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 1-5 of 5 results.

A100047 A Chebyshev transform of the Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, -1, -1, 0, -1, -1, 1, 1, 0, 1, 1, -1, -1, 0, -1, -1, 1, 1, 0, 1, 1, -1, -1, 0, -1, -1, 1, 1, 0, 1, 1, -1, -1, 0, -1, -1, 1, 1, 0, 1, 1, -1, -1, 0, -1, -1, 1, 1, 0, 1, 1, -1, -1, 0, -1, -1, 1, 1, 0, 1, 1, -1, -1, 0, -1, -1, 1, 1, 0, 1, 1, -1, -1, 0, -1, -1, 1, 1, 0, 1, 1, -1, -1, 0, -1, -1, 1, 1
Offset: 0

Views

Author

Paul Barry, Oct 31 2004

Keywords

Comments

Multiplicative with a(p^e) = (-1)^(e+1) if p = 2, 0 if p = 5, 1 if p == 1 or 9 (mod 10), (-1)^e if p == 3 or 7 (mod 10). - David W. Wilson, Jun 10 2005
This sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. Case P1 = 1, P2 = -1, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 24 2014
From Peter Bala, Mar 24 2014: (Start)
This is the particular case P1 = 1, P2 = -1, Q = 1 of the following results:
Let P1, P2 and Q be integers. Let alpha and beta denote the roots of the quadratic equation x^2 - 1/2*P1*x + 1/4*P2 = 0. Let T(n,x;Q) denote the bivariate Chebyshev polynomial of the first kind defined by T(n,x;Q) = 1/2*( (x + sqrt(x^2 - Q))^n + (x - sqrt(x^2 - Q))^n ) (when Q = 1, T(n,x;Q) reduces to the ordinary Chebyshev polynomial of the first kind T(n,x)). Then we have
1) The sequence A(n) := ( T(n,alpha;Q) - T(n,beta;Q) )/(alpha - beta) is a linear divisibility sequence of the fourth order.
2) A(n) belongs to the 3-parameter family of fourth-order divisibility sequences found by Williams and Guy.
3) The o.g.f. of the sequence A(n) is the rational function x*(1 - Q*x^2)/(1 - P1*x + (P2 + 2*Q)*x^2 - P1*Q*x^3 + Q^2*x^4).
4) The o.g.f. is the Chebyshev transform of the rational function x/(1 - P1*x + P2*x^2), where the Chebyshev transform takes the function A(x) to the function (1 - Q*x^2)/(1 + Q*x^2)*A(x/(1 + Q*x^2)).
5) Let q = sqrt(Q) and set a = sqrt( q + (P2)/(4*q) + (P1)/2 ) and b = sqrt( q + (P2)/(4*q) - (P1)/2 ). Then the o.g.f. of the sequence A(n) is the Hadamard product of the rational functions x/(1 - (a + b)*x + q*x^2) and x/(1 - (a - b)*x + q*x^2). Thus A(n) is the product of two (usually, non-integer) Lucas-type sequences.
6) A(n) = the bottom left entry of the 2 X 2 matrix 2*T(n,1/2*M;Q), where M is the 2 X 2 matrix [0, -P2; 1, P1].
For examples of the above see A006238, A054493, A078070, A092184, A098306, A100048, A108196, A138573, A152090 and A218134. (End)

Examples

			A Chebyshev transform of the Fibonacci numbers A000045: if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)).
The denominator is the 10th cyclotomic polynomial.
G.f. = x + x^2 - x^3 - x^4 - x^6 - x^7 + x^8 + x^9 + x^11 + x^12 - x^13 + ...
		

Crossrefs

Programs

  • Maple
    a := n -> (-1)^iquo(n, 5)*signum(mods(n, 5)):
    seq(a(n), n=0..89); # after Michael Somos, Peter Luschny, Dec 30 2018
  • Mathematica
    a[ n_] := {1, 1, -1, -1, 0, -1, -1, 1, 1, 0}[[Mod[ n, 10, 1]]]; (* Michael Somos, May 24 2015 *)
    a[ n_] := (-1)^Quotient[ n, 5] Sign[ Mod[ n, 5, -2]]; (* Michael Somos, May 24 2015 *)
    a[ n_] := (-1)^Quotient[n, 5] {1, 1, -1, -1, 0}[[Mod[ n, 5, 1]]]; (* Michael Somos, May 24 2015 *)
    LinearRecurrence[{1, -1, 1, -1}, {0, 1, 1, -1}, 90] (* Jean-François Alcover, Jun 11 2019 *)
  • PARI
    {a(n) = (-1)^(n\5) * [0, 1, 1, -1, -1][n%5+1]}; /* Michael Somos, May 24 2015 */
    
  • PARI
    {a(n) = (-1)^(n\5) * sign( centerlift( Mod(n, 5)))}; /* Michael Somos, May 24 2015 */

Formula

G.f.: x*(1 - x^2)/(1 - x + x^2 - x^3 + x^4).
a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4).
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k *binomial(n-k, k)*A000045(n-2*k)/(n -k).
From Peter Bala, Mar 24 2014: (Start)
a(n) = (T(n,alpha) - T(n,beta))/(alpha - beta), where alpha = (1 + sqrt(5))/4 and beta = (1 - sqrt(5))/4 and T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = bottom left entry of the matrix T(n, M), where M is the 2 X 2 matrix [0, 1/4; 1, 1/2].
The o.g.f. is the Hadamard product of the rational functions x/(1 - 1/2*(sqrt(5) + 1)*x + x^2) and x/(1 - 1/2*(sqrt(5) - 1)*x + x^2). (End)
Euler transform of length 10 sequence [ 1, -2, 0, 0, -1, 0, 0, 0, 0, 1]. - Michael Somos, May 24 2015
a(n) = a(-n) = -a(n + 5) for all n in Z. - Michael Somos, May 24 2015
|A011558(n)| = |A080891(n)| = |a(n)| = A244895(n). - Michael Somos, May 24 2015

A100051 A Chebyshev transform of 1,1,1,...

Original entry on oeis.org

1, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1
Offset: 0

Views

Author

Paul Barry, Oct 31 2004

Keywords

Comments

1, followed by period 6: repeat [1, -1, -2, -1, 1, 2]. - Joerg Arndt, Aug 28 2024
A Chebyshev transform of 1/(1-x): if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)).
Transform of 1/(1+x) under the mapping g(x)->((1+x)/(1-x))g(x/(1-x)^2). - Paul Barry, Dec 01 2004
Multiplicative with a(p^e) = -1 if p = 2; -2 if p = 3; 1 otherwise. - David W. Wilson, Jun 10 2005

Examples

			G.f. = 1 + x - x^2 - 2*x^3 - x^4 + x^5 + 2*x^6 + x^7 - x^8 - 2*x^9 - x^10 + ...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1 - x^2)/(1 - x + x^2), {x,0,50}], x] (* G. C. Greubel, May 03 2017 *)
    LinearRecurrence[{1,-1},{1,1,-1},80] (* Harvey P. Dale, Mar 25 2019 *)
  • PARI
    {a(n) = - (n == 0) + [2, 1, -1, -2, -1, 1][n%6 + 1]}; /* Michael Somos, Mar 21 2011 */

Formula

From Paul Barry, Dec 01 2004: (Start)
G.f.: (1-x^2)/(1-x+x^2).
a(n) = a(n-1) - a(n-2), n>2.
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)/(n-k).
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*(2n/(n+k))*(-1)^k, n>1. (End)
Moebius transform is length 6 sequence [1, -2, -3, 0, 0, 6].
Euler transform of length 6 sequence [1, -2, -1, 0, 0, 1].
a(n) = a(-n). a(n) = c_6(n) if n>1 where c_k(n) is Ramanujan's sum. - Michael Somos, Mar 21 2011
a(n) = A087204(n), n>0. - R. J. Mathar, Sep 02 2008
a(n) = A057079(n+1), n>0. Dirichlet g.f. zeta(s) *(1-2^(1-s)-3^(1-s)+6^(1-s)). - R. J. Mathar, Apr 11 2011

A100050 A Chebyshev transform of n.

Original entry on oeis.org

0, 1, 2, 0, -4, -5, 0, 7, 8, 0, -10, -11, 0, 13, 14, 0, -16, -17, 0, 19, 20, 0, -22, -23, 0, 25, 26, 0, -28, -29, 0, 31, 32, 0, -34, -35, 0, 37, 38, 0, -40, -41, 0, 43, 44, 0, -46, -47, 0, 49, 50, 0, -52, -53, 0, 55, 56, 0, -58, -59, 0, 61, 62, 0, -64, -65, 0, 67, 68, 0, -70, -71, 0, 73, 74, 0, -76, -77, 0, 79, 80, 0, -82, -83, 0, 85, 86, 0
Offset: 0

Views

Author

Paul Barry, Oct 31 2004

Keywords

Comments

A Chebyshev transform of x/(1-x)^2: if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)).

Examples

			x + 2*x^2 - 4*x^4 - 5*x^5 + 7*x^7 + 8*x^8 - 10*x^10 - 11*x^11 + 13*x^13 + ...
		

Crossrefs

Cf. A165202 (partial sums).

Programs

  • Mathematica
    LinearRecurrence[{2, -3, 2, -1}, {0, 1, 2, 0},50] (* G. C. Greubel, Aug 08 2017 *)
  • PARI
    {a(n) = n * (-1)^(n\3) * sign( n%3)} /* Michael Somos, Mar 19 2011 */
    
  • PARI
    {a(n) = local(A, p, e); if( abs(n)<1, 0, A = factor(abs(n)); prod( k=1, matsize(A)[1], if( p=A[k,1], e=A[k,2]; if( p==2, -(-2)^e, (kronecker( -12, p) * p)^e))))} /* Michael Somos, Mar 19 2011 */
  • Sage
    [lucas_number1(n,2,1)*lucas_number1(n,1,1) for n in range(0,88)] # Zerinvary Lajos, Jul 06 2008
    

Formula

Euler transform of length 6 sequence [ 2, -3, -2, 0, 0, 2]. - Michael Somos, Mar 19 2011
a(n) is multiplicative with a(2^e) = -(-2)^e if e>0, a(3^e) = 0^e, a(p^e) = p^e if p == 1 (mod 6), a(p^e) = (-p)^e if p == 5 (mod 6). - Michael Somos, Mar 19 2011
G.f.: x*(1 - x^2)^3 *(1 - x^3)^2 / ((1 - x)^2 *(1 - x^6)^2) = x *(1 + x)^2 *(1 - x^2) / (1 + x^3)^2. - Michael Somos, Mar 19 2011
a(3*n) = 0, a(3*n + 1) = (-1)^n * (3*n + 1), a(3*n + 2) = (-1)^n * (3*n + 2). a(-n) = a(n). - Michael Somos, Mar 19 2011
G.f.: x(1-x^2)/(1-x+x^2)^2.
a(n) = 2*a(n-1) -3*a(n-2) +2*a(n-3) -a(n-4).
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k,k)*(n-2k)/(n-k).

A100049 A Chebyshev transform of the Padovan numbers.

Original entry on oeis.org

1, 0, -1, 1, -1, -3, 3, 3, -6, 2, 10, -13, -9, 29, -9, -43, 55, 32, -126, 48, 183, -243, -121, 541, -241, -765, 1082, 450, -2326, 1171, 3179, -4803, -1617, 9993, -5601, -13168, 21250, 5552, -42849, 26489, 54351, -93763, -17765, 183347, -124086, -223422, 412698, 49827, -782881, 576541, 914279
Offset: 0

Views

Author

Paul Barry, Oct 31 2004

Keywords

Comments

A Chebyshev transform of the Padovan numbers A000931(n+3): if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{0, -2, 1, -2, 0, -1}, {1, 0, -1, 1, -1, -3, 3}, 50] (* G. C. Greubel, Aug 08 2017 *)
  • PARI
    x='x+O('x^50); Vec((1-x^2)*(1+x^2)^2/(1+2*x^2-x^3+2*x^4+x^6)) \\ G. C. Greubel, Aug 08 2017

Formula

G.f.: (1-x^2)*(1+x^2)^2/(1+2*x^2-x^3+2*x^4+x^6).
a(n) = -2*a(n-2) +a(n-3) -2*a(n-4) -a(n-6).
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k,k)*A000931(n-2*k+3)/(n-k).

A140824 Expansion of (x-x^3)/(1-3*x+2*x^2-3*x^3+x^4).

Original entry on oeis.org

0, 1, 3, 6, 15, 41, 108, 281, 735, 1926, 5043, 13201, 34560, 90481, 236883, 620166, 1623615, 4250681, 11128428, 29134601, 76275375, 199691526, 522799203, 1368706081, 3583319040, 9381251041, 24560434083, 64300051206, 168339719535, 440719107401, 1153817602668
Offset: 0

Views

Author

N. J. A. Sloane, Sep 07 2009, based on email from R. K. Guy, Mar 09 2009

Keywords

Comments

Case P1 = 3, P2 = 0, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 25 2014

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3, -2, 3, -1}, {0, 1, 3, 6}, 50] (* G. C. Greubel, Aug 08 2017 *)
  • PARI
    x='x+O('x^50); concat([0], Vec((x-x^3)/(1-3*x+2*x^2-3*x^3+x^4))) \\ G. C. Greubel, Aug 08 2017

Formula

a(0) = 0, a(1) = 1, a(2) = 3, a(3) = 6, a(n) - 3 a(n + 1) + 2 a(n + 2) - 3 a(n + 3) + a(n + 4) = 0.
From Peter Bala, Mar 25 2014: (Start)
a(n) = 2/3*( T(n,3/2) - T(n,0) ), where T(n,x) is a Chebyshev polynomial of the first kind.
a(n) = 1/3 * (A005248(n) - (i^n + (-i)^n)) = 1/3 * (Fibonacci(2*n-1) + Fibonacci(2*n+1) - (i^n + (-i)^n)).
a(n) = bottom left entry of the 2 X 2 matrix 2*T(n, 1/2*M), where M is the 2 X 2 matrix [0, 0; 1, 3].
The o.g.f. is the Hadamard product of the rational functions x/(1 - 1/sqrt(2)*(sqrt(5) + i)*x + x^2) and x/(1 - 1/sqrt(2)*(sqrt(5) - i)*x + x^2). See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. (End)
a(n) = A099483(n) - A099483(n-2). - R. J. Mathar, Feb 10 2016
Showing 1-5 of 5 results.