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.

Previous Showing 31-36 of 36 results.

A185331 Riordan array ((1-x+x^2)/(1+x^2), x/(1+x^2)).

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 1, -1, -1, 1, 0, 2, -2, -1, 1, -1, 1, 3, -3, -1, 1, 0, -3, 3, 4, -4, -1, 1, 1, -1, -6, 6, 5, -5, -1, 1, 0, 4, -4, -10, 10, 6, -6, -1, 1, -1, 1, 10, -10, -15, 15, 7, -7, -1, 1, 0, -5, 5, 20, -20, -21, 21, 8, -8, -1, 1
Offset: 0

Views

Author

Philippe Deléham, Feb 08 2012

Keywords

Comments

Triangle T(n,k), read by rows, given by (-1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Examples

			Triangle begins:
   1;
  -1,  1;
   0, -1,   1;
   1, -1,  -1,   1;
   0,  2,  -2,  -1,   1;
  -1,  1,   3,  -3,  -1,   1;
   0, -3,   3,   4,  -4,  -1,   1;
   1, -1,  -6,   6,   5,  -5,  -1,  1;
   0,  4,  -4, -10,  10,   6,  -6, -1,  1;
  -1,  1,  10, -10, -15,  15,   7, -7, -1,  1;
   0, -5,   5,  20, -20, -21,  21,  8, -8, -1,  1;
   1, -1, -15,  15,  35, -35, -28, 28,  9, -9, -1, 1;
		

Crossrefs

Cf. A206474 (unsigned version).

Programs

  • Mathematica
    CoefficientList[Series[CoefficientList[Series[(1 - x + x^2)/(1 - y*x + x^2), {x, 0, 10}], x], {y, 0, 10}], y] // Flatten (* G. C. Greubel, Jun 27 2017 *)

Formula

T(n,k) = T(n-1,k-1) - T(n-2,k), T(0,0) = 1, T(0,1) = -1, T(0,2) = 0.
G.f.: (1-x+x^2)/(1-y*x+x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = (-1)^n*A184334(n), A163805(n), A000007(n), A028310(n), A025169(n-1), A005320(n) (n>0) for x = -1, 0, 1, 2, 3, 4 respectively.
T(n,n) = 1, T(n+1,n) = -1, T(n+2,n) = -n, T(n+3,n) = n+1, T(n+4,n) = n(n+1)/2 = A000217(n).
T(2n,2k) = (-1)^(n-k) * A128908(n,k), T(2n+1,2k+1) = -T(2n+1,2k) = A129818(n,k), T(2n+2,2k+1) = (-1)*A053122(n,k). - Philippe Deléham, Feb 09 2012

A217478 Triangle of coefficients of polynomials providing the second term of the numerator for the generating function for odd powers (2*m+1) of Chebyshev S-polynomials. The present polynomials are called P(m;1,x^2).

Original entry on oeis.org

-2, 3, -4, -4, 10, -6, 5, -20, 21, -8, -6, 35, -56, 36, -10, 7, -56, 126, -120, 55, -12, -8, 84, -252, 330, -220, 78, -14, 9, -120, 462, -792, 715, -364, 105, -16, -10, 165, -792, 1716, -2002, 1365, -560, 136, -18, 11, -220, 1287, -3432, 5005, -4368, 2380, -816, 171, -20
Offset: 1

Views

Author

Wolfdieter Lang, Nov 14 2012

Keywords

Comments

The o.g.f. for S(n,x)^(2*m+1), m >= 0, with Chebyshev's S-polynomials (see A049310), is
G(m;z,x) := sum(S(n,x)^(2*m+1)*z^n, n=0..infinity)= sum(T(m,k)*S(2*k,x)/(1-z*x*tau(k,x)+z^2), k=0..m)/(x^2-4)^m, with the signed Riordan triangle
T(m,k) = binomial(2*m+1,m-k)*(-1)^(m-k) = A113187(m,k),
and tau(k,x):= R(2*k+1,x)/x with R the monic integer Chebyshev T-polynomials (see A127672). The proof uses the de Moivre-Binet formula: S(n,x) = (q^(n+1) - 1/q^(n+1))/(q-1/q), with q:=(x+sqrt(x^2-4))/2, and the one for tau(n,x) = (q^(2*n+1) + 1/q^(2*n+1))/(q+1/q). This can be written as G(m;z,x) = Z(m;z,x)/N(m;z,x) with N(m;z,x) = product((1+z^2) - z*x*tau(k,x),k=0..m), and Z(m;z,x) = sum((1+z^2)^(m-l)*(-z*x)^l*P(m;l,x^2),l=0..m), where P(m;l,x^2) = sum(T(m,k)*S(2*k,x)*sigma(m;k,l,x^2), k=0..m)/(x^2-4)^m, with sigma(m;k,l,x^2) the elementary symmetric function of a product of l factors from tau(j,x), for j=0..m, with tau(k,x) missing. E.g., sigma(3;1,2,x^2) = tau(0,x)*tau(2,x) + tau(0,x)*tau(3,x) + tau(2,x)*tau(3,x), (tau(1,x) is missing). P(m;0,x^2) = 1 due to the identity Id(0;m,x^2) := sum(T(m,k)*S(2*k,x), k=0..m) = (x^2-4)^m (proof by using the de Moivre-Binet formula and the formula mentioned in a comment on A113187). Also the other P(m;l,x^2) turn out to be polynomials in x^2.
The present triangle a(m,k) provides the P(m;1,x^2) coefficients: P(m;1,x^2) = sum(a(m,k)*(x^2)^k, k=0..m-1), m>=1.
Using inclusion-exclusion one can write (x^2-4)^m*P(m;1,x^2) = sum(T(m,k)*S(2*k,x)*(sum(tau(k,x),k=0..m) - tau(k,x)), k=0..m) = sum(tau(k,x),k=0..m)*(x^2-4)^m - sum(T(m,k)*S(2*k,x)* tau(k,x),k=0..m), using the mentioned identity Id(0;m,x^2). In the second term S(2*k,x)*tau(k,x) = S(4*k+1,x)/x (de Moivre-Binet formulas for S and tau). This leads to the l=1 identity Id(1;m,x^2) := sum(T(m,k)*S(4*k+1,x)/x,k=0..m) =
((x^2-4)*x^2)^m, using again de Moivre-Binet and the identity
given in a comment on A113187. Therefore, after dividing by (x^2-4)^m, P(m;1,x^2) = sum(tau(k,x),k=0..m) - x^(2*m).

Examples

			The triangle a(m,k) begins:
m\k   0    1    2     3     4      5     6     7    8    9 ...
1:   -2
2:    3   -4
3:   -4   10   -6
4:    5  -20   21    -8
5:   -6   35  -56    36   -10
6:    7  -56  126  -120    55    -12
7:    8   84 -252   330  -220     78   -14
8:    9 -120  462  -792   715   -364   105   -16
9:  -10  165 -792  1716 -2002   1365  -560   136  -18
10:  11 -220 1287 -3432  5005  -4368  2380  -816  171  -20
...
P(2;1,x^2) = 3 - 4*x^2, appears in the second term of the numerator of the o.g.f. for S(n,x)^5 which is  Z(2;z,x) = (1+z^2)^2 + (1+z^2)*(-x*z)*(3-4*x^2) + ((-x*z)^2)*2*(-4 +3*x^2). The last term is taken from row m=2 of A217479. The denominator is  N(2;z,x) = product((1+z^2)-z*x*tau(k,x), k=0..2). This checks with [1,x^5,-1+5*x^2-10*x^4+10*x^6-5*x^8
+x^10,-32*x^5+80*x^7-80*x^9+40*x^11-10* x^13+x^15,...] for S(n,x)^5, n=0,1,2,3,...
		

Crossrefs

Cf. A049310, A217479 (P(m;2,x^2)).

Formula

a(m,k) = [x^(2*k)] P(m;1,x^2) = [x^(2*k)](sum(tau(k,x),k=0..m) - x^(2*m)) (see the comment above), m>=1, k = 0..m-1.
a(m,k) = (-1)^(m-k)*binomial(m+k+1,2*k+1). For the proof one uses the identity sum(tau(j,x),j=0..m) = S(m,x^2-2) which holds by comparing the o.g.f.s of both sides (see a Nov 13 2012 comment on Riordan A053122 where tau is called r).

A285072 Triangle read by rows: coefficients of the Laplacian polynomial of the n-path graph P_n.

Original entry on oeis.org

0, -1, 0, -2, 1, 0, -3, 4, -1, 0, -4, 10, -6, 1, 0, -5, 20, -21, 8, -1, 0, -6, 35, -56, 36, -10, 1, 0, -7, 56, -126, 120, -55, 12, -1, 0, -8, 84, -252, 330, -220, 78, -14, 1, 0, -9, 120, -462, 792, -715, 364, -105, 16, -1, 0, -10, 165, -792, 1716, -2002, 1365, -560, 136, -18, 1
Offset: 1

Views

Author

Eric W. Weisstein, Apr 09 2017

Keywords

Comments

Version of A053122 with row-ending 0's and differing signs.

Examples

			Table starts:
  0
 -1    0
 -2    1    0
 -3    4   -1     0
 -4   10   -6     1     0
 -5   20  -21     8    -1    0
 -6   35  -56    36   -10    1     0
 -7   56  -126  120   -55   12    -1   0
 -8   84  -252  330  -220   78   -14   1   0
 -9  120  -462  792  -715  364  -105  16  -1   0
		

Crossrefs

Cf. A053122 (version lacking row-ending 0's and with differing signs).
Cf. A321620.

Programs

  • Maple
    S := proc(n, k) option remember;
    if n <= k then 0 elif k = 0 then (-1)^n*n
    else S(n-1, k-1) - S(n-2, k) - 2*S(n-1, k) fi end:
    T := (n, k) -> (-1)^(n+1)*S(n, k):
    seq(seq(T(n, k), k=0..n), n=0..10); # Peter Luschny, Apr 03 2020
  • Mathematica
    CoefficientList[Table[CharacteristicPolynomial[KirchhoffMatrix[PathGraph[Range[n]]], x], {n, 10}], x] // Flatten
    CoefficientList[LinearRecurrence[{2 - x, -1}, {-x, (-2 + x) x}, 10], x] // Flatten
    CoefficientList[Table[(-1)^(n + 1) x^(1/2) ChebyshevU[2 n - 1, -Sqrt[x]/2], {n, 10}], x] // Flatten
    CoefficientList[Table[(2^-n ((2 - Sqrt[-4 + x] Sqrt[x] - x)^n - (2 + Sqrt[-4 + x] Sqrt[x] - x)^n))/Sqrt[(-4 + x)/x], {n, 10}] // Expand // FullSimplify, x] // Flatten
    T[n_,k_]:=(-1)^(k+1)*Binomial[n+k,2*k+1];Flatten[Table[T[n,k],{n,0,10},{k,0,n}]] (* Detlef Meya, Oct 09 2023 *)
  • Sage
    # uses[riordan_square from A321620]
    # Returns the triangle as a matrix.
    riordan_square(-x/(1 - x)^2, 9) # Peter Luschny, Apr 03 2020

Formula

T(n,k) = (-1)^(k+1)*binomial[n+k,2*k+1]; 0 <= n <= k - Detlef Meya, Oct 09 2023

A136321 Triangular sequence of coefficients of a polynomial recursion for C_n and B_n Cartan matrices: p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2) p(x,n)=x2-4*x+4-m:m=5;(related sequence: A_n:m=1,G_n,m=3,B_n,C_n,m=2) This triangular sequence is an extension to the Cartan pattern of matrices.

Original entry on oeis.org

1, -2, 1, -1, -4, 1, 4, 6, -6, 1, -7, -4, 17, -8, 1, 10, -5, -32, 32, -10, 1, -13, 24, 42, -88, 51, -12, 1, 16, -56, -28, 186, -180, 74, -14, 1, -19, 104, -42, -312, 495, -316, 101, -16, 1, 22, -171, 216, 396, -1122, 1053, -504, 132, -18, 1, -25, 260, -561, -264, 2145, -2912, 1960, -752, 167, -20, 1
Offset: 1

Views

Author

Roger L. Bagula, Apr 12 2008

Keywords

Comments

Row sums are:
{1, -1, -4, 5, -1, -4, 5, -1, -4, 5, -1}
This sequence is also related to different p(x,2) start:
1) A_n like sequence A053122 ( sign change)
2) my G_n matrix A136674
3) B_n,C_n A110162

Examples

			{1},
{-2, 1},
{-1, -4, 1},
{4, 6, -6, 1},
{-7, -4, 17, -8, 1},
{10, -5, -32, 32, -10, 1},
{-13, 24, 42, -88,51, -12, 1},
{16, -56, -28,186, -180, 74, -14, 1},
{-19, 104, -42, -312, 495, -316, 101, -16, 1},
{22, -171, 216, 396, -1122, 1053, -504, 132, -18, 1},
{-25, 260, -561, -264,2145, -2912, 1960, -752, 167, -20, 1}
		

Crossrefs

Programs

  • Mathematica
    Clear[p, a] p[x, 0] = 1; p[x, 1] = -2 + x; p[x, 2] = x^2 - 4*x - 1; p[x_, n_] := p[x, n] = (-2 + x)*p[x, n - 1] - p[x, n - 2]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[CoefficientList[p[x, n], x], {n, 0, 10}] Flatten[a]

Formula

p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2) Three start vectors necessary: p(x,0)=1;p(x,1)=2-x; p(x,2)=x^2-4*x-1=CharacteristicPolynomial[{{2, -5}, {-1, 2}}, x] or CharacteristicPolynomial[{{2, -1}, {-5, 2}}, x]

A136329 Triangular sequence of coefficients of a polynomial recursion for C_n and B_n Cartan matrices: p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2) p(x,n)=x2-4*x+4-m:m=4;(related sequence: A_n:m=1,G_n,m=3,B_n,C_n,m=2) This triangular sequence is an extension to the Cartan pattern of matrices.

Original entry on oeis.org

1, -2, 1, 0, -4, 1, 2, 7, -6, 1, -4, -8, 18, -8, 1, 6, 5, -38, 33, -10, 1, -8, 4, 63, -96, 52, -12, 1, 10, -21, -84, 222, -190, 75, -14, 1, -12, 48, 84, -432, 550, -328, 102, -16, 1, 14, -87, -36, 726, -1342, 1131, -518, 133, -18, 1, -16, 140, -99, -1056, 2860, -3276, 2065, -768, 168, -20, 1
Offset: 1

Views

Author

Roger L. Bagula, Apr 12 2008

Keywords

Comments

Row sums are:
{1, -1, -3, 4, -1, -3, 4, -1, -3, 4, -1}
This sequence is also related to different p(x,2) start:
1) A_n like sequence A053122 ( sign change)
2) my G_n matrix A136674
3) B_n,C_n A110162

Examples

			{1},
{-2, 1},
{0, -4, 1},
{2, 7, -6, 1},
{-4, -8, 18, -8, 1},
{6, 5, -38, 33, -10,1},
{-8, 4, 63, -96, 52, -12, 1},
{10, -21, -84, 222, -190, 75, -14, 1},
{-12, 48, 84, -432, 550, -328, 102, -16, 1},
{14, -87, -36, 726, -1342, 1131, -518, 133, -18, 1},
{-16, 140, -99, -1056, 2860, -3276, 2065, -768, 168, -20, 1}
		

Crossrefs

Programs

  • Mathematica
    Clear[p, a] p[x, 0] = 1; p[x, 1] = -2 + x; p[x, 2] = x^2 - 4*x ; p[x_, n_] := p[x, n] = (-2 + x)*p[x, n - 1] - p[x, n - 2]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[CoefficientList[p[x, n], x], {n, 0, 10}] Flatten[a]

Formula

p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2) Three start vectors necessary: p(x,0)=1;p(x,1)=2-x; p(x,2)=x^2-4*x=CharacteristicPolynomial[{{2, -4}, {-1, 2}}, x] or CharacteristicPolynomial[{{2, -1}, {-4, 2}}, x]

A136672 Riordan array ((1+x^2)/(1-x)^2, -x/(1-x)^2).

Original entry on oeis.org

1, 2, -1, 4, -4, 1, 6, -11, 6, -1, 8, -24, 22, -8, 1, 10, -45, 62, -37, 10, -1, 12, -76, 147, -128, 56, -12, 1, 14, -119, 308, -366, 230, -79, 14, -1, 16, -176, 588, -912, 770, -376, 106, -16, 1, 18, -249, 1044, -2046, 2222, -1443, 574, -137, 18, -1, 20, -340, 1749, -4224, 5720, -4732, 2485, -832, 172, -20, 1
Offset: 1

Views

Author

Roger L. Bagula, Apr 04 2008, Apr 12 2008

Keywords

Comments

Row sums are: {1, 1, 1, 0, -1, -1, 0, 1, 1, 0, -1, ...}

Examples

			{1},
{2, -1},
{4, -4, 1},
{6, -11, 6, -1},
{8, -24, 22, -8, 1},
{10, -45, 62, -37, 10, -1},
{12, -76, 147, -128, 56, -12, 1},
{14, -119,308, -366, 230, -79, 14, -1},
{16, -176, 588, -912, 770, -376, 106, -16, 1},
{18, -249, 1044, -2046, 2222, -1443, 574, -137, 18, -1},
{20, -340, 1749, -4224, 5720, -4732, 2485, -832, 172, -20, 1}
		

Crossrefs

Programs

  • Mathematica
    T[n_, m_, d_] := If[ n == m, 2, If[n == d && m == d - 1, 0, If[(n == m - 1 || n == m + 1), -1, 0]]]; M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}]; Table[Det[M[d]], {d, 1, 10}]; Table[Det[M[d] - x*IdentityMatrix[d]], {d, 1, 10}]; a = Join[{{1}}, Table[CoefficientList[Det[M[d] - x*IdentityMatrix[d]],x], {d, 1, 10}]]; Flatten[a] (* polynomial recursion*) Clear[p] p[x, 0] = 1; p[x, 1] = (2 - x); p[x, 2] = 4 - 4 x + x^2; p[x_, n_] := p[x, n] = (2 - x)*p[x, n - 1] - p[x, n - 2]; Table[ExpandAll[p[x, n]], {n, 0, Length[g] - 1}]
    p[t_] = (2 - x)/(1 - (2 - x)*t + t^2); Table[ ExpandAll[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[ CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a]
    {1} ~ Join ~ CoefficientList[Table[(2 - x) ChebyshevU[n - 2, 1 - x/2], {n, 2, 10}], x] // Flatten  (* Eric W. Weisstein, May 10 2017 *)

Formula

G.f.: g(x,t)=(2 - x)/(1 - (2 - x)*t + t^2).

Extensions

Edited by Ralf Stephan, Feb 07 2014
Previous Showing 31-36 of 36 results.