cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A127672 Monic integer version of Chebyshev T-polynomials (increasing powers).

Original entry on oeis.org

2, 0, 1, -2, 0, 1, 0, -3, 0, 1, 2, 0, -4, 0, 1, 0, 5, 0, -5, 0, 1, -2, 0, 9, 0, -6, 0, 1, 0, -7, 0, 14, 0, -7, 0, 1, 2, 0, -16, 0, 20, 0, -8, 0, 1, 0, 9, 0, -30, 0, 27, 0, -9, 0, 1, -2, 0, 25, 0, -50, 0, 35, 0, -10, 0, 1, 0, -11, 0, 55, 0, -77, 0, 44, 0, -11, 0, 1, 2, 0, -36, 0, 105, 0, -112, 0, 54, 0, -12, 0, 1, 0, 13, 0, -91
Offset: 0

Views

Author

Wolfdieter Lang, Mar 07 2007

Keywords

Comments

The row polynomials R(n,x) := Sum_{m=0..n} a(n,m)*x^m have been called Chebyshev C_n(x) polynomials in the Abramowitz-Stegun handbook, p. 778, 22.5.11 (see A049310 for the reference, and note that on p. 774 the S and C polynomials have been mixed up in older printings). - Wolfdieter Lang, Jun 03 2011
This is a signed version of triangle A114525.
The unsigned column sequences (without zeros) are, for m=1..11: A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.
The row polynomials R(n,x) := Sum_{m=0..n} a(n,m)*x*m, give for n=2,3,...,floor(N/2) the positive zeros of the Chebyshev S(N-1,x)-polynomial (see A049310) in terms of its largest zero rho(N):= 2*cos(Pi/N) by putting x=rho(N). The order of the positive zeros is falling: n=1 corresponds to the largest zero rho(N) and n=floor(N/2) to the smallest positive zero. Example N=5: rho(5)=phi (golden section), R(2,phi)= phi^2-2 = phi-1, the second largest (and smallest) positive zero of S(4,x). - Wolfdieter Lang, Dec 01 2010
The row polynomial R(n,x), for n >= 1, factorizes into minimal polynomials of 2*cos(Pi/k), called C(k,x), with coefficients given in A187360, as follows.
R(n,x) = Product_{d|oddpart(n)} C(2*n/d,x)
= Product_{d|oddpart(n)} C(2^(k+1)*d,x),
with oddpart(n)=A000265(n), and 2^k is the largest power of 2 dividing n, where k=0,1,2,...
(Proof: R and C are monic, the degree on both sides coincides, and the zeros of R(n,x) appear all on the r.h.s.) - Wolfdieter Lang, Jul 31 2011 [Theorem 1B, eq. (43) in the W. Lang link. - Wolfdieter Lang, Apr 13 2018]
The zeros of the row polynomials R(n,x) are 2*cos(Pi*(2*k+1)/(2*n)), k=0,1, ..., n-1; n>=1 (from those of the Chebyshev T-polynomials). - Wolfdieter Lang, Sep 17 2011
The discriminants of the row polynomials R(n,x) are found under A193678. - Wolfdieter Lang, Aug 27 2011
The determinant of the N X N matrix M(N) with entries M(N;n,m) = R(m-1,x[n]), 1 <= n,m <= N, N>=1, and any x[n], is identical with twice the Vandermondian Det(V(N)) with matrix entries V(N;n,m) = x[n]^(m-1). This is an instance of the general theorem given in the Vein-Dale reference on p. 59. Note that R(0,x) = 2 (not 1). See also the comments from Aug 26 2013 under A049310 and from Aug 27 2013 under A000178. - Wolfdieter Lang, Aug 27 2013
This triangle a(n,m) is also used to express in the regular (2*(n+1))-gon, inscribed in a circle of radius R, the length ratio side/R, called s(2*(n+1)), as a polynomial in rho(2*(n+1)), the length ratio (smallest diagonal)/side. See the bisections ((-1)^(k-s))*A111125(k,s) and A127677 for comments and examples. - Wolfdieter Lang, Oct 05 2013
From Tom Copeland, Nov 08 2015: (Start)
These are the characteristic polynomials a_n(x) = 2*T_n(x/2) for the adjacency matrix of the Coxeter simple Lie algebra B_n, related to the Cheybshev polynomials of the first kind, T_n(x) = cos(n*q) with x = cos(q) (see p. 20 of Damianou). Given the polynomial (x - t)*(x - 1/t) = 1 - (t + 1/t)*x + x^2 = e2 - e1*x + x^2, the symmetric power sums p_n(t,1/t) = t^n + t^(-n) of the zeros of this polynomial may be expressed in terms of the elementary symmetric polynomials e1 = t + 1/t = y and e2 = t*1/t = 1 as p_n(t,1/t) = a_n(y) = F(n,-y,1,0,0,...), where F(n,b1,b2,...,bn) are the Faber polynomials of A263916.
The partial sum of the first n+1 rows given t and y = t + 1/t is PS(n,t) = Sum_{k=0..n} a_n(y) = (t^(n/2) + t^(-n/2))*(t^((n+1)/2) - t^(-(n+1)/2)) / (t^(1/2) - t^(-1/2)). (For n prime, this is related simply to the cyclotomic polynomials.)
Then a_n(y) = PS(n,t) - PS(n-1,t), and for t = e^(iq), y = 2*cos(q), and, therefore, a_n(2*cos(q)) = PS(n,e^(iq)) - PS(n-1,e^(iq)) = 2*cos(nq) = 2*T_n(cos(q)) with PS(n,e^(iq)) = 2*cos(nq/2)*sin((n+1)q/2) / sin(q/2).
(End)
R(45, x) is the famous polynomial used by Adriaan van Roomen (Adrianus Romanus) in his Ideae mathematicae from 1593 to pose four problems, solved by Viète. See, e.g., the Havil reference, pp. 69-74. - Wolfdieter Lang, Apr 28 2018
From Wolfdieter Lang, May 05 2018: (Start)
Some identities for the row polynomials R(n, x) following from the known ones for Chebyshev T-polynomials (A053120) are:
(1) R(-n, x) = R(n, x).
(2) R(n*m, x) = R(n, R(m, x)) = R(m, R(n, x)).
(3) R(2*k+1, x) = (-1)^k*x*S(2*k, sqrt(4-x^2)), k >= 0, with the S row polynomials of A049310.
(4) R(2*k, x) = R(k, x^2-2), k >= 0.
(End)
For y = z^n + z^(-n) and x = z + z^(-1), Hirzebruch notes that y(z) = R(n,x) for the row polynomial of this entry. - Tom Copeland, Nov 09 2019

Examples

			Row n=4: [2,0,-4,0,1] stands for the polynomial 2*y^0 - 4*y^2 + 1*y^4. With y^m replaced by 2^(m-1)*x^m this becomes T(4,x) = 1 - 8*x^2 + 8*x^4.
Triangle begins:
n\m   0   1   2   3   4   5   6   7   8   9  10 ...
0:    2
1:    0   1
2:   -2   0   1
3:    0  -3   0   1
4:    2   0  -4   0   1
5:    0   5   0  -5   0   1
6:   -2   0   9   0  -6   0   1
7:    0  -7   0  14   0  -7   0   1
8:    2   0 -16   0  20   0  -8   0   1
9:    0   9   0 -30   0  27   0  -9   0   1
10:  -2   0  25   0 -50   0  35   0 -10   0   1 ...
Factorization into minimal C-polynomials:
R(12,x) = R((2^2)*3,x) = C(24,x)*C(8,x) = C((2^3)*1,x)*C((2^3)*3,x). - _Wolfdieter Lang_, Jul 31 2011
		

References

  • Julian Havil, The Irrationals, A Story of the Numbers You Can't Count On, Princeton University Press, Princeton and Oxford, 2012, pp. 69-74.
  • F. Hirzebruch et al., Manifolds and Modular Forms, Vieweg 1994 pp. 77, 105.
  • R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer, 1999.

Crossrefs

Row sums (signed): A057079(n-1). Row sums (unsigned): A000032(n) (Lucas numbers). Alternating row sums: A099837(n+3).
Bisection: A127677 (even n triangle, without zero entries), ((-1)^(n-m))*A111125(n, m) (odd n triangle, without zero entries).

Programs

  • Maple
    seq(seq(coeff(2*orthopoly[T](n,x/2),x,j),j=0..n),n=0..20); # Robert Israel, Aug 04 2015
  • Mathematica
    a[n_, k_] := SeriesCoefficient[(2 - t*x)/(1 - t*x + x^2), {x, 0, n}, {t, 0, k}]; Flatten[Table[a[n, k], {n, 0, 12}, {k, 0, n}]] (* L. Edson Jeffery, Nov 02 2017 *)

Formula

a(n,0) = 0 if n is odd, a(n,0) = 2*(-1)^(n/2) if n is even, else a(n,m) = t(n,m)/2^(m-1) with t(n,m):=A053120(n,m) (coefficients of Chebyshev T-polynomials).
G.f. for m-th column (signed triangle): 2/(1+x^2) if m=0 else (x^m)*(1-x^2)/(1+x^2)^(m+1).
Riordan type matrix ((1-x^2)/(1+x^2),x/(1+x^2)) if one puts a(0,0)=1 (instead of 2).
O.g.f. for row polynomials: R(x,z) := Sum_{n>=0} R(n,x)*z^n = (2-x*z)*S(x,z), with the o.g.f. S(x,z) = 1/(1 - x*z + z^2) for the S-polynomials (see A049310).
Note that R(n,x) = R(2*n,sqrt(2+x)), n>=0 (from the o.g.f.s of both sides). - Wolfdieter Lang, Jun 03 2011
a(n,m) := 0 if n < m or n+m odd; a(n,0) = 2*(-1)^(n/2) (n even); else a(n,m) = ((-1)^((n+m)/2 + m))*n*binomial((n+m)/2-1,m-1)/m.
Recursion for n >= 2 and m >= 2: a(n,m) = a(n-1,m-1) - a(n-2,m), a(n,m) = 0 if n < m, a(2*k,1) = 0, a(2*k+1,1) = (2*k+1)*(-1)^k. In addition, for column m=0: a(2*k,0) = 2*(-1)^k, a(2*k+1,0) = 0, k>=0.
Chebyshev T(n,x) = Sum{m=0..n} a(n,m)*2^(m-1)*x^m. - Wolfdieter Lang, Jun 03 2011
R(n,x) = 2*T(n,x/2) = S(n,x) - S(n-2,x), n>=0, with Chebyshev's T- and S-polynomials, showing that they are integer and monic polynomials. - Wolfdieter Lang, Nov 08 2011
From Tom Copeland, Nov 08 2015: (Start)
a(n,x) = sqrt(2 + a(2n,x)), or 2 + a(2n,x) = a(n,x)^2, is a reflection of the relation of the Chebyshev polynomials of the first kind to the cosine and the half-angle formula, cos(q/2)^2 = (1 + cos(q))/2.
Examples: For n = 2, -2 + x^2 = sqrt(2 + 2 - 4*x^2 + x^4).
For n = 3, -3*x + x^3 = sqrt(2 - 2 + 9*x^2 - 6*x^4 + x^6).
(End)
L(x,h1,h2) = -log(1 - h1*x + h2*x^2) = Sum_{n>0} F(n,-h1,h2,0,...,0) x^n/n = h1*x + (-2*h2 + h1^2) x^2/2 + (-3*h1*h2 + h1^3) x^3/3 + ... is a log series generator of the bivariate row polynomials where T(0,0) = 0 and F(n,b1,b2,...,bn) are the Faber polynomials of A263916. exp(L(x,h1,h2)) = 1 / (1 - h1*x + h2*x^2) is the o.g.f. of A049310. - Tom Copeland, Feb 15 2016

Extensions

Name changed and table rewritten by Wolfdieter Lang, Nov 08 2011

A114525 Triangle of coefficients of the Lucas (w-)polynomials.

Original entry on oeis.org

2, 0, 1, 2, 0, 1, 0, 3, 0, 1, 2, 0, 4, 0, 1, 0, 5, 0, 5, 0, 1, 2, 0, 9, 0, 6, 0, 1, 0, 7, 0, 14, 0, 7, 0, 1, 2, 0, 16, 0, 20, 0, 8, 0, 1, 0, 9, 0, 30, 0, 27, 0, 9, 0, 1, 2, 0, 25, 0, 50, 0, 35, 0, 10, 0, 1, 0, 11, 0, 55, 0, 77, 0, 44, 0, 11, 0, 1, 2, 0, 36, 0, 105, 0, 112, 0, 54, 0, 12, 0, 1
Offset: 0

Views

Author

Eric W. Weisstein, Dec 06 2005

Keywords

Comments

Unsigned version of A108045.
The row reversed triangle is A162514. - Paolo Bonzini, Jun 23 2016

Examples

			2, x, 2 + x^2, 3*x + x^3, 2 + 4*x^2 + x^4, 5*x + 5*x^3 + x^5, ... give triangle
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  0:    2
  1:    0  1
  2:    2  0  1
  3:    0  3  0  1
  4:    2  0  4  0  1
  5:    0  5  0  5  0  1
  6:    2  0  9  0  6  0  1
  7:    0  7  0 14  0  7  0  1
  8:    2  0 16  0 20  0  8  0  1
  9:    0  9  0 30  0 27  0  9  0  1
  10:   2  0 25  0 50  0 35  0 10  0  1
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  .... reformatted by _Wolfdieter Lang_, Feb 10 2023
		

Crossrefs

Cf. A108045 (signed version).
Cf. Sequences L(n,x): A000032(x = 1), A002203 (x = 2), A006497 (x = 3), A014448 (x = 4), A087130 (x = 5), A085447 (x = 6), A086902 (x = 7), A086594 (x = 8), A087798 (x = 9), A086927 (x = 10), A001946 (x = 11), A086928 (x = 12), A088316 (x = 13), A090300 (x = 14), A090301 (x = 15), A090305 (x = 16), A090306 (x = 17), A090307 (x = 18), A090308 (x = 19), A090309 (x = 20), A090310 (x = 21), A090313 (x = 22), A090314 (x = 23), A090316 (x = 24), A087281 (x = 29), A087287 (x = 76), A089772 (x = 199).

Programs

  • Maple
    Lucas := proc(n,x)
        option remember;
        if  n=0 then
            2;
        elif n =1 then
            x ;
        else
            x*procname(n-1,x)+procname(n-2,x) ;
        end if;
    end proc:
    A114525 := proc(n,k)
        coeftayl(Lucas(n,x),x=0,k) ;
    end proc:
    seq(seq(A114525(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Aug 16 2019
  • Mathematica
    row[n_] := CoefficientList[LucasL[n, x], x];
    Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Aug 11 2018 *)

Formula

From Peter Bala, Mar 18 2015: (Start)
The Lucas polynomials L(n,x) satisfy the recurrence L(n+1,x) = x*L(n,x) + L(n-1,x) with L(0,x) = 2 and L(1,x) = x.
O.g.f.: Sum_{n >= 0} L(n,x)*t^n = (2 - x*t)/(1 - t^2 - x*t) = 2 + x*t + (x^2 + 2)*t^2 + (3*x + x^3)*t^3 + ....
L(n,x) = trace( [ x, 1; 1, 0 ]^n ).
exp( Sum_{n >= 1} L(n,x)*t^n/n ) = Sum_{n >= 0} F(n+1,x)*t^n, where F(n,x) denotes the n-th Fibonacci polynomial. (see Appendix A3 in Johnson).
exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*t^n/n ) = 1/( F(1,x)*F(2*x)*F(3,x) ) * Sum_{n >= 0} F(n+1,x)*F(n+2,x)*F(n+3,x)*t^n.
exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n.
L(n,1) = Lucas(n) = A000032(n); L(n,4) = Lucas(3*n) = A014448(n); L(n,11) = Lucas(5*n) = A001946(n); L(n,29) = Lucas(7*n) = A087281(n); L(n,76) = Lucas(9*n) = A087287(n); L(n,199) = Lucas(11*n) = A089772(n). The general result is L(n,Lucas(2*k + 1)) = Lucas((2*k + 1)*n). (End)
From Jeremy Dover, Jun 10 2016: (Start)
Read as a triangle T(n,k), n >= 0, n >= k >= 0, T(n,k) = (Binomial((n+k)/2,k) + Binomial((n+k-2)/2,k))*(1+(-1)^(n-k))/2.
T(n,k) = A046854(n-1,k-1) + A046854(n-1,k) + A046854(n-2,k) for even n+k with n+k > 0, assuming A046854(n,k) = 0 for n < 0, k < 0, k > n.
T(n,k) is the number of binary strings of length n with exactly k pairs of consecutive 0's and no pair of consecutive 1's, where the first and last bits are considered consecutive. (End)
From Peter Bala, Sep 03 2022: (Start)
L(n,x) = 2*(i)^n*T(n,-i*x/2), where i = sqrt(-1) and T(n,x) is the n-th Chebyshev polynomial of the first kind.
d/dx(L(n,x)) = n*F(n,x), where F(n,x) denotes the n-th Fibonacci polynomial.
Let P_n(x,y) = (L(n,x) - L(n,y))/(x - y). Then {P_n(x,y): n >= 1} is a fourth-order linear divisibility sequence of polynomials in the ring Z[x,y]: if m divides n in Z then P_m(x,y) divides P_n(x,y) in Z[x,y].
P_n(1,1) = A045925(n); P_n(1,4) = A273622; P_n(2,2) = A093967(n).
L(2*n,x)^2 - L(2*n-1,x)*L(2*n+1,x) = x^2 + 4 for n >= 1.
Sum_{n >= 1} L(2*n,x)/( L(2*n-1,x) * L(2*n+1,x) ) = 1/x^2 and
Sum_{n >= 1} (-1)^(n+1)/( L(2*n,x) + x^2/L(2*n,x) ) = 1/(x^2 + 4), both valid for all nonzero real x. (End)
From Peter Bala, Nov 18 2022: (Start)
L(n,x) = Sum_{k = 0..floor(n/2)} (n/(n-k))*binomial(n-k,k)*x^(n-2*k) for n >= 1.
For odd m, L(n, L(m,x)) = L(n*m, x).
For integral x, the sequence {u(n)} := {L(n,x)} satisfies the Gauss congruences: u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all positive integers m and r and all primes p.
Let p be an odd prime and let 0 <= k <= p - 1. Let alpha_k = the p-adic limit_{n -> oo} L(p^n,k). Then alpha_k is a well-defined p-adic-integer and the polynomial L(p,x) - x of degree p factorizes as L(p,x) - x = Product_{k = 0..p-1} (x - alpha_k). For example, L(5,x) - x = x^5 + 5*x^3 + 4*x = x*(x - A269591)*(x - A210850)*(x - A210851)*(x - A269592) in the ring of 5-adic integers. (End)
The formula for L(n,x) given in the first line of the preceding section, with L(0, x) = 2, is rewritten L(n, x) = Sum_{k = 0..floor(n/2)} A034807(n, k)*x^(n - 2*k). See the formula by Alexander Elkins in A034807. - Wolfdieter Lang, Feb 10 2023

A108044 Triangle read by rows: right half of Pascal's triangle (A007318) interspersed with 0's.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 3, 0, 1, 6, 0, 4, 0, 1, 0, 10, 0, 5, 0, 1, 20, 0, 15, 0, 6, 0, 1, 0, 35, 0, 21, 0, 7, 0, 1, 70, 0, 56, 0, 28, 0, 8, 0, 1, 0, 126, 0, 84, 0, 36, 0, 9, 0, 1, 252, 0, 210, 0, 120, 0, 45, 0, 10, 0, 1, 0, 462, 0, 330, 0, 165, 0, 55, 0, 11, 0, 1, 924, 0, 792, 0, 495, 0, 220, 0, 66
Offset: 0

Views

Author

N. J. A. Sloane, Jun 02 2005

Keywords

Comments

Column k has e.g.f. Bessel_I(k,2x). - Paul Barry, Mar 10 2010
T(n,k) is the number of binary sequences of length n in which the number of ones minus the number of zeros is k. If 2 divides(n+k), such a sequence will have (n+k)/2 ones and (n-k)/2 zeros. Since there are C(n,(n+k)/2) ways to choose the sequence entries that get a one, T(n,k)=binomial(n,(n+k)/2) whenever (n+k) is even and T(n,k)= 0 otherwise. See the example below in the example section. - Dennis P. Walsh, Apr 11 2012
T(n,k) is the number of walks on the semi-infinite integer line with n steps that end at k. The walks start at 0, move at each step either one to the left or one to the right, and never enter the region of negative k. [Walks with impenetrable wall at -1/2. Dyck excursions of n steps that end at level k.] The variant without the restriction of negative positions is A053121. - R. J. Mathar, Nov 02 2023

Examples

			Triangle begins:
  1
  0 1
  2 0 1
  0 3 0 1
  6 0 4 0 1
  0 10 0 5 0 1
  20 0 15 0 6 0 1
From _Paul Barry_, Mar 10 2010: (Start)
Production matrix is
  0, 1,
  2, 0, 1,
  0, 1, 0, 1,
  0, 0, 1, 0, 1,
  0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 0, 1, 0, 1 (End)
T(5,1)=10 since there are 10 binary sequences of length 5 in which the number of ones exceed the number of zeros by exactly 1, namely, 00111, 01011, 01101, 01110, 10011, 10101, 10110, 11001, 11010, and 11100. Also, T(5,2)=0 since there are no binary sequences in which the number of ones exceed the number of zeros by exactly 2. - _Dennis P. Walsh_, Apr 11 2012
		

Crossrefs

Cf. A007318, A108045 (matrix inverse),
Cf. A204293, A357136, A000984 (column 0), A001700 (column 1), A001791 (column 2), A002054 (column 3)

Programs

  • Haskell
    import Data.List (intersperse)
    a108044 n k = a108044_tabl !! n !! k
    a108044_row n = a108044_tabl !! n
    a108044_tabl = zipWith drop [0..] $ map (intersperse 0) a007318_tabl
    -- Reinhard Zumkeller, May 18 2013
  • Maple
    T:=proc(n,k) if n+k mod 2 = 0 then binomial(n,(n+k)/2) else 0 fi end: for n from 0 to 13 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form; Emeric Deutsch, Jun 05 2005
  • Mathematica
    b[n_,k_]:=If[EvenQ[n+k],Binomial[n,(n+k)/2],0]; Flatten[Table[b[n,k],{n,0,20},{k,0,n}]] (* Harvey P. Dale, May 05 2013 *)

Formula

Each entry is the sum of those in the previous row that are to its left and to its right.
Riordan array (1/sqrt(1-4*x^2), (1-sqrt(1-4*x^2))/(2*x)).
T(n, k) = binomial(n, (n+k)/2) if n+k is even, T(n, k)=0 if n+k is odd. G.f.=f/(1-tg), where f=1/sqrt(1-4x^2) and g=(1-sqrt(1-4x^2))/(2x). - Emeric Deutsch, Jun 05 2005
From Peter Bala, Jun 29 2015: (Start)
Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = ( 1 - sqrt(1 - 4*x) )/(2*x) and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan).
T(n,k) = [x^(n-k)] f(x)^n with f(x) = 1 + x^2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ).
The inverse array is A108045 (a hitting time array with h(x) = x/(1 + x^2)). (End)

Extensions

More terms from Emeric Deutsch and Christian G. Bower, Jun 05 2005
Showing 1-3 of 3 results.