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.

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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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