|
Search: id:A160464
|
|
| |
|
| -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
(list; table; graph; listen)
|
|
|
OFFSET
|
2,2
|
|
|
COMMENT
|
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 the 'look-a-like' of the denominators in Taylor series for tan(x), i.e. A156769(n).
Some of our results are conjectures based on numerical evidence, see especially A160466.
|
|
LINKS
|
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.
J.W. Meijer and N.H.G. Baken, The Exponential Integral Distribution, Statistics and Probability Letters, Volume 5, No.3, April 1987. pp 209-211.
Eric. W. Weisstein, Dirichlet Eta Function from Wolfram MathWorld.
|
|
FORMULA
|
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(ln(n-1)/ln(2))-floor(ln(n-2)/ln(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(ES1[2*m-1,n] * z^(2*m-2), m=1..infinity), with n = 1, 2, 3, .. . This leads to
GF(z;n=1) = (2*ln(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) = ln(1/(2^(-(2*(n-1)-1-floor(ln(n-1)/ ln(2))))))/ln(2)
|
|
EXAMPLE
|
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
|
|
MAPLE
|
restart; nmax:=8; mmax:=nmax: 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) od: for n from 2 to nmax do p(n):=2^(-GCS(n-1))*(2*n-1)! od: for n from 2 to nmax do ETA(n, 1):=p(n)*c(n) end do: for n from 2 to mmax do ETA(n, n):=0 end do: 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: for n from 2 to nmax do for m from 1 to n-1 do b((((n-2)*(n-1)/2))+m+1) :=ETA(n, m) end do end do: a:=n-> b(n): seq(a(n), n=2..(nmax-1)*(nmax)/2+1);
restart; nmax:=20; m:=1; ES1row:=1-2*m; jn:=nmax: im:=nmax: for n from 1 to nmax do for j from 1 to jn do cfn1[1, j]:=1 end do: for i from 2 to im do cfn1[i, 1]:=0 end do: for j from 2 to jn do for i from 2 to im do cfn1[i, j]:=cfn1[i-1, j-1]*(j-1)^2+cfn1[i, j-1] end do end do: end do: mmax:=nmax: for m1 from 1 to mmax do M(m1-1):=2^(2*m1-2)/((2*m1-1)!) end do: for m1 from 1 to mmax do ES1[ -2*m1+1, 1]:=2*(1-2^(1-(1-2*m1)))*(-bernoulli(2*m1)/(2*m1)) od: for n from 2 to nmax do for m1 from 1 to mmax-n+1 do ES1[1-2*m1, n]:= (-1)^(n-1)*M(n-1)*sum((-1)^(k+1)*cfn1[k, n]* ES1[2*k -2*n-2*m1+1, 1], k=1..n) od: od: a:=n-> ES1[1-2*m, n]: seq(a(n), n=1..nmax-m+1);
|
|
CROSSREFS
|
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 M(n-1) sequence equals A001316(n-1)/A156769(n) (n>=1).
The q(n) sequence leads to A081729 and the 'gossip sequence' A007456.
The first right hand column equals A053644 (n>=1).
The first left hand column equals A160465.
The row sums equal A160466.
The CFN1(z, n) and the cfn1(m, n) lead to A008955.
Cf. A094665 and A160468.
Cf. The Zeta, Beta and Lambda triangles A160474, A160480 and A160487.
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Jul 06 2009: (Start)
Cf. A162440 (EG1 matrix)
(End)
Adjacent sequences: A160461 A160462 A160463 this_sequence A160465 A160466 A160467
Sequence in context: A089365 A130217 A096044 this_sequence A038316 A139311 A140749
|
|
KEYWORD
|
easy,sign,tabl
|
|
AUTHOR
|
Johannes W. Meijer (meijgia(AT)hotmail.com), May 24 2009
|
|
|
Search completed in 0.003 seconds
|