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 1-4 of 4 results.

A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

Original entry on oeis.org

2, 5, 27, 140, 727, 3775, 19602, 101785, 528527, 2744420, 14250627, 73997555, 384238402, 1995189565, 10360186227, 53796120700, 279340789727, 1450500069335, 7531841136402, 39109705751345, 203080369893127
Offset: 0

Views

Author

Paul Barry, Aug 16 2003

Keywords

Comments

Sequence is related to the fifth metallic mean [5;5,5,5,5,...] (see A098318).
The solution to the general recurrence b(n) = (2*k+1)*b(n-1)+b(n-2) with b(0)=2, b(1) = 2*k+1 is b(n) = ((2*k+1)+sqrt(4*k^2+4*k+5))^n+(2*k+1)-sqrt(4*k^2+4*k+5))^n)/2; b(n) = 2^(1-n)*Sum_{j=0..n} C(n, 2*j)*(4*k^2+4*k+5)^j*(2*k+1)^(n-2*j); b(n) = 2*T(n, (2*k+1)*x/2)(-1)^i with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
Primes in this sequence include a(0) = 2; a(1) = 5; a(4) = 727; a(8) = 528527 (3) semiprimes in this sequence include a(7) = 101785; a(13) = 1995189565; a(16) = 279340789727; a(19) = 39109705751345; a(20) = 203080369893127 - Jonathan Vos Post, Feb 09 2005
a(n)^2 - 29*A052918(n-1)^2 = 4*(-1)^n, with n>0 - Gary W. Adamson, Oct 07 2008
For more information about this type of recurrence follow the Khovanova link and see A054413 and A086902. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A072263. - Johannes W. Meijer, Aug 01 2010

Links

Crossrefs

Cf. A006497, A014448, A085447.
Cf. A086902, A000032, A052918.

Programs

  • Magma
    I:=[2,5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016
  • Mathematica
    RecurrenceTable[{a[0] == 2, a[1] == 5, a[n] == 5 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)
  • PARI
    {a(n) = if( n<0, (-1)^n * a(-n), polsym(x^2 - 5*x -1, n) [n + 1])} /* Michael Somos, Nov 04 2008 */
  • Sage
    [lucas_number2(n,5,-1) for n in range(0, 21)] # Zerinvary Lajos, May 14 2009

Formula

a(n) = ((5+sqrt(29))/2)^n+((5-sqrt(29))/2)^n.
a(n) = A100236(n) + 1.
E.g.f. : 2*exp(5*x/2)*cosh(sqrt(29)*x/2); a(n) = 2^(1-n)*Sum_{k=0..floor(n/2)} C(n, 2k)*29^k*5^(n-2*k). a(n) = 2T(n, 5i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
O.g.f.: (-2+5*x)/(-1+5*x+x^2). - R. J. Mathar, Dec 02 2007
a(-n) = (-1)^n * a(n). - Michael Somos, Nov 01 2008
A090248(n) = a(2*n). 5 * A097834(n) = a(2*n + 1). - Michael Somos, Nov 01 2008
Limit(a(n+k)/a(k), k=infinity) = (A087130(n) + A052918(n-1)*sqrt(29))/2. Limit(A087130(n)/A052918(n-1), n= infinity) = sqrt(29). - Johannes W. Meijer, Jun 12 2010
a(3n+1) = A041046(5n), a(3n+2) = A041046(5n+3) and a(3n+3) = 2*A041046 (5n+4). - Johannes W. Meijer, Jun 12 2010
a(n) = 2*A052918(n) - 5*A052918(n-1). - R. J. Mathar, Oct 02 2020
From Peter Bala, Jul 09 2025 : (Start)
The following series telescope (Cf. A000032):
For k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

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)

A097835 First differences of Chebyshev polynomials S(n,27) = A097781(n) with Diophantine property.

Original entry on oeis.org

1, 26, 701, 18901, 509626, 13741001, 370497401, 9989688826, 269351100901, 7262490035501, 195817879857626, 5279820266120401, 142359329305393201, 3838422070979496026, 103495036587140999501, 2790527565781827490501
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(5*b(n))^2 - 29*a(n)^2 = -4 with b(n)=A097834(n) give all positive solutions of this Pell equation.

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

Cf. similar sequences listed in A238379.

Programs

  • GAP
    a:=[1,26];; for n in [3..30] do a[n]:=27*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)/(1-27*x+x^2) )); // G. C. Greubel, Jan 12 2019
  • Mathematica
    LinearRecurrence[{27,-1},{1,26},30] (* Harvey P. Dale, May 31 2013 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-x)/(1-27*x+x^2)) \\ G. C. Greubel, Jan 12 2019
  • Sage
    ((1-x)/(1-27*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 12 2019

Formula

a(n) = ((-1)^n)*S(2*n, 5*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.
G.f.: (1-x)/(1-27*x+x^2).
a(n) = S(n, 27) - S(n-1, 27) = T(2*n+1, sqrt(29)/2)/(sqrt(29)/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) = 27*a(n-1) - a(n-2), a(0)=1, a(1)=26. - Philippe Deléham, Nov 18 2008
Showing 1-4 of 4 results.

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