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A160464
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The Eta triangle.
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22
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-1, -11, 2, -114, 29, -2, -3963, 1156, -122, 4, -104745, 32863, -4206, 222, -4, -3926745, 1287813, -184279, 12198, -366, 4, -198491580, 67029582, -10317484, 781981, -30132, 562, -4
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OFFSET
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2,2
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COMMENTS
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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 .. .
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.
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).
The CFN1(z,n) polynomials depend on the central factorial numbers A008955.
The ETA(z,n) are the Eta polynomials which lead to the Eta triangle.
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.
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, .. .
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).
Some of our results are conjectures based on numerical evidence, see especially A160466.
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REFERENCES
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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.
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 23, pp. 811-812.
Johannes W. Meijer, The zeros of the Eta, Zeta, Beta and Lambda polynomials, jpg and pdf, Mar 03 2013.
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FORMULA
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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
q(n) = 1 + (-1)^(n-3)*(floor(log(n-1)/log(2)) - floor(log(n-2)/log(2))) for n = 3, 4, ....
See A160465 for ETA(n,m=1) and furthermore ETA(n,n) = 0 for n = 2, 3, ....
The generating functions GF(z;n) of the coefficients in the matrix columns are defined by
GF(z;n) = sum_{m>=1} ES1[2*m-1,n] * z^(2*m-2), with n = 1, 2, 3, .... This leads to
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.
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)).
We found for GF(z;n), for n = 2, 3, ..., the following general expression:
GF(z;n) = ((-1)^(n-1)*r(n)*CFN1(z,n)*GF(z;n=1) + ETA(z,n) )/p(n) with
r(n) = 2^floor(log(n-1)/log(2)+1) and
p(n) = 2^(-GCS(n))*(2*n-1)! with
GCS(n) = log(1/(2^(-(2*(n-1)-1-floor(log(n-1)/ log(2))))))/log(2).
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EXAMPLE
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The first few rows of the triangle ETA(n,m) with n=2,3,.. and m=1,2,... are
[ -1]
[ -11, 2]
[ -114, 29, -2]
[ -3963, 1156, -122, 4].
The first few ETA(z,n) polynomials are
ETA(z,n=2) = -1;
ETA(z,n=3) = -11+2*z^2;
ETA(z,n=4) = -114 + 29*z^2 - 2*z^4.
The first few CFN1(z,n) polynomials are
CFN1(z,n=2) = (z^2-1);
CFN1(z,n=3) = (z^4 - 5*z^2 + 4);
CFN1(z,n=4) = (z^6 - 14*z^4 + 49*z^2 - 36).
The first few generating functions GF(z;n) are:
GF(z;n=2) = ((-1)*2*(z^2 - 1)*GF(z;n=1) + (- 1))/3;
GF(z;n=3) = (4*(z^4 - 5*z^2+4) *GF(z;n=1) + (-11 + 2*z^2))/30;
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.
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MAPLE
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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);
# End first program.
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);
# End second program.
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CROSSREFS
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The r(n) sequence equals A062383 (n>=1).
The p(n) sequence equals A160473(n) (n>=2).
The GCS(n) sequence equals the Geometric Connell sequence A049039(n).
The first right hand column equals A053644 (n>=1).
The first left hand column equals A160465.
The CFN1(z, n) and the cfn1(n, k) lead to A008955.
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KEYWORD
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AUTHOR
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STATUS
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approved
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