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 1-9 of 9 results.

A001542 a(n) = 6*a(n-1) - a(n-2) for n > 1, a(0)=0 and a(1)=2.

Original entry on oeis.org

0, 2, 12, 70, 408, 2378, 13860, 80782, 470832, 2744210, 15994428, 93222358, 543339720, 3166815962, 18457556052, 107578520350, 627013566048, 3654502875938, 21300003689580, 124145519261542, 723573111879672
Offset: 0

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Author

Keywords

Comments

Consider the equation core(x) = core(2x+1) where core(x) is the smallest number such that x*core(x) is a square: solutions are given by a(n)^2, n > 0. - Benoit Cloitre, Apr 06 2002
Terms > 0 give numbers k which are solutions to the inequality |round(sqrt(2)*k)/k - sqrt(2)| < 1/(2*sqrt(2)*k^2). - Benoit Cloitre, Feb 06 2006
Also numbers m such that A125650(6*m^2) is an even perfect square, where A124650(m) is a numerator of m*(m+3)/(4*(m+1)*(m+2)) = Sum_{k=1..m} 1/(k*(k+1)*(k+2)). Sequence A033581 is a bisection of A125651. - Alexander Adamchuk, Nov 30 2006
The upper principal convergents to 2^(1/2), beginning with 3/2, 17/12, 99/70, 577/408, comprise a strictly decreasing sequence; essentially, numerators = A001541 and denominators = {a(n)}. - Clark Kimberling, Aug 26 2008
Even Pell numbers. - Omar E. Pol, Dec 10 2008
Numbers k such that 2*k^2+1 is a square. - Vladimir Joseph Stephan Orlovsky, Feb 19 2010
These are the integer square roots of the Half-Squares, A007590(k), which occur at values of k given by A001541. Also the numbers produced by adding m + sqrt(floor(m^2/2) + 1) when m is in A002315. See array in A227972. - Richard R. Forberg, Aug 31 2013
A001541(n)/a(n) is the closest rational approximation of sqrt(2) with a denominator not larger than a(n), and 2*a(n)/A001541(n) is the closest rational approximation of sqrt(2) with a numerator not larger than 2*a(n). These rational approximations together with those obtained from the sequences A001653 and A002315 give a complete set of closest rational approximations of sqrt(2) with restricted numerator as well as denominator. - A.H.M. Smeets, May 28 2017
Conjecture: Numbers k such that c/m < k for all natural a^2 + b^2 = c^2 (Pythagorean triples), a < b < c and a+b+c = m. Numbers which correspondingly minimize c/m are A002939. - Lorraine Lee, Jan 31 2020
All of the positive integer solutions of a*b + 1 = x^2, a*c + 1 = y^2, b*c + 1 = z^2, x + z = 2*y, 0 < a < b < c are given by a=a(n), b=A005319(n), c=a(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022

Examples

			G.f. = 2*x + 12*x^2 + 70*x^3 + 408*x^4 + 2378*x^5 + 13860*x^6 + ...
		

References

  • Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002; pp. 480-481.
  • Thomas Koshy, Fibonacci and Lucas Numbers with Applications, 2001, Wiley, pp. 77-79.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-258.
  • P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

Crossrefs

Bisection of Pell numbers A000129: {a(n)} and A001653(n+1), n >= 0.

Programs

  • GAP
    a:=[0,2];; for n in [3..20] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
  • Haskell
    a001542 n = a001542_list !! n
    a001542_list =
       0 : 2 : zipWith (-) (map (6 *) $ tail a001542_list) a001542_list
    -- Reinhard Zumkeller, Aug 14 2011
    
  • Magma
    I:=[0,2]; [n le 2 select I[n] else 6*Self(n-1) -Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 23 2019
    
  • Maple
    A001542:=2*z/(1-6*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation
    seq(combinat:-fibonacci(2*n, 2), n = 0..20); # Peter Luschny, Jun 28 2018
  • Mathematica
    LinearRecurrence[{6, -1}, {0, 2}, 30] (* Harvey P. Dale, Jun 11 2011 *)
    Fibonacci[2*Range[0,20], 2] (* G. C. Greubel, Dec 23 2019 *)
    Table[2 ChebyshevU[-1 + n, 3], {n, 0, 20}] (* Herbert Kociemba, Jun 05 2022 *)
  • Maxima
    a[0]:0$
    a[1]:2$
    a[n]:=6*a[n-1]-a[n-2]$
    A001542(n):=a[n]$
    makelist(A001542(x),x,0,30); /* Martin Ettl, Nov 03 2012 */
    
  • PARI
    {a(n) = imag( (3 + 2*quadgen(8))^n )}; /* Michael Somos, Jan 20 2017 */
    
  • PARI
    vector(21, n, 2*polchebyshev(n-1, 2, 33) ) \\ G. C. Greubel, Dec 23 2019
    
  • Python
    l=[0, 2]
    for n in range(2, 51): l+=[6*l[n - 1] - l[n - 2], ]
    print(l) # Indranil Ghosh, Jun 06 2017
    
  • Sage
    [2*chebyshev_U(n-1,3) for n in (0..20)] # G. C. Greubel, Dec 23 2019
    

Formula

a(n) = 2*A001109(n).
a(n) = ((3+2*sqrt(2))^n - (3-2*sqrt(2))^n) / (2*sqrt(2)).
G.f.: 2*x/(1-6*x+x^2).
a(n) = sqrt(2*(A001541(n))^2 - 2)/2. - Barry E. Williams, May 07 2000
a(n) = (C^(2n) - C^(-2n))/sqrt(8) where C = sqrt(2) + 1. - Gary W. Adamson, May 11 2003
For all terms x of the sequence, 2*x^2 + 1 is a square. Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 10 2002
For n > 0: a(n) = A001652(n) + A046090(n) - A001653(n); e.g., 70 = 119 + 120 - 169. Also a(n) = A001652(n - 1) + A046090(n - 1) + A001653(n - 1); e.g., 70 = 20 + 21 + 29. Also a(n)^2 + 1 = A001653(n - 1)*A001653(n); e.g., 12^2 + 1 = 145 = 5*29. Also a(n + 1)^2 = A084703(n + 1) = A001652(n)*A001652(n + 1) + A046090(n)*A046090(n + 1). - Charlie Marion, Jul 01 2003
a(n) = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))/(2*sqrt(2)). - Antonio Alberto Olivares, Dec 24 2003
2*A001541(k)*A001653(n)*A001653(n+k) = A001653(n)^2 + A001653(n+k)^2 + a(k)^2; e.g., 2*3*5*29 = 5^2 + 29^2 + 2^2; 2*99*29*5741 = 29^2 + 5741^2 + 70^2. - Charlie Marion, Oct 12 2007
a(n) = sinh(2*n*arcsinh(1))/sqrt(2). - Herbert Kociemba, Apr 24 2008
For n > 0, a(n) = A001653(n) + A002315(n-1). - Richard R. Forberg, Aug 31 2013
a(n) = 3*a(n-1) + 2*A001541(n-1); e.g., a(4) = 70 = 3*12 + 2*17. - Zak Seidov, Dec 19 2013
a(n)^2 + 1^2 = A115598(n)^2 + (A115598(n)+1)^2. - Hermann Stamm-Wilbrandt, Jul 27 2014
E.g.f.: exp(3*x)*sinh(2*sqrt(2)*x)/sqrt(2). - Ilya Gutkovskiy, Dec 07 2016
A007814(a(n)) = A001511(n). See Mathematical Reflections link. - Michel Marcus, Jan 06 2017
a(n) = -a(-n) for all n in Z. - Michael Somos, Jan 20 2017
From A.H.M. Smeets, May 28 2017: (Start)
A051009(n) = a(2^(n-2)).
a(2n) = 2*a(2)*A001541(n).
A001541(n)/a(n) > sqrt(2) > 2*a(n)/A001541(n). (End)
a(A298210(n)) = A002349(2*n^2). - A.H.M. Smeets, Jan 25 2018
a(n) = A000129(n)*A002203(n). - Adam Mohamed, Jul 20 2024

A001601 a(n) = 2*a(n-1)^2 - 1, if n>1. a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 17, 577, 665857, 886731088897, 1572584048032918633353217, 4946041176255201878775086487573351061418968498177
Offset: 0

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Author

Keywords

Comments

Reduced numerators of Newton's iteration for sqrt(2). - Eric W. Weisstein
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Evaluation of the 2^n - 1 degree interpolating polynomial of 1/x at Chebyshev nodes in the interval (1,2): v = 1.0; for(i = 1, n, v *= 4*(a(i) - x*v)); v *= 2/a(n+1). - Jose Hortal, Apr 07 2012
Smallest positive integer x satisfying the Pell equation x^2 - 2^(2*n+1) * y^2 = 1, for n > 0. - A.H.M. Smeets, Sep 29 2017

References

  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 376.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = A001333(2^n).

Programs

  • Mathematica
    Table[Simplify[Expand[(1/2) ((1 + Sqrt[2])^(2^n) + (1 - Sqrt[2])^(2^n))]], {n, 0, 7}]  (* Artur Jasinski, Oct 10 2008 *)
    Join[{1},NestList[2#^2-1&,3,7]]  (* Harvey P. Dale, Mar 24 2011 *)
  • PARI
    a(n)=if(n<1,n==0,2*a(n-1)^2-1)

Formula

From Mario Catalani (mario.catalani(AT)unito.it), May 27 2003, May 30 2003: (Start)
a(n) = a(n-1)^2 + 2*A051009(n)^2 for n > 0.
a(n)^2 = 2*A051009(n+1)^2 + 1.
a(n) = Sum_{r=0..2^(n-1)} binomial(2^n, 2*r)*2^r. (End)
Expansion of 1/sqrt(2) as an infinite product: 1/sqrt(2) = Product_{k>=1} (1 - 1/(a(n)+1)). a(1)=3; a(n) = floor(1/(1-1/(sqrt(2)*Product_{k=1..n-1} 1-1/(a(k)+1)))). - Thomas Baruchel, Nov 06 2003
2*a(n+1) = A003423(n).
a(n) = (1/2)*((1 + sqrt(2))^(2^n) + (1 - sqrt(2))^(2^n)). - Artur Jasinski, Oct 10 2008
For n > 1: a(n) - 1 = 4^n * Product_{i=1..n-2} a(i)^2. - Jose Hortal, Apr 13 2012
From Peter Bala, Nov 11 2012: (Start)
4*sqrt(2)/7 = Product_{n>=1} (1 - 1/(2*a(n))).
sqrt(2) = Product_{n>=1} (1 + 1/a(n)).
a(n) = (1/2)*A003423(n-1). (End)
a(n) = cos(2^(n-1) * arccos(3)) = cosh(2^(n-1) * log(3 + 2*sqrt(2))) for n >= 1. - Daniel Suteu, Jul 28 2016
a(n+1) = T(2^n,3), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Feb 01 2017
a(n) = A001541(2^(n-1)). - A.H.M. Smeets, May 28 2017

A244012 Numerators of rational approximations to sqrt(7) obtained from Newton's method.

Original entry on oeis.org

2, 11, 233, 108497, 23543191457, 1108563727961872518977, 2457827077905448997994482872789298261401217, 12081827889770476116093110581355561229584727594431650162181251776430351279198649072897
Offset: 0

Views

Author

N. J. A. Sloane, Jun 18 2014

Keywords

Examples

			2, 11/4, 233/88, 108497/41008, 23543191457/8898489952, ...
		

Crossrefs

Cf. A244013 (denominators).
The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

Programs

  • Maple
    N:=7;
    s:=[floor(sqrt(N))];
    M:=8;
    for n from 1 to M do
    x:=s[n];
    h:=(N-x^2)/(2*x);
    s:=[op(s),x+h]; od:
    lprint(s);
    s1:=map(numer,s);
    s2:=map(denom,s);

A244013 Denominators of rational approximations to sqrt(7) obtained from Newton's method.

Original entry on oeis.org

1, 4, 88, 41008, 8898489952, 418997705236253480128, 928971316248341903257187589777603944778112, 4566501711345281867283814391125123371716411674583075407993026856131137508750543524608
Offset: 0

Views

Author

N. J. A. Sloane, Jun 18 2014

Keywords

Examples

			2, 11/4, 233/88, 108497/41008, 23543191457/8898489952, ...
		

Crossrefs

Cf. A244012 (numerators).
The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

Programs

  • Maple
    N:=7;
    s:=[floor(sqrt(N))];
    M:=8;
    for n from 1 to M do
    x:=s[n];
    h:=(N-x^2)/(2*x);
    s:=[op(s),x+h]; od:
    lprint(s);
    s1:=map(numer,s);
    s2:=map(denom,s);

A244014 Numerators of rational approximations to sqrt(6) obtained from Newton's method.

Original entry on oeis.org

2, 5, 49, 4801, 46099201, 4250272665676801, 36129635465198759610694779187201, 2610701117696295981568349760414651575095962187244375364404428801
Offset: 0

Views

Author

N. J. A. Sloane, Jun 18 2014

Keywords

Examples

			2, 5/2, 49/20, 4801/1960, 46099201/18819920, ...
		

Crossrefs

The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

Programs

  • Maple
    N:=6;
    s:=[floor(sqrt(N))];
    M:=8;
    for n from 1 to M do
    x:=s[n];
    h:=(N-x^2)/(2*x);
    s:=[op(s),x+h]; od:
    lprint(s);
    s1:=map(numer,s);
    s2:=map(denom,s);

A244015 Denominators of rational approximations to sqrt(6) obtained from Newton's method.

Original entry on oeis.org

1, 2, 20, 1960, 18819920, 1735166549767840, 14749861913749949808286047759680, 1065814268211609269094400465471990022332221793124358274759711360
Offset: 0

Views

Author

N. J. A. Sloane, Jun 18 2014

Keywords

Examples

			2, 5/2, 49/20, 4801/1960, 46099201/18819920, ...
		

Crossrefs

Cf. A244014 (numerators).
The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

Programs

  • Magma
    m:=9; f:=[n eq 1 select 2 else (Self(n-1)+6/Self(n-1))/2: n in [1..m]]; [Denominator(f[n]): n in [1..m]]; // Vincenzo Librandi, Jan 12 2016
  • Maple
    N:=6;
    s:=[floor(sqrt(N))];
    M:=8;
    for n from 1 to M do
    x:=s[n];
    h:=(N-x^2)/(2*x);
    s:=[op(s),x+h]; od:
    lprint(s);
    s1:=map(numer,s);
    s2:=map(denom,s);

A163261 Numerators of fractions in the approximation of the square root of 2 by means of: f(n)= 3f(n-1)/(f(n-1)^2+1); with f(1)= 1.

Original entry on oeis.org

1, 3, 18, 702, 1038258, 2292015792222, 11135478955084481409955218, 263108590829588406010634295716681354685770554450302
Offset: 1

Views

Author

Mark Dols, Jul 23 2009

Keywords

Comments

For root c: f(n) = (1+c)*f(n-1)/(f(n-1)^2+1).

Crossrefs

Cf. A002193 (sqrt(2)), A163262 (denominators).
Cf. A001601 and A051009: Newton's method.

Programs

  • PARI
    f(n) = if (n==1, 1, 3*f(n-1)/(f(n-1)^2+1));
    a(n) = numerator(f(n)); \\ Michel Marcus, Mar 05 2019

Formula

a(n+1) = 3*a(n)*A163262(n). - Philippe Deléham, Oct 11 2011

Extensions

More terms from Michel Marcus, Mar 05 2019

A163262 Denominators of fractions in the approximation of the square root of 2 by means of: f(n) = 3*f(n-1)/(f(n-1)^2+1); with f(1)= 1.

Original entry on oeis.org

1, 2, 13, 493, 735853, 1619459312173, 7875984855578888541679213, 186030029004437379749629399827828117533654561726893
Offset: 1

Views

Author

Mark Dols, Jul 23 2009

Keywords

Comments

For root of c: f(n) = (1+c)*f(n-1)/(f(n-1)^2+1).
a(9) has 102 digits. - Emeric Deutsch, Jul 29 2009

Crossrefs

Cf. A002193 (sqrt(2)), A163261 (numerators).
Cf. A001601 and A051009: Newton's method.

Programs

  • Maple
    f[1] := 1: for n from 2 to 10 do f[n] := 3*f[n-1]/(1+f[n-1]^2) end do: seq(denom(f[n]), n = 1 .. 8); # Emeric Deutsch, Jul 29 2009
  • PARI
    f(n) = if (n==1, 1, 3*f(n-1)/(f(n-1)^2+1));
    a(n) = denominator(f(n)); \\ Michel Marcus, Mar 04 2019

Extensions

a(7) and a(8) from Emeric Deutsch, Jul 29 2009
Name edited by Michel Marcus, Mar 04 2019

A179823 Denominators in the approximation of sqrt(2) satisfying the recurrence: a(n)= [a(n-1)*a(n-2)+2]/[a(n-1)+a(n-2)] with a(1)=a(2)=1.

Original entry on oeis.org

1, 1, 2, 5, 29, 408, 33461, 38613965, 3654502875938, 399133058537705128729, 4125636888562548868221559797461449, 4657508918199804645965719872781284840798220312648198320
Offset: 1

Views

Author

Mark Dols, Jul 28 2010

Keywords

Comments

The recurrence is a transform of the Babylonian (Newton's) method for square root computation: a(n+1)= N/2a(n)+a(n)/2 = (a(n)^2+N)/2a(n).

Examples

			1/1=1.0, 1/1=1.0, 3/2=1.5, 7/5=1.4, 42/29=1.41379.., 577/408=1.4142156,... - _R. J. Mathar_, Nov 03 2016
		

Crossrefs

Programs

  • Mathematica
    a[1] = a[2] = 1; a[n_] := (a[n - 1] a[n - 2] + 2)/(a[n - 1] + a[n - 2]); Denominator@ Array[ a, 12] (* Robert G. Wilson v, Aug 03 2010 *)

Extensions

a(10) - a(12) from Robert G. Wilson v, Aug 03 2010
Showing 1-9 of 9 results.