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 1-3 of 3 results.

A086902 a(n) = 7*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 7.

Original entry on oeis.org

2, 7, 51, 364, 2599, 18557, 132498, 946043, 6754799, 48229636, 344362251, 2458765393, 17555720002, 125348805407, 894997357851, 6390330310364, 45627309530399, 325781497023157, 2326097788692498, 16608466017870643, 118585359913786999, 846705985414379636
Offset: 0

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Author

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Sep 18 2003

Keywords

Comments

a(n+1)/a(n) converges to (7+sqrt(53))/2 = 7.14005... = A176439.
Lim a(n)/a(n+1) as n approaches infinity = 0.1400549... = 2/(7+sqrt(53)) = (sqrt(53)-7)/2 = 1/A176439 = A176439 - 7.
From Johannes W. Meijer, Jun 12 2010: (Start)
In general sequences with recurrence a(n) = k*a(n-1)+a(n-2) with a(0)=2 and a(1)=k [and a(-1)=0] have generating function (2-k*x)/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n+1) = b(5n), a(3n+2)=b(5*n+3), a(3n+3)=2*b(5n+4) where b(n) is the sequence of numerators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n)/2, for n>=1, is the sequence of numerators of continued fraction convergents to sqrt(k^2/4+1).]
For the sequence given above k=7 which implies that it is associated with A041090.
For a similar statement about sequences with recurrence a(n) = k*a(n-1)+a(n-2) but with a(0)=1 [and a(-1)=0] see A054413; a sequence that is associated with A041091.
For more information follow the Khovanova link and see A087130, A140455 and A178765.
(End)

Examples

			a(4) = 7*a(3) + a(2) = 7*364 + 51 = 2599.

Links

Crossrefs

Cf. A058316, A176439.
Cf. A000032 (k=1), A006497 (k=3), A087130 (k=5), A086902 (k=7), A087798 (k=9), A001946 (k=11), A088316 (k=13), A090301 (k=15), A090306 (k=17). - Johannes W. Meijer, Jun 12 2010

Programs

  • Magma
    I:=[2,7]; [n le 2 select I[n] else 7*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016
  • Mathematica
    RecurrenceTable[{a[0] == 2, a[1] == 7, a[n] == 7 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)
LinearRecurrence[{7,1},{2,7},30] (* Harvey P. Dale, May 25 2023 *)
  • PARI
    a(n)=([0,1; 1,7]^n*[2;7])[1,1] \\ Charles R Greathouse IV, Apr 06 2016

Formula

a(n) = ((7+sqrt(53))/2)^n + ((7-sqrt(53))/2)^n.
E.g.f. : 2exp(7x/2)cosh(sqrt(53)x/2); a(n)=2^(1-n)sum{k=0..floor(n/2), C(n, 2k)53^k7^(n-2k)}. a(n)=2T(n, 7i/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
G.f.: (2-7x)/(1-7x-x^2). - Philippe Deléham, Nov 16 2008
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 7*A097837(n), a(2n) = A099368(n).
a(3n+1) = A041090(5n), a(3n+2) = A041090(5*n+3), a(3n+3) = 2*A041090(5n+4).
Limit(a(n+k)/a(k), k=infinity) = (A086902(n) + A054413(n-1)*sqrt(53))/2.
Limit(A086902(n)/A054413(n-1), n=infinity) = sqrt(53). (End)

A099367 a(n) = A054413(n-1)^2, n >= 1.

Original entry on oeis.org

0, 1, 49, 2500, 127449, 6497401, 331240000, 16886742601, 860892632649, 43888637522500, 2237459621014849, 114066552034234801, 5815156694124960000, 296458924848338725201, 15113590010571150025249, 770496631614280312562500
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

See the comment in A099279. This is example a=7.

Links

Crossrefs

Cf. A054413.
Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, this sequence, A099369, A099372, A099374.

Programs

  • Mathematica
    LinearRecurrence[{50,50,-1},{0,1,49},20] (* Harvey P. Dale, Jul 27 2023 *)

Formula

a(n) = A054413(n-1)^2, n >= 1. a(0)=0.
a(n) = 50*a(n-1) + 50*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=49.
a(n) = 51*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = 2*(T(n, 51/2) - (-1)^n)/53 with twice the Chebyshev polynomials of the first kind: 2*T(n, 51/2) = A099368(n).
G.f.: x*(1-x)/((1-51*x+x^2)*(1+x)) = x*(1-x)/(1-50*x-50*x^2+x^3).
a(n+1) = (1 + (-1)^n)/2 + 49*Sum_{k=1..n} k*a(n+1-k). - Michael A. Allen, Feb 21 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (7 + sqrt(53))/14 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A097836 Chebyshev polynomials S(n,51).

Original entry on oeis.org

1, 51, 2600, 132549, 6757399, 344494800, 17562477401, 895341852651, 45644872007800, 2326993130545149, 118631004785794799, 6047854250944989600, 308321935793408674801, 15718370871212897425251, 801328592496064360013000
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

Used for all positive integer solutions of Pell equation x^2 - 53*y^2 = -4. See A097837 with A097838.
a(n-1), with a(-1) := 0, and b(n) := A099368(n) give the proper and improper nonnegative solutions of the Pell equation b(n)^2 - 53*(7*a(n-1))^2 = +4, n >= 0. - Wolfdieter Lang, Jun 27 2013

Links

Programs

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

Formula

a(n) = S(n, 51)=U(n, 51/2)= S(2*n+1, sqrt(53))/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).
a(n) = 51*a(n-1) - a(n-2), n >= 1, a(-1)=0, a(0)=1, a(1)=51.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap := (51+7*sqrt(53))/2 and am := (51-7*sqrt(53))/2 = 1/ap.
G.f.: 1/(1-51*x+x^2).
Showing 1-3 of 3 results.

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