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%I #34 Apr 15 2021 21:48:08
%S -1,-11,2,-114,29,-2,-3963,1156,-122,4,-104745,32863,-4206,222,-4,
%T -3926745,1287813,-184279,12198,-366,4,-198491580,67029582,-10317484,
%U 781981,-30132,562,-4
%N The Eta triangle.
%C The ES1 matrix coefficients are defined by ES1[2*m-1,n] = 2^(2*m-1) * int(y^(2*m-1)/(cosh(y))^(2*n),y=0..infinity)/(2*m-1)! for m = 1, 2, 3, .. and n = 1, 2, 3 .. .
%C This definition leads to ES1[2*m-1,n=1] = 2*eta(2*m-1) and the recurrence relation ES1[2*m-1,n] = ((2*n-2)/(2*n-1))*(ES1[2*m-1,n-1] - ES1[2*m-3,n-1]/(n-1)^2) which we used to extend our definition of the ES1 matrix coefficients to m = 0, -1, -2, .. . We discovered that ES1[ -1,n] = 0.5 for n = 1, 2, .. . As usual eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function.
%C The coefficients in the columns of the ES1 matrix, for m = 1, 2, 3, .. , and n = 2, 3, 4 .. , can be generated with the polynomials GF(z,n) for which we found the following general expression GF(z;n) = ((-1)^(n-1)*r(n)*CFN1(z,n)*GF(z;n=1) + ETA(z,n))/p(n).
%C The CFN1(z,n) polynomials depend on the central factorial numbers A008955.
%C The ETA(z,n) are the Eta polynomials which lead to the Eta triangle.
%C The zero patterns of the Eta polynomials resemble a UFO. These patterns resemble those of the Zeta, Beta and Lambda polynomials, see A160474, A160480 and A160487.
%C The first Maple algorithm generates the coefficients of the Eta triangle. The second Maple algorithm generates the ES1[2*m-1,n] coefficients for m= 0, -1, -2, -3, .. .
%C The M(n) sequence, see the second Maple algorithm, leads to Gould's sequence A001316 and a sequence that resembles the denominators of the Taylor series for tan(x), A156769(n).
%C Some of our results are conjectures based on numerical evidence, see especially A160466.
%D Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 23, pp. 811-812.
%H Johannes W. Meijer, The zeros of the Eta, Zeta, Beta and Lambda polynomials, <a href="/A160464/a160464.jpg">jpg</a> and <a href="/A160464/a160464.pdf">pdf</a>, Mar 03 2013.
%H J. W. Meijer and N.H.G. Baken, <a href="http://dx.doi.org/10.1016/0167-7152(87)90041-1">The Exponential Integral Distribution</a>, Statistics and Probability Letters, Volume 5, No.3, April 1987. pp 209-211.
%H Eric. W. Weisstein, <a href="http://mathworld.wolfram.com/DirichletEtaFunction.html">Dirichlet Eta Function</a>, Wolfram MathWorld.
%F We discovered an interesting relation between the Eta triangle coefficients ETA(n,m) = q(n)*((-1)*ETA(n-1,m-1)+(n-1)^2*ETA(n-1,m)), for n = 3, 4, ... and m = 2, 3, ... , with
%F q(n) = 1 + (-1)^(n-3)*(floor(log(n-1)/log(2)) - floor(log(n-2)/log(2))) for n = 3, 4, ....
%F See A160465 for ETA(n,m=1) and furthermore ETA(n,n) = 0 for n = 2, 3, ....
%F The generating functions GF(z;n) of the coefficients in the matrix columns are defined by
%F GF(z;n) = sum_{m>=1} ES1[2*m-1,n] * z^(2*m-2), with n = 1, 2, 3, .... This leads to
%F GF(z;n=1) = (2*log(2) - Psi(z) - Psi(-z) + Psi(1/2*z) + Psi(-1/2*z)); Psi(z) is the digamma-function.
%F GF(z;n) = ((2*n-2)/(2*n-1)-2*z^2/((n-1)*(2*n-1)))*GF(z;n-1)-1/((n-1)*(2*n-1)).
%F We found for GF(z;n), for n = 2, 3, ..., the following general expression:
%F GF(z;n) = ((-1)^(n-1)*r(n)*CFN1(z,n)*GF(z;n=1) + ETA(z,n) )/p(n) with
%F r(n) = 2^floor(log(n-1)/log(2)+1) and
%F p(n) = 2^(-GCS(n))*(2*n-1)! with
%F GCS(n) = log(1/(2^(-(2*(n-1)-1-floor(log(n-1)/ log(2))))))/log(2).
%e The first few rows of the triangle ETA(n,m) with n=2,3,.. and m=1,2,... are
%e [ -1]
%e [ -11, 2]
%e [ -114, 29, -2]
%e [ -3963, 1156, -122, 4].
%e The first few ETA(z,n) polynomials are
%e ETA(z,n=2) = -1;
%e ETA(z,n=3) = -11+2*z^2;
%e ETA(z,n=4) = -114 + 29*z^2 - 2*z^4.
%e The first few CFN1(z,n) polynomials are
%e CFN1(z,n=2) = (z^2-1);
%e CFN1(z,n=3) = (z^4 - 5*z^2 + 4);
%e CFN1(z,n=4) = (z^6 - 14*z^4 + 49*z^2 - 36).
%e The first few generating functions GF(z;n) are:
%e GF(z;n=2) = ((-1)*2*(z^2 - 1)*GF(z;n=1) + (- 1))/3;
%e GF(z;n=3) = (4*(z^4 - 5*z^2+4) *GF(z;n=1) + (-11 + 2*z^2))/30;
%e GF(z;n=4) = ((-1)*4*(z^6 - 14*z^4 + 49*z^2 - 36)*GF(z;n=1) + (-114 + 29*z^2 - 2*z^4))/315.
%p nmax:=8; c(2 ):= -1/3: for n from 3 to nmax do c(n) := (2*n-2)*c(n-1)/(2*n-1)-1/((n-1)*(2*n-1)) end do: for n from 2 to nmax do GCS(n-1) := ln(1/(2^(-(2*(n-1)-1-floor(ln(n-1)/ ln(2))))))/ln(2); p(n) := 2^(-GCS(n-1))*(2*n-1)!; ETA(n, 1) := p(n)*c(n); ETA(n, n) := 0 end do: mmax:=nmax: for m from 2 to mmax do for n from m+1 to nmax do q(n) := (1+(-1)^(n-3)*(floor(ln(n-1)/ln(2)) - floor(ln(n-2)/ln(2)))): ETA(n, m) := q(n)*((-1)*ETA(n-1, m-1)+(n-1)^2*ETA(n-1, m)) end do end do: seq(seq(ETA(n,m), m=1..n-1), n=2..nmax);
%p # End first program.
%p nmax1:=20; m:=1; ES1row:=1-2*m; with (combinat): cfn1 := proc(n, k): sum((-1)^j*stirling1(n+1, n+1-k+j) * stirling1(n+1, n+1-k-j), j=-k..k) end proc: mmax1:=nmax1: for m1 from 1 to mmax1 do M(m1-1) := 2^(2*m1-2)/((2*m1-1)!); ES1[-2*m1+1,1] := 2*(1-2^(1-(1-2*m1)))*(-bernoulli(2*m1)/(2*m1)) od: for n from 2 to nmax1 do for m1 from 1 to mmax1-n+1 do ES1[1-2*m1, n] := (-1)^(n-1)*M(n-1)*sum((-1)^(k+1)*cfn1(n-1,k-1)* ES1[2*k-2*n-2*m1+1, 1], k=1..n) od: od: seq(ES1[1-2*m, n], n=1..nmax1-m+1);
%p # End second program.
%Y The r(n) sequence equals A062383 (n>=1).
%Y The p(n) sequence equals A160473(n) (n>=2).
%Y The GCS(n) sequence equals the Geometric Connell sequence A049039(n).
%Y The M(n-1) sequence equals A001316(n-1)/A156769(n) (n>=1).
%Y The q(n) sequence leads to A081729 and the 'gossip sequence' A007456.
%Y The first right hand column equals A053644 (n>=1).
%Y The first left hand column equals A160465.
%Y The row sums equal A160466.
%Y The CFN1(z, n) and the cfn1(n, k) lead to A008955.
%Y Cf. A094665 and A160468.
%Y Cf. the Zeta, Beta and Lambda triangles A160474, A160480 and A160487.
%Y Cf. A162440 (EG1 matrix).
%K easy,sign,tabl,uned
%O 2,2
%A _Johannes W. Meijer_, May 24 2009
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