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.

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A053120 Triangle of coefficients of Chebyshev's T(n,x) polynomials (powers of x in increasing order).

Original entry on oeis.org

1, 0, 1, -1, 0, 2, 0, -3, 0, 4, 1, 0, -8, 0, 8, 0, 5, 0, -20, 0, 16, -1, 0, 18, 0, -48, 0, 32, 0, -7, 0, 56, 0, -112, 0, 64, 1, 0, -32, 0, 160, 0, -256, 0, 128, 0, 9, 0, -120, 0, 432, 0, -576, 0, 256, -1, 0, 50, 0, -400, 0, 1120, 0, -1280, 0, 512, 0, -11, 0, 220, 0, -1232, 0, 2816, 0, -2816, 0, 1024
Offset: 0

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Comments

Row sums (signed triangle): A000012 (powers of 1). Row sums (unsigned triangle): A001333(n).
From Wolfdieter Lang, Oct 21 2013: (Start)
The row polynomials T(n,x) equal (S(n,2*x) - S(n-2,2*x))/2, n >= 0, with the row polynomials S from A049310, with S(-1,x) = 0, and S(-2,x) = -1.
The zeros of T(n,x) are x(n,k) = cos((2*k+1)*Pi/(2*n)), k = 0, 1, ..., n-1, n >= 1. (End)
From Wolfdieter Lang, Jan 03 2020 and Paul Weisenhorn: (Start)
The (sub)diagonal sequences {D_{2*k}(m)}{m >= 0}, for k >= 0, have o.g.f. GD{2*k}(x) = (-1)^k*(1-x)/(1-2*x)^(k+1), for k >= 0, and GD_{2*k+1}(x) = 0, for k >= 0. This follows from their o.g.f. GGD(z, x) := Sum_{k>=0} GD_k(x)*z^n which is obtained from the o.g.f. of the T-triangle GT(z, x) = (1-x*z)/(1 - 2*x + z^2) (see the formula section) by GGD(z, x) = GT(z, x/z).
The explicit form is then D_{2*k}(m) = (-1)^k, for m = 0, and
(-1)^k*(2*k+m)*2^(m-1)*risefac(k+1, m-1)/m!, for m >= 1, with the rising factorial risefac(x, n). (End)

Examples

			The triangle a(n,m) begins:
n\m  0  1   2    3     4    5     6     7      8    9   10...
0:   1
1:   0  1
2:  -1  0   2
3:   0 -3   0    4
4:   1  0  -8    0     8
5:   0  5   0  -20     0   16
6:  -1  0  18    0   -48    0    32
7:   0 -7   0   56     0 -112     0    64
8:   1  0 -32    0   160    0  -256     0    128
9:   0  9   0 -120     0  432     0  -576      0  256
10: -1  0  50    0  -400    0  1120     0  -1280    0  512
... Reformatted and extended - _Wolfdieter Lang_, Oct 21 2013
E.g., the fourth row (n=3) corresponds to the polynomial T(3,x) = -3*x + 4*x^3.
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964. Tenth printing, Wiley, 2002 (also electronically available), p. 795.
  • F. Hirzebruch et al., Manifolds and Modular Forms, Vieweg 1994 pp. 77, 105.
  • Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 22, page 196.
  • TableCurve 2D, Automated curve fitting and equation discovery, Version 5.01 for Windows, User's Manual, Chebyshev Series Polynomials and Rationals, pages 12-21 - 12-24, SYSTAT Software, Inc., Richmond, WA, 2002.

Crossrefs

The first nonzero (sub)diagonal sequences are A011782, -A001792, A001793(n+1), -A001794, A006974, -A006975, A006976, -A209404.

Programs

  • Julia
    using Nemo
    function A053120Row(n)
        R, x = PolynomialRing(ZZ, "x")
        p = chebyshev_t(n, x)
        [coeff(p, j) for j in 0:n] end
    for n in 0:6 A053120Row(n) |> println end # Peter Luschny, Mar 13 2018
    
  • Magma
    &cat[ Coefficients(ChebyshevT(n)): n in [0..11] ]; // Klaus Brockhaus, Mar 08 2008
    
  • Maple
    with(orthopoly) ;
    A053120 := proc(n,k)
        T(n,x) ;
        coeftayl(%,x=0,k) ;
    end proc: # R. J. Mathar, Jun 30 2013
    T := (n, x) -> `if`(n = 0, 1, add((-1)^(n - k) * (n/(2*k))*binomial(k, n - k) *(2*x)^(2*k - n), k = 1 ..n)):
    seq(seq(coeff(T(n, x), x, k), k = 0..n), n = 0..11); # Peter Luschny, Sep 20 2022
  • Mathematica
    t[n_, k_] := Coefficient[ ChebyshevT[n, x], x, k]; Flatten[ Table[ t[n, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, Jan 16 2012 *)
  • PARI
    for(n=0,5,P=polchebyshev(n);for(k=0,n,print1(polcoeff(P,k)", "))) \\ Charles R Greathouse IV, Jan 16 2012
    
  • SageMath
    def f(n,k): # f = A039991
        if (n<2 and k==0): return 1
        elif (k<0 or k>n): return 0
        else: return 2*f(n-1, k) - f(n-2, k-2)
    def A053120(n,k): return f(n, n-k)
    flatten([[A053120(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Aug 10 2022

Formula

T(n, m) = A039991(n, n-m).
G.f. for row polynomials T(n,x) (signed triangle): (1-x*z)/(1-2*x*z+z^2). If unsigned: (1-x*z)/(1-2*x*z-z^2).
T(n, m) := 0 if n < m or n+m odd; T(n, m) = (-1)^(n/2) if m=0 (n even); otherwise T(n, m) = ((-1)^((n+m)/2 + m))*(2^(m-1))*n*binomial((n+m)/2-1, m-1)/m.
Recursion for n >= 2: T(n, m) = T*a(n-1, m-1) - T(n-2, m), T(n, m)=0 if n < m, T(n, -1) := 0, T(0, 0) = T(1, 1) = 1.
G.f. for m-th column (signed triangle): 1/(1+x^2) if m=0, otherwise (2^(m-1))*(x^m)*(1-x^2)/(1+x^2)^(m+1).
From G. C. Greubel, Aug 10 2022: (Start)
Sum_{k=0..floor(n/2)} T(n-k, k) = A000007(n).
T(2*n, n) = i^n * A036909(n/2) * (1+(-1)^n)/2 + [n=0]/3. (End)
T(n, k) = [x^k] T(n, x) for n >= 1, where T(n, x) = Sum_{k=1..n}(-1)^(n - k)*(n/ (2*k))*binomial(k, n - k)*(2*x)^(2*k - n). - Peter Luschny, Sep 20 2022

A001793 a(n) = n*(n+3)*2^(n-3).

Original entry on oeis.org

1, 5, 18, 56, 160, 432, 1120, 2816, 6912, 16640, 39424, 92160, 212992, 487424, 1105920, 2490368, 5570560, 12386304, 27394048, 60293120, 132120576, 288358400, 627048448, 1358954496, 2936012800, 6325010432, 13589544960, 29125246976
Offset: 1

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Comments

Coefficients of Chebyshev T polynomials: the subdiagonal A053120(n+3, n-1), for n > = 1. [rewritten by Wolfdieter Lang, Nov 25 2019]
Number of 132-avoiding permutations of [n+3] containing exactly two 123 patterns. - Emeric Deutsch, Jul 13 2001
Number of Dyck paths of semilength n+2 having pyramid weight n+1 (for pyramid weight see Denise and Simion). Example: a(2)=5 because the Dyck paths of semilength 4 having pyramid weight 3 are: (ud)u(ud)(ud)d, u(ud)(ud)d(ud), u(ud)(ud)(ud)d, u(ud)(uudd)d and u(uudd)(ud)d [here u=(1,1), d=(1,-1) and the maximal pyramids, of total length 3, are shown between parentheses]. - Emeric Deutsch, Mar 10 2004
a(n) is the number of dissections of a regular (n+3)-gon using n-1 noncrossing diagonals such that every piece of the dissection contains at least one non-base side of the (n+3)-gon. (One side of the (n+3)-gon is designated the base.) - David Callan, Mar 23 2004
If X_1,X_2,...,X_n are 2-blocks of a (2n+1)-set X then a(n) is the number of (n+2)-subsets of X intersecting each X_i, (i=1..n). - Milan Janjic, Nov 18 2007
The second corrector line for transforming 2^n offset 0 with a leading 1 into the Fibonacci sequence. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
Sum of all nodes of all integer compositions of n, see example. - Olivier Gérard, Oct 22 2011
Number of compositions of 2n with exactly two odd summands (see example). - Mamuka Jibladze, Sep 04 2013
4*a(n) is the number of North-East paths from (0,0) to (n+2,n+2) with exactly two east steps below y = x-1 or above y = x+1. It is related to paired pattern P_1 and P_6 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
From Paul Weisenhorn, Oct 18 2019: (Start)
The polynomials S(n,x)= Sum_(k>=1) b(n,k)*x^k has the recurrence relation S(n+2,x)=2*S(n+1,x))-x*S(n) with S(1,x)=1, S(2,x)=2-x and are generated by the coefficients b(n,k). b(n,k) is defined by b(n,k)=Sum_(j=1..k) binomials(k+1,j)*b(n-j,k) or by b(n,k)=((n-2+k)!*(n-1+2k)*2^n)/(4*(n-1)!*k!). b(n,1)=A001792, b(n,2)=A001793, b(n,3)=A001794, b(n,4)=A006974, b(n,5)=A006975, b(n,6)=A006976, b(n,7)=A209404.
The general formula for the sequences with k>=1: a(n)=((n-2+k)!*(n-1+2k)*2^n)/(4*(n-1)!*k!) with n >= 1. (End) [See a comment in A053120 on subdiagonal sequences. - Wolfdieter Lang, Jan 03 2020]

Examples

			a(2)=5 since 32415, 32451, 34125, 42135 and 52134 are the only 132-avoiding permutations of 12345 containing exactly two increasing subsequences of length 3.
a(4)=56: the compositions of 4 are 4, 3+1, 1+3, 2+2, 2+1+1, 1+2+1, 1+1+2, 1+1+1+1, the corresponding nodes (partial sums) are {0, 4}, {0, 3, 4}, {0, 1, 4}, {0, 2, 4}, {0, 2, 3, 4}, {0, 1, 3, 4}, {0, 1, 2, 4}, {0, 1, 2, 3, 4}, with individual sums {4, 7, 5, 6, 9, 8, 7, 10} and total 56. - _Olivier Gérard_, Oct 22 2011
The a(3)=18 compositions of 2*3=6 with two odd summands are 5+1, 1+5, 3+3, 4+1+1, 1+4+1, 1+1+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2. - _Mamuka Jibladze_, Sep 04 2013
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = A039991(n+3, 4) = A055252(n, 1).
Cf. A053120.

Programs

Formula

G.f.: x*(1-x)/(1-2*x)^3. Binomial transform of squares [1, 4, 9, ...].
a(n) = Sum_{k=0..floor((n+4)/2)} C(n+4, 2k)*C(k, 2). - Paul Barry, May 15 2003
With two leading zeros, binomial transform of quarter-squares A002620. - Paul Barry, May 27 2003
a(n) = Sum_{k=0..n+2} C(n+2, k) * floor(k^2/4). - Paul Barry, May 27 2003
a(n) = Sum_{i=0..j} binomial(i+1, 2)*binomial(j, i). - Jon Perry, Feb 26 2004
With one leading zero, binomial transform of triangular numbers A000217. - Philippe Deléham, Aug 02 2005
a(n) = Sum_{k=0..n+1} (-1)^(n-k+1)*C(k, n-k+1)*k*C(2k, k)/2. - Paul Barry, Oct 07 2005
Left-shifted sequence is binomial transform of left-shifted squares (A000290). - Franklin T. Adams-Watters, Nov 29 2006
Binomial transform of a(n) = n^2 offset 1. a(3)=18. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
a(n) = (1/n) * Sum_{k=0..n} binomial(n,k)*k^3. - Gary Detlefs, Nov 26 2011
For n > 1, a(n) = Sum_{k=0..n-1} Sum_{i=0..n} (k+2) * C(n-2,i). - Wesley Ivan Hurt, Sep 20 2017
a(n) = a(-3-n)*2^(2*n+3), a(n)*(n+3) = -A058645(-3-n)*2^(2*n+3) for all n in Z. - Michael Somos, Apr 19 2019
E.g.f.: (1/2)*exp(2*x)*x*(2 + x). - Stefano Spezia, Aug 17 2019
From Amiram Eldar, Jan 05 2022: (Start)
Sum_{n>=1} 1/a(n) = 128/9 - 56*log(2)/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 24*log(3/2) - 80/9. (End)
Showing 1-2 of 2 results.