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.

A097838 First differences of Chebyshev polynomials S(n,51) = A097836(n) with Diophantine property.

Original entry on oeis.org

1, 50, 2549, 129949, 6624850, 337737401, 17217982601, 877779375250, 44749530155149, 2281348258537349, 116304011655249650, 5929223246159194801, 302274081542463685201, 15410048935419488750450, 785610221624851462587749
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(7*b(n))^2 - 53*a(n)^2 = -4 with b(n)=A097837(n) give 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

Programs

  • GAP
    a:=[1,50];; for n in [3..20] do a[n]:=51*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)/(1-51*x+x^2) )); // G. C. Greubel, Jan 13 2019
  • Mathematica
    LinearRecurrence[{51,-1}, {1,50}, 20] (* G. C. Greubel, Jan 13 2019 *)
  • PARI
    my(x='x+O('x^20)); Vec((1-x)/(1-51*x+x^2)) \\ G. C. Greubel, Jan 13 2019
  • Sage
    ((1-x)/(1-51*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019

Formula

a(n) = ((-1)^n)*S(2*n, 7*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.
G.f.: (1-x)/(1 - 51*x + x^2).
a(n) = S(n, 51) - S(n-1, 51) = T(2*n+1, sqrt(53)/2)/(sqrt(53)/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) = 51*a(n-1) - a(n-2), a(0)=1, a(1)=50. - Philippe Deléham, Nov 18 2008

A054413 a(n) = 7*a(n-1) + a(n-2), with a(0)=1 and a(1)=7.

Original entry on oeis.org

1, 7, 50, 357, 2549, 18200, 129949, 927843, 6624850, 47301793, 337737401, 2411463600, 17217982601, 122937341807, 877779375250, 6267392968557, 44749530155149, 319514104054600, 2281348258537349, 16288951913816043, 116304011655249650, 830417033500563593
Offset: 0

Views

Author

Henry Bottomley, May 10 2000

Keywords

Comments

In general, sequences with recurrence a(n) = k*a(n-1) + a(n-2) and a(0)=1 (and a(-1)=0) have the generating function 1/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n) = b(5n), a(3n+1) = b(5n+3), a(3n+2) = 2*b(5n+4) where b(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2/4+1).]
a(p-1) == 53^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
From Johannes W. Meijer, Jun 12 2010: (Start)
For the sequence given above k=7 which implies that it is associated with A041091.
For a similar statement about sequences with recurrence a(n) = k*a(n-1) + a(n-2) but with a(0) = 2, and a(-1) = 0, see A086902; a sequence that is associated with A041090.
For more information follow the Khovanova link and see A087130, A140455 and A178765.
(End)
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 7's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,7} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 21 2023: (Start)
Also called the 7-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 7 kinds of squares available. (End)

Links

Crossrefs

Cf. A000045, A000129, A001076, A005668, A006190, A052918, A243399.
Row n=7 of A073133, A172236 and A352361.
Cf. A099367 (squares).

Programs

Formula

a(3n) = A041091(5n), a(3n+1) = A041091(5n+3), a(3n+2) = 2*A041091(5n+4).
G.f.: 1/(1 - 7x - x^2).
a(n) = U(n, 7*i/2)*(-i)^n with i^2=-1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See A049310.
a(n) = F(n, 7), the n-th Fibonacci polynomial evaluated at x=7. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = (sigma^n - (-sigma)^(-n))/(sqrt(53)) with sigma = (7+sqrt(53))/2;
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*7^(n-1-2i). (End)
a(n) = ((7 + sqrt(53))^n - (7 - sqrt(53))^n)/(2^n*sqrt(53)). Offset 1. a(3)=50. - Al Hakanson (hawkuu(AT)gmail.com), Jan 17 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 7*A097836(n), a(2n) = A097838(n).
Lim_{k->oo} a(n+k)/a(k) = (A086902(n) + A054413(n-1)*sqrt(53))/2.
Lim_{n->oo} A086902(n)/A054413(n-1) = sqrt(53).
(End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(53)-7)/2. - Vladimir Shevelev, Feb 23 2013
From Kai Wang, Feb 24 2020: (Start)
Sum_{m>=0} 1/(a(m)*a(m+2)) = 1/49.
Sum_{m>=0} 1/(a(2*m)*a(2*m+2)) = (sqrt(53)-7)/14.
In general, for sequences with recurrence f(n)= k*f(n-1)+f(n-2) and f(0)=1,
Sum_{m>=0} 1/(f(m)*f(m+2)) = 1/(k^2).
Sum_{m>=0} 1/(f(2*m)*f(2*m+2)) = (sqrt(k^2+4) - k)/(2*k). (End)
E.g.f.: (1/53)*exp(7*x/2)*(53*cosh(sqrt(53)*x/2) + 7*sqrt(53)*sinh(sqrt(53)*x/2)). - Stefano Spezia, Feb 26 2020
G.f.: x/(1 - 7*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 7 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

Extensions

Formula corrected by Johannes W. Meijer, May 30 2010, Jun 02 2010
Extended by T. D. Noe, May 23 2011

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)

A099368 Twice Chebyshev polynomials of the first kind, T(n,x), evaluated at x=51/2.

Original entry on oeis.org

2, 51, 2599, 132498, 6754799, 344362251, 17555720002, 894997357851, 45627309530399, 2326097788692498, 118585359913786999, 6045527257814444451, 308203304788622880002, 15712323016961952435651, 801020270560270951338199
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

a(n) and b(n):= A097836(n-1) with b(0) = 0 are the improper and proper nonnegative solutions of the Pell equation a(n)^2 - 53*(7*b(n))^2 = +4. - Wolfdieter Lang, Jun 27 2013

Links

Programs

  • Mathematica
    LinearRecurrence[{51, -1}, {2, 51}, 15] (* or *) CoefficientList[Series[(2 - 51 x)/(1 - 51 x + x^2), {x, 0, 14}], x] (* Michael De Vlieger, Feb 08 2017 *)

Formula

a(n) = 51*a(n-1) - a(n-2), n >= 1; a(-1)=51, a(0)=2.
a(n) = S(n, 51) - S(n-2, 51) = 2*T(n, 51/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 51)=A097836(n). U-, resp. T-, are Chebyshev polynomials of the second, resp. first, case. See A049310 and A053120.
a(n)= ap^n + am^n, with ap:=(51 + 7*sqrt(53))/2 and am:=(51 - 7*sqrt(53))/2.
G.f.: (2-51*x)/(1-51*x+x^2).
Showing 1-4 of 4 results.

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