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.

A041226 Numerators of continued fraction convergents to sqrt(125).

Original entry on oeis.org

11, 56, 67, 123, 682, 15127, 76317, 91444, 167761, 930249, 20633239, 104096444, 124729683, 228826127, 1268860318, 28143753123, 141987625933, 170131379056, 312119004989, 1730726404001, 38388099893011, 193671225869056, 232059325762067, 425730551631123
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

From Johannes W. Meijer, Jun 12 2010: (Start)
The a(n) terms of this sequence can be constructed with the terms of sequence A001946.
For the terms of the periodical sequence of the continued fraction for sqrt(125) see A010186. We observe that its period is five. (End)

Links

Crossrefs

Cf. A041227, A041018, A041046, A041090, A041150, A041226, A041318, A041426, A041550.

Programs

  • Mathematica
    Numerator[Convergents[Sqrt[125], 30]] (* Vincenzo Librandi, Oct 31 2013 *)

Formula

From Johannes W. Meijer, Jun 12 2010: (Start)
a(5n) = A001946(3n+1),
a(5n+1) = (A001946(3n+2) - A001946(3n+1))/2,
a(5n+2) = (A001946(3n+2) + A001946(3n+1))/2,
a(5n+3) = A001946(3n+2),
a(5n+4) = A001946(3n+3)/2. (End)
G.f.: -(x^9 -11*x^8 +56*x^7 -67*x^6 +123*x^5 +682*x^4 +123*x^3 +67*x^2 +56*x +11) / ((x^2 +4*x -1)*(x^4 -7*x^3 +19*x^2 -3*x +1)*(x^4 +3*x^3 +19*x^2 +7*x +1)). - Colin Barker, Nov 08 2013

Extensions

More terms from Colin Barker, Nov 08 2013

A041227 Denominators of continued fraction convergents to sqrt(125).

Original entry on oeis.org

1, 5, 6, 11, 61, 1353, 6826, 8179, 15005, 83204, 1845493, 9310669, 11156162, 20466831, 113490317, 2517253805, 12699759342, 15217013147, 27916772489, 154800875592, 3433536035513, 17322481053157, 20756017088670, 38078498141827, 211148507797805
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The a(n) terms of this sequence can be constructed with the terms of sequence A049666. For the terms of the periodical sequence of the continued fraction for sqrt(125) see A010186. We observe that its period is five. - Johannes W. Meijer, Jun 12 2010

Links

Crossrefs

Cf. A041226, A010186, A041019, A041047, A041091, A041151, A041227, A041319, A041427, A041551.

Programs

  • Magma
    I:=[1, 5, 6, 11, 61, 1353, 6826, 8179, 15005, 83204]; [n le 10 select I[n] else 1364*Self(n-5)+Self(n-10): n in [1..40]]; // Vincenzo Librandi, Dec 13 2013
  • Mathematica
    Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[125], n]]], {n, 1, 50}] (* Vladimir Joseph Stephan Orlovsky, Jun 23 2011 *)
Denominator[Convergents[Sqrt[125], 30]]  (* Vincenzo Librandi, Dec 13 2013 *)
LinearRecurrence[{0,0,0,0,1364,0,0,0,0,1},{1,5,6,11,61,1353,6826,8179,15005,83204},30] (* Harvey P. Dale, Apr 29 2022 *)

Formula

a(5*n) = A049666(3*n+1), a(5*n+1) = (A049666(3*n+2) - A049666(3*n+1))/2, a(5*n+2) = (A049666(3*n+2)+A049666(3*n+1))/2, a(5*n+3):= A049666(3*n+2) and a(5*n+4) = A049666(3*n+3)/2. - Johannes W. Meijer, Jun 12 2010
G.f.: -(x^8 -5*x^7 +6*x^6 -11*x^5 +61*x^4 +11*x^3 +6*x^2 +5*x +1) / ((x^2 +4*x -1)*(x^4 -7*x^3 +19*x^2 -3*x +1)*(x^4 +3*x^3 +19*x^2 +7*x +1)). - Colin Barker, Nov 12 2013
a(n) = 1364*a(n-5) + a(n-10). - Vincenzo Librandi, Dec 13 2013

A190567 Continued fraction expansion of 46*sqrt(46).

Original entry on oeis.org

311, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622, 1, 76, 1, 622
Offset: 0

Views

Author

Bruno Berselli, May 13 2011

Keywords

Links

Crossrefs

Cf. A010136; A040005, A040021, A010186, A040201, A040324, A040489, A040968.

Programs

  • Magma
    [311] cat &cat[ [1,76,1,622]: n in [1..18] ];
  • Magma
    I:=[311,1,76,1,622]; [n le 5 select I[n] else Self(n-4): n in [1..80]]; // Vincenzo Librandi, Jun 14 2013
  • Mathematica
    ContinuedFraction[46 Sqrt[46], 80] (* or *) PadRight[{311}, 80, {622, 1, 76, 1}]
CoefficientList[Series[(311 + x + 76 x^2 + x^3 + 311 x^4) / (1 - x^4), {x, 0, 100}], x] (* Vincenzo Librandi, Jun 14 2013 *)
  • PARI
    a(n)=if(n,[622,1,76,1][n%4+1],311) \\ Charles R Greathouse IV, May 13 2011

Formula

G.f.: (311+x+76*x^2+x^3+311*x^4)/(1-x^4).
a(n) = 1+3*(1+(-1)^n)*(116+91*i^n)/2 with n>0, i=sqrt(-1) and a(0)=311.
a(n) = (-1513*(n mod 4)+575*((n+1) mod 4)+125*((n+2) mod 4)+2213*((n+3) mod 4))/12 for n>0.
a(n) = a(n-4), n>=5. - Vincenzo Librandi, Jun 14 2013

A172074 Continued fraction expansion of sqrt(12500)+50 = 100*phi, where phi=(sqrt(5)+1)/2 is the golden ratio.

Original entry on oeis.org

161, 1, 4, 11, 1, 1, 3, 6, 1, 13, 8, 1, 6, 1, 4, 1, 1, 2, 1, 1, 1, 1, 13, 2, 1, 3, 8, 1, 2, 19, 1, 54, 1, 19, 2, 1, 8, 3, 1, 2, 13, 1, 1, 1, 1, 2, 1, 1, 4, 1, 6, 1, 8, 13, 1, 6, 3, 1, 1, 11, 4, 1, 222
Offset: 0

Views

Author

Shane Findley, Jan 25 2010

Keywords

Comments

The 62 trailing terms are repeated infinitely.
This is just one of an infinite set of continued fractions, related to the golden ratio, and more specifically to the square root of 125, 12500, 1250000...
Taking phi*10^k, one can look at sqrt(125) + 5, sqrt(12500) + 50 (this sequence), sqrt(1250000) + 500, etc.
This is not an efficient way to calculate phi. - Franklin T. Adams-Watters, Sep 10 2011
Periodic with a period of length 62, starting right after the initial term. Moreover, the sequence is symmetric when any 54 or 222 is taken as central value (cf. formula). - M. F. Hasler, Sep 09 2011

Crossrefs

Cf. A001622, A010186.

Programs

  • Mathematica
    ContinuedFraction[N[Sqrt[12500], 50000], 63]
ContinuedFraction[100*GoldenRatio,100] (* Harvey P. Dale, Dec 30 2018 *)
  • PARI
    default(realprecision, 199); contfrac((sqrt(5)+1)/.02)  \\ M. F. Hasler, Sep 09 2011
  • PARI
    a(n)=[222-61*!n, 1, 4, 11, 1, 1, 3, 6, 1, 13, 8, 1, 6, 1, 4, 1, 1, 2, 1, 1, 1, 1, 13, 2, 1, 3, 8, 1, 2, 19, 1, 54][32-abs(n%62-31)]  \\ M. F. Hasler, Sep 09 2011

Formula

a(31*k - n) = a(31*k + n), for all n < 31k, k > 0. - M. F. Hasler, Sep 09 2011

Extensions

Clarified the definition, following an observation by Franklin T. Adams-Watters. M. F. Hasler, Sep 09 2011
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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