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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A116201 a(n) = a(n-1) + a(n-2) + a(n-3) - a(n-4); a(0)=0, a(1)=1, a(2)=1, a(3)=1.

Original entry on oeis.org

0, 1, 1, 1, 3, 4, 7, 13, 21, 37, 64, 109, 189, 325, 559, 964, 1659, 2857, 4921, 8473, 14592, 25129, 43273, 74521, 128331, 220996, 380575, 655381, 1128621, 1943581, 3347008, 5763829, 9925797, 17093053, 29435671, 50690692, 87293619, 150326929, 258875569
Offset: 0

Views

Author

R. K. Guy, Mar 23 2008

Keywords

Comments

This is a divisibility sequence; that is, if n divides m then a(n) divides a(m). - T. D. Noe, Dec 22 2008
This is the case P1 = 1, P2 = -3, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 31 2014
Also, the inverse radii of a family of spheres defined as follows: the first three spheres have radius of 1 and touch each other and the common plane, while each subsequent sphere touches the three immediately preceding ones and the same plane. - Ivan Neretin, Sep 11 2018

Examples

			G.f. = x + x^2 + x^3 + 3*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + 21*x^8 + ... - _Michael Somos_, Feb 26 2019

Links

Crossrefs

Cf. A100047, A116732, A135431.

Programs

  • Maple
    a[0]:=0: a[1]:=1: a[2]:=1: a[3]:=1: for n from 4 to 35 do a[n]:= a[n-1]+a[n-2]+a[n-3]-a[n-4] end do: seq(a[n],n=0..35); # Emeric Deutsch, Apr 12 2008
  • Mathematica
    a = {0, 1, 1, 1, 3}; Do[AppendTo[a, a[[ -1]]+a[[ -2]]+a[[ -3]]-a[[ -4]]], {80}]; a (* Stefan Steinerberger, Mar 24 2008 *)
CoefficientList[Series[(- x^3 + x)/(x^4 - x^3 - x^2 - x + 1), {x, 0, 50}], x] (* Vincenzo Librandi, Apr 02 2014 *)
a[ n_] := 1 - SeriesCoefficient[ (1 - 2 x) / (1 - 2 x + 2 x^4 - x^5), {x, 0, Abs@n}]; (* Michael Somos, Feb 26 2019 *)
LinearRecurrence[{1,1,1,-1},{0,1,1,1},50] (* Harvey P. Dale, Mar 26 2019 *)
  • PARI
    {a(n) = n=abs(n); 1 - polcoeff( (1 - 2*x) / (1 - 2*x + 2*x^4 - x^5) + x * O(x^n), n)}; /* Michael Somos, Feb 26 2019 */

Formula

From R. J. Mathar, Mar 31 2008: (Start)
O.g.f: -x*(x-1)*(x+1)/(1 - x - x^2 - x^3 + x^4).
a(n) = A135431(n) - A135431(n-1). (End)
From Peter Bala, Mar 31 2014: (Start)
a(n) = ( T(n,alpha) - T(n,beta) )/(alpha - beta), where alpha = (1 + sqrt(13))/4 and beta = (1 - sqrt(13))/4 and T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 3/4; 1, 1/2].
a(n) = U(n-1,(sqrt(3) + i)/4)*U(n-1,(sqrt(3) - i)/4), where U(n,x) denotes the Chebyshev polynomial of the second kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. (End)
a(n) = a(-n) = A116732(n+2) - A116732(n), 0 = a(n) - 2*a(n+1) + 2*a(n+4) - a(n+5) for all n in Z. - Michael Somos, Feb 26 2019

A097843 First differences of Chebyshev polynomials S(n,123) = A049670(n+1) with Diophantine property.

Original entry on oeis.org

1, 122, 15005, 1845493, 226980634, 27916772489, 3433536035513, 422297015595610, 51939099382224517, 6388086926998019981, 785682752921374233146, 96632590522402032656977, 11885022951502528642575025, 1461761190444288621004071098, 179784741401695997854858170029
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(11*b(n))^2 - 5*(5*a(n))^2 = -4 with b(n)=A097842(n) give all positive solutions of this Pell equation.

Examples

			All positive solutions of Pell equation x^2 - 125*y^2 = -4 are (11 = 11*1,1), (1364 = 11*124,122), (167761 = 11*15251,15005), (20633239 = 11*1875749,1845493), ...

Links

Programs

  • GAP
    a:=[1,122];; for n in [3..20] do a[n]:=123*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2019
  • Magma
    m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-123*x+x^2) )); // G. C. Greubel, Jan 14 2019
  • Mathematica
    LinearRecurrence[{123,-1}, {1,122}, 20] (* G. C. Greubel, Jan 14 2019 *)
  • PARI
    Vec((1-x)/(1-123*x+x^2) + O(x^30)) \\ Colin Barker, Jun 15 2015
  • Sage
    ((1-x)/(1-123*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 14 2019

Formula

a(n) = ((-1)^n)*S(2*n, 11*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.
G.f.: (1-x)/(1-123*x+x^2).
a(n) = S(n, 123) - S(n-1, 123) = T(2*n+1, 5*sqrt(5)/2)/(5*sqrt(5)/2), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x) and T(n, x) Chebyshev's polynomials of the first kind, A053120.
a(n) = 123*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=122. - Philippe Deléham, Nov 18 2008
a(n) = (F(10*(n+1)) - F(10*n))/F(10), with F=A000045 (Fibonacci). F(10*n)/F(10) = A049670. - Wolfdieter Lang, Oct 11 2012
a(n) = (1/5)*F(10*n + 5). Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/11^2. Compare with A001519 and A007805. - Peter Bala, Nov 29 2013
From Peter Bala, Mar 23 2015: (Start)
a(n) = A049666(2*n + 1).
a(n) = ( Fibonacci(10*n + 10 - 2*k) - Fibonacci(10*n + 2*k) )/( Fibonacci(10 - 2*k) - Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(10*n + 10 - 2*k - 1) + Fibonacci(10*n + 2*k + 1) )/( Fibonacci(10 - 2*k - 1) + Fibonacci(2*k + 1) ), for k an arbitrary integer.
The aerated sequence (b(n))n>=1 = [1, 0, 122, 0, 15005, 0, 1845493, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -125, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)

A078070 Expansion of (1-x)/(1 + 2*x + 2*x^2 + x^3).

Original entry on oeis.org

1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3, 4, -3, 1, 0, 1, -3
Offset: 0

Views

Author

N. J. A. Sloane, Nov 17 2002

Keywords

Comments

Period 6: repeat [1, -3, 4, -3, 1, 0].
The unsigned sequence is the r=3 member in the r-family of sequences S_r(n) defined in A092184 where more information can be found. |a(n)| = 2-2*T(n,1/2), with twice the Chebyshev polynomials of the first kind 2*T(n,x=1/2) = A057079(n+1) = S(n+1,1) + S(n,1) with S(n,1)= A010892(n).
Working with an offset of 1, this sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. Case P1 = -3, P2 = 2, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 25 2014

Examples

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

Links

Crossrefs

Cf. A100047, A254745.

Programs

  • Mathematica
    a[ n_] := {-3, 4, -3, 1, 0, 1}[[Mod[ n, 6, 1]]]; (* Michael Somos, Aug 05 2015 *)
CoefficientList[Series[(1-x)/(1+2x+2x^2+x^3),{x,0,120}],x] (* or *) PadRight[ {},120,{1,-3,4,-3,1,0}] (* or *) LinearRecurrence[{-2,-2,-1},{1,-3,4},120] (* Harvey P. Dale, Jan 06 2016 *)
  • PARI
    Vec((1-x)/(1+2*x+2*x^2+x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012
  • PARI
    {a(n) = [1, -3, 4, -3, 1, 0][n%6 + 1]}; /* Michael Somos, Aug 05 2015 */

Formula

abs(a(n)) = 2 + 2*cos(Pi*n/3 - 2*Pi/3). - Paul Barry, Mar 14 2004
Euler transform of finite sequence [-3, 1, 1]. - Michael Somos, Sep 17 2004
a(n) = (n+1)*(Sum_{k=0..floor((n+1)/2)} (-1)^k*binomial(n-k+1, k)*(-1)^(n-2k+1)/(n-k+1)) + 2*(-1)^n; a(n) = 2*T(n+1, -1/2) + 2(-1)^n. - Paul Barry, Dec 12 2004
From Peter Bala, Mar 25 2014: (Start)
The following formulas assume an offset of 1.
Let {u_j(n)}, j = 0 or j = 1, be two Lucas sequences in the quadratic integer ring Z[w], where w = exp(2*Pi*i/3), defined by the recurrences u_j(0) = 0, u_j(1) = 1 and u_j(n) = (-1)^j*sqrt(3)*u(n-1) - u(n-2) for n >= 2. Then a(n) = u_0(n)*u_1(n).
Equivalently, a(n) = U(n-1,sqrt(3)/2)*U(n-1,-sqrt(3)/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = - ( ((sqrt(3) + i)/2)^n - ((sqrt(3) - i)/2)^n )*( ((-sqrt(3) + i)/2)^n - ((-sqrt(3) - i)/2)^n ) = w^n + w^(2*n) - 2*(-1)^n = 2*cos(2*n*Pi/3) - 2*(-1)^n.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -1/2; 1, -3/2] and T(n,x) denotes the Chebyshev polynomial of the first kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(n) = a(n+6) = a(-2-n) for all n in Z. - Michael Somos, Aug 05 2015
a(n) = (-1)^n * A254745(n). - Michael Somos, Jul 16 2017

Extensions

Chebyshev comment and related formulas from Wolfdieter Lang, Sep 10 2004

A097733 Pell equation solutions (7*b(n))^2 - 2*(5*a(n))^2 = -1 with b(n)=A097732(n), n >= 0. Note that D=50=2*5^2 is not squarefree.

Original entry on oeis.org

1, 197, 39005, 7722793, 1529074009, 302748930989, 59942759261813, 11868363584907985, 2349876047052519217, 465263588952813896981, 92119840736610099083021, 18239263202259846804541177
Offset: 0

Views

Author

Wolfdieter Lang, Aug 31 2004

Keywords

Examples

			(x,y) = (7,1), (1393,197), (275807,39005), ... give the positive integer solutions to x^2 - 50*y^2 =-1.

Links

Crossrefs

Cf. A097731 for S(n, 198).
Row 7 of array A188647.
Cf. A000129, A100047.

Programs

  • GAP
    a:=[1,197];; for n in [3..20] do a[n]:=198*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 01 2019
  • Magma
    I:=[1,197]; [n le 2 select I[n] else 198*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Aug 01 2019
  • Mathematica
    LinearRecurrence[{198, -1},{1, 197},20] (* Ray Chandler, Aug 11 2015 *)
  • PARI
    my(x='x+O('x^20)); Vec((1-x)/(1-198*x+x^2)) \\ G. C. Greubel, Aug 01 2019
  • Sage
    ((1-x)/(1-198*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019

Formula

a(n) = S(n, 2*99) - S(n-1, 2*99) = T(2*n+1, 5*sqrt(2))/(5*sqrt(2)), with Chebyshev polynomials of the 2nd and first kind. See A049310 for the triangle of S(n, x)= U(n, x/2) coefficients. S(-1, x) := 0 =: U(-1, x); and A053120 for the T-triangle.
a(n) = ((-1)^n)*S(2*n, 14*i) with the imaginary unit i and Chebyshev polynomials S(n, x) with coefficients shown in A049310.
G.f.: (1-x)/(1-198*x+x^2).
a(n) = 198*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=197. - Philippe Deléham, Nov 18 2008
a(n) = k^n + k^(-n) - a(n-1) = A003499(3n) - a(n-1), where k = (sqrt(2)+1)^6 = 99 + 70*sqrt(2) and a(0)=1. - Charles L. Hohn, Apr 05 2011
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Pell(6*n + 6 - 2*k) - Pell(6*n + 2*k) )/( Pell(6 - 2*k) - Pell(2*k) ), for k an arbitrary integer.
a(n) = ( Pell(6*n + 6 - 2*k - 1) + Pell(6*n + 2*k + 1) )/( Pell(6 - 2*k - 1) + Pell(2*k + 1) ), for k an arbitrary integer.
The aerated sequence (b(n))n>=1 = [1, 0, 197, 0, 39005, 0, 7722793, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -200, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
Sum_{n >= 1} 1/( a(n) - 1/a(n) ) = 1/196. - Peter Bala, Mar 26 2015

A097837 Chebyshev polynomials S(n,51) + S(n-1,51) with Diophantine property.

Original entry on oeis.org

1, 52, 2651, 135149, 6889948, 351252199, 17906972201, 912904330052, 46540213860451, 2372638002552949, 120957997916339948, 6166485255730784399, 314369790044353664401, 16026692807006306100052, 817046963367277257438251
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(7*a(n))^2 - 53*b(n)^2 = -4 with b(n)=A097838(n) gives all positive solutions of this Pell equation.

Examples

			All positive solutions of Pell equation x^2 - 53*y^2 = -4 are (7=7*1,1), (364=7*52,50), (18557=7*2651,2549), (946043=7*135149,129949), ...

Links

Crossrefs

Cf. A097783, A097834, A097836, A097840.

Programs

  • GAP
    a:=[1,52];; for n in [3..30] do a[n]:=51*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2019
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-51*x+x^2) )); // G. C. Greubel, Jan 12 2019
  • Mathematica
    LinearRecurrence[{51,-1}, {1,52}, 30] (* G. C. Greubel, Jan 12 2019 *)
  • PARI
    my(x='x+O('x^30)); Vec((1+x)/(1-51*x+x^2)) \\ G. C. Greubel, Jan 12 2019
  • Sage
    ((1+x)/(1-51*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 12 2019

Formula

a(n) = S(n, 51) + S(n-1, 51) = S(2*n, sqrt(53)), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 51)=A097836(n).
a(n) = (-2/7)*i*((-1)^n)*T(2*n+1, 7*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-51*x+x^2).
a(n) = 51*a(n-1) - a(n-2); a(0)=1, a(1)=52. - Philippe Deléham, Nov 18 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/7)*(7/2 o 7/2 o ... o 7/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 52, 0, 2651, 0, 135149, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 51*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -49, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = (1/2)*( (-1)^n - 1 )*F(n,7) + (1/7)*( 1 + (-1)^(n+1) )*F(n+1,7), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 14*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 14*A054413(n)*x^n.
Exp( Sum_{n >= 1} (-14)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 14*A054413(n)*(-x)^n. (End)

A097840 Chebyshev polynomials S(n,83) + S(n-1,83) with Diophantine property.

Original entry on oeis.org

1, 84, 6971, 578509, 48009276, 3984191399, 330639876841, 27439125586404, 2277116783794691, 188973253929372949, 15682502959354160076, 1301458772372465913359, 108005395603955316648721
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(9*a(n))^2 - 85*b(n)^2 = -4 with b(n)=A097841(n) give all positive solutions of this Pell equation.

Examples

			All positive solutions of Pell equation x^2 - 85*y^2 = -4 are (9=9*1,1), (756=9*84,82), (62739=9*6971,6805), (5206581=9*578509,564733), ...

Links

Crossrefs

Cf. A097783, A097834, A097837, A097839.

Programs

  • GAP
    a:=[1,84];; for n in [3..20] do a[n]:=83*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2019
  • Magma
    m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-83*x+x^2) )); // G. C. Greubel, Jan 13 2019
  • Mathematica
    CoefficientList[Series[(1+x)/(1-83x+x^2), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)
  • PARI
    my(x='x+O('x^20)); Vec((1+x)/(1-83*x+x^2)) \\ G. C. Greubel, Jan 13 2019
  • Sage
    ((1+x)/(1-83*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019

Formula

a(n) = S(n, 83) + S(n-1, 83) = S(2*n, sqrt(85)), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 83) = A097839(n).
a(n) = (-2/9)*i*((-1)^n)*T(2*n+1, 9*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1 - 83*x + x^2).
a(n) = 83*a(n-1) - a(n-2) for n > 1; a(0)=1, a(1)=84. - Philippe Deléham, Nov 18 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/9)*(9/2 o 9/2 o ... o 9/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 84, 0, 6971, 0, 578509, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 83*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -81, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = 1/2*( (-1)^n - 1 )*F(n,9) + 1/9*( 1 + (-1)^(n+1) )*F(n+1,9), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 18*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 18*A099371(n)*x^n.
Exp( Sum_{n >= 1} (-18)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 18*A099371(n)*(-x)^n. (End)

A100048 A Chebyshev transform of the Pell numbers.

Original entry on oeis.org

0, 1, 2, 2, 4, 9, 16, 29, 56, 106, 198, 373, 704, 1325, 2494, 4698, 8848, 16661, 31376, 59089, 111276, 209554, 394634, 743177, 1399552, 2635641, 4963450, 9347186, 17602652, 33149377, 62427024, 117562789, 221394656, 416931194, 785166286
Offset: 0

Views

Author

Paul Barry, Oct 31 2004

Keywords

Comments

A Chebyshev transform of the Pell numbers A000129: if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)).
This sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. Case P1 = 2, P2 = -1, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. - Peter Bala, Mar 24 2014

Links

Crossrefs

Cf. A099443, A011655, A100047.

Programs

  • Mathematica
    LinearRecurrence[{2,-1,2,-1},{0,1,2,2},40] (* Harvey P. Dale, Jun 07 2015 *)
  • PARI
    my(x='x+O('x^50)); Vec(x*(1-x^2)/(1-2*x+x^2-2*x^3+x^4)) \\ G. C. Greubel, Aug 08 2017

Formula

G.f.: x(1-x^2)/(1-2x+x^2-2x^3+x^4).
a(n) = 2*a(n-1) - 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)*A000129(n-2*k)/(n-k).
From Peter Bala, Mar 24 2014: (Start)
a(n) = |u(n)|^2, where {u(n)} is the Lucas-type sequence defined by the recurrence u(0) = 0, u(1) = 1 and u(n) = P*u(n-1) - u(n-2) for n >= 2, where P = 1/2*(sqrt(7) + i).
a(n) = ( T(n,alpha) - T(n,beta) )/(alpha - beta), where alpha = (1 + sqrt(2))/2 and beta = (1 - sqrt(2))/2 and T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 1/4; 1, 1].
The o.g.f. is the Hadamard product of the rational functions x/(1 - 1/2*(sqrt(7) + i)*x + x^2) and x/(1 - 1/2*(sqrt(7) - i)*x + x^2). (End)

A143699 a(n) = 19*a(n-1) - 41*a(n-2) + 19*a(n-3) - a(n-4).

Original entry on oeis.org

0, 1, 19, 319, 5301, 88000, 1460701, 24245719, 402446619, 6680076601, 110880352000, 1840465787401, 30549274537419, 507077165538919, 8416803858813901, 139707705280792000, 2318961358994380101
Offset: 0

Views

Author

N. J. A. Sloane, based on email from R. K. Guy, Feb 08 2009

Keywords

Comments

This is a divisibility sequence; that is, if n divides m, then a(n) divides a(m).
A003733 = 5 * (A143699)^2. - R. K. Guy, Mar 11 2010
The sequence is the case P1 = 19, P2 = 39, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014

Links

Crossrefs

Cf. A003733, A100047.

Programs

  • Magma
    I:=[0,1,19,319]; [n le 4 select I[n] else 19*Self(n-1) -41*Self(n-2) +19*Self(n-3) -Self(n-4): n in [1..30]]; // G. C. Greubel, May 31 2021
  • Mathematica
    LinearRecurrence[{19,-41,19,-1}, {0,1,19,319}, 20] (* Jean-François Alcover, Dec 12 2016 *)
  • PARI
    {a(n) = n = abs(n); polcoeff( x*(1-x^2)/(1 -19*x +41*x^2 -19*x^3 +x^4) + x*O(x^n), n)} \\ Michael Somos, Feb 24 2009
  • Sage
    def A143699_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1-x^2)/(1 -19*x +41*x^2 -19*x^3 +x^4) ).list()
A143699_list(30) # G. C. Greubel, May 31 2021

Formula

Equals sqrt(A003733(n)/5).
G.f.: x*(1+x)*(1-x)/(1 - 19*x + 41*x^2 - 19*x^3 + x^4). - R. J. Mathar, Feb 09 2009
a(-n) = a(n). - Michael Somos, Feb 24 2009
a(n) = (r1^n + r2^n - r3^n - r4^n) / s1 where s1 = sqrt(205), s2 = sqrt(550 + 38*s1), s3 = 36 * sqrt(5) / s2, r1 = (19 + s1 + s2) / 4, r2 = 1/r1, r3 = (19 - s1 + s3) / 4, r4 = 1/r3. - Michael Somos, Feb 12 2012
From Peter Bala, Apr 03 2014: (Start)
a(n) = ( T(n,alpha) - T(n,beta) )/(alpha - beta), n >= 1, where alpha = (1/4)*(19 + sqrt(205)), beta = (1/4)*(19 - sqrt(205)) and where T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n)= U(n-1, (sqrt(5) - 9)/4)*U(n-1, -(sqrt(5) + 9)/4) for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -39/4; 1, 19/2]. See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)

A097834 Chebyshev polynomials S(n,27) + S(n-1,27) with Diophantine property.

Original entry on oeis.org

1, 28, 755, 20357, 548884, 14799511, 399037913, 10759224140, 290100013867, 7821941150269, 210902311043396, 5686540457021423, 153325690028535025, 4134107090313424252, 111467565748433919779, 3005490168117402409781
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(5*a(n))^2 - 29*b(n)^2 = -4 with b(n) = A097835(n) give all positive solutions of this Pell equation.
From Klaus Purath, Sep 24 2024: (Start)
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)), where (t) is any sequence satisfying t(i) = 28t(i-1) - 28t(i-2) + t(i-3) or t(i) = 27t(i-1) - t(i-2) without regard to initial values and including this sequence itself, as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0.
a(n) = (Sum_{i .. i+2n} t(i))/t(i+n), where (t) is any recurrence of the form (27,-1) without regard to initial values and including this sequence itself, as long as t(i+n) != 0 for integer i and n >= 0.
a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/28, where (t) is any third order recurrence with constant coefficients (28,-28,1) and initial values t(0) = x, t(1) = x + 1, t(2) = x + 29 for any integer x.
a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/26, where (t) is any third order recurrence with constant coefficients (26,26,-1) and initial values t(0) = x, t(1) = 1 - x, t(2) = x + 27 for any integer x.
a(n) = (t(i+4n+2) - t(i))/(t(i+2n+2) - t(i+2n)), where (t) is any recurrence of the form (5,1) without regard to initial values, as long as t(i+2n+2) - t(i+2n) != 0 for nonnegative integer i and n. (End)

Examples

			All positive solutions of Pell equation x^2 - 29*y^2 = -4 are
(5=5*1,1), (140=5*28,26), (3775=5*755,701), (101785=5*20357,18901), ...

Links

Crossrefs

A087130(2*n + 1) = 5 * a(n). - Michael Somos, Nov 01 2008
Cf. A097781, A097783, A097837, A097840.

Programs

  • Mathematica
    a[n_] := -2/5*I*(-1)^n*ChebyshevT[2*n + 1, 5*I/2]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jun 21 2013, from 2nd formula *)
  • PARI
    {a(n) = (-1)^n * subst(2 * I / 5 * poltchebi(2*n + 1), 'x, -5/2 * I)}; /* Michael Somos, Nov 04 2008 */

Formula

a(n) = S(n, 27) + S(n-1, 27) = S(2*n, sqrt(29)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 27)=A097781(n).
a(n) = (-2/5)*i*((-1)^n)*T(2*n+1, 5*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-27*x+x^2).
a(n) = - a(-1-n) for all n in Z. - Michael Somos, Nov 01 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/5)*(5/2 o 5/2 o ... o 5/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 28, 0, 755, 0, 20357, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 27*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -25, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = (1/2)*( (-1)^n - 1 )*F(n,5) + (1/5)*( 1 + (-1)^(n+1) )*F(n+1,5), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 10*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*x^n.
Exp( Sum_{n >= 1} (-10)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*(-x)^n.
(End)
a(n) = (a(n-1)*a(n-2) - 783)/a(n-3) for n >= 3. - Klaus Purath, Sep 24 2024

A097842 Chebyshev polynomials S(n,123) + S(n-1,123) with Diophantine property.

Original entry on oeis.org

1, 124, 15251, 1875749, 230701876, 28374454999, 3489827263001, 429220378894124, 52790616776714251, 6492816643156958749, 798563656491529211876, 98216836931814936101999, 12079872378956745611334001, 1485726085774747895257980124, 182732228677915034371120221251, 22474578401297774479752529233749
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(11*a(n))^2 - 5*(5*b(n))^2 = -4 with b(n)=A097843(n) give all positive solutions of this Pell equation.

Examples

			All positive solutions of Pell equation x^2 - 125*y^2 = -4 are (11=11*1,1), (1364=11*124,122), (167761=11*15251,15005), (20633239=11*1875749,1845493), ...

Links

Programs

  • GAP
    a:=[1,124];; for n in [3..20] do a[n]:=123*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2019
  • Magma
    m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-123*x+x^2) )); // G. C. Greubel, Jan 13 2019
  • Mathematica
    CoefficientList[Series[(1+x)/(1-123x+x^2), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)
LinearRecurrence[{123,-1}, {1,124}, 20] (* G. C. Greubel, Jan 13 2019 *)
  • PARI
    a(n)=polchebyshev(n, 2, 123/2) + polchebyshev(n - 1, 2, 123/2); \\ Michel Marcus, Aug 04 2017
  • PARI
    my(x='x+O('x^20)); Vec((1+x)/(1-123*x+x^2)) \\ G. C. Greubel, Jan 13 2019
  • Sage
    ((1+x)/(1-123*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019

Formula

a(n) = S(n, 123) + S(n-1, 123) = S(2*n, 5*sqrt(5)), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x). S(n, 123) = A049670(n+1).
a(n) = (-2/11)*i*((-1)^n)*T(2*n+1, 11*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-123*x+x^2).
a(n) = 123*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=124. - Philippe Deléham, Nov 18 2008
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Fibonacci(10*n + 10 - 2*k) + Fibonacci(10*n + 2*k) )/( Fibonacci(10 - 2*k) + Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(10*n + 10 - 2*k - 1) - Fibonacci(10*n + 2*k + 1) )/( Fibonacci(10 - 2*k - 1) - Fibonacci(2*k + 1) ), for k an arbitrary integer, k != 2.
The aerated sequence (b(n))n>=1 = [1, 0, 124, 0, 15251, 0, 1875749, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -121, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = Lucas(10*n + 5)/11. - Ehren Metcalfe, Jul 29 2017
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