment on asymptotics under A084950. This leads then to the limit for Phat(n,x)/n! given in the formula section. The limit for the continued fraction mentioned in the name and above is also found in this comment on A084950. - _Wolfdieter Lang_, Mar 08 2013
%C This is the unsigned Lah triangle read by ascending antidiagonals. Conversely, reading the given triangle beginning at the left in descending steps yields a row of the unsigned Lah triangle. This can be verified immediately by means of the explicit formulas. For example, [T(5,0), T(6,1), T(7,2), T(8,3), T(9,4)] is row 5 of A105278. - _Peter Luschny_, Dec 07 2019
%F Recurrence (short version): a(n,m) = n*a(n-1,m) + a(n-2,m-1), n>=2, a(1,1) =1, a(n,-1) = 0, a(n,m) = 0 if n < 2*m+1. From the recurrence for the Phat(n,x) polynomials given in a comment above.
%F Recurrence (long version): a(n,m) = 2*(n-1-m)*a(n-1,m) + a(n-2,m-1) - (n-1-m)*(n-2-m)*a(n-2,m), n >= 1, a(1,0) = 1, a(n,-1) = 0, a(n,m) = 0 if n < 2*m + 1. From the recurrence for the unsigned generalized Laguerre polynomial with parameter alpha = 1. This recurrence can be simplified to the preceding short version, because the following explicit form follows from the one for the generalized Laguerre coefficients (which, in turn, derives from the Rodrigues formula and the Leibniz rule). This proves the relation a(n,m) = |Lhat(1,n-1-m,m)|, with the coefficients |Lhat(1,n,m)| = |A066667(n,m)| of the unsigned n!*L(1,n,x) Laguerre polynomials (parameter alpha = 1).
%F a(n,m) = binomial(n-1-m,m)*(n-m)!/(m+1)!, n >= 1, 0 <= m <= floor((n-1)/2).
%F For the e.g.f.s of the column sequences see A105278 (here with different offset, which could be obtained by integration).
%F E.g.f. for row polynomials gPhat(z,x) := Sum_{z>=0} Phat(n,x)*z^n = Pi*(BesselJ(1, 2*i*sqrt(x)*sqrt(1-z))*BesselY(1, 2*i*sqrt(x)) - BesselY(1, (2*i)*sqrt(x)*sqrt(1-z))*BesselJ(1, 2*i*sqrt(x)))/sqrt(1-z) with Bessel functions and the imaginary unit i = sqrt(-1). Phat(0,x) = 0.
%F From _Wolfdieter Lang_, Mar 06 2013 (Start)
%F For the row polynomials one finds Phat(n,x) = Pi*(z/2)^n*(BesselY(1,z)* BesselJ(n+1,z) - BesselJ(1,z)*BesselY(n+1,z)) where z := -i*2*sqrt(x) and the i is the imaginary unit. An alternative form is Phat(n,x) = 2*(w/2)^n*(BesselI(1,w)*BesselK(n+1,w) + BesselK(1,w)*BesselI(n+1,w)*(-1)^(n+1)), n >= 1, where w := -2*sqrt(x). See a comment above for the derivation. (End)
%F Limit_{n -> oo} Phat(n,x)/n! = BesselI(1,2*sqrt(x))/sqrt(x). See a comment above. - _Wolfdieter Lang_, Mar 08 2013
%e The irregular triangle a(n,m) begins:
%e n\m 0 1 2 3 4 5 6
%e 1: 1
%e 2: 2
%e 3: 6 1
%e 4: 24 6
%e 5: 120 36 1
%e 6: 720 240 12
%e 7: 5040 1800 120 1
%e 8: 40320 15120 1200 20
%e 9: 362880 141120 12600 300 1
%e 10: 3628800 1451520 141120 4200 30
%e 11: 39916800 16329600 1693440 58800 63 1
%e 12: 479001600 19958400 21772800 846720 11760 42
%e 13: 6227020800 2634508800 299376000 12700800 211680 1176 1
%e ...
%e Recurrence (short version): a(6,1) = 6*36 + 24 = 240.
%e Recurrence (long version): a(6,1) = 2*4*36 + 24 - 4*3*6 = 240.
%e a(6,1) = binomial(4,1)*5!/2! = 4*3*4*5 = 240.
%t row[n_] := x/ContinuedFractionK[x, i, {i, 0, n}] // Simplify // Together // Numerator // CoefficientList[#, x]& // Rest;
%t row /@ Range[12] // Flatten (* _Jean-François Alcover_, Oct 28 2019 *)
%Y Cf. |A066667|, A105278, A084950, A221915.
%K nonn,tabf,easy
%O 1,2
%A _Wolfdieter Lang_, Feb 23 2013
|
