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.

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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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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

A002530 a(n) = 4*a(n-2) - a(n-4) for n > 1, a(n) = n for n = 0, 1.

Original entry on oeis.org

0, 1, 1, 3, 4, 11, 15, 41, 56, 153, 209, 571, 780, 2131, 2911, 7953, 10864, 29681, 40545, 110771, 151316, 413403, 564719, 1542841, 2107560, 5757961, 7865521, 21489003, 29354524, 80198051, 109552575, 299303201, 408855776, 1117014753, 1525870529, 4168755811
Offset: 0

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Comments

Denominators of continued fraction convergents to sqrt(3), for n >= 1.
Also denominators of continued fraction convergents to sqrt(3) - 1. See A048788 for numerators. - N. J. A. Sloane, Dec 17 2007. Convergents are 1, 2/3, 3/4, 8/11, 11/15, 30/41, 41/56, 112/153, ...
Consider the mapping f(a/b) = (a + 3*b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 2/1, 5/3, 7/4, 19/11, ... converging to 3^(1/2). Sequence contains the denominators. The same mapping for N, i.e., f(a/b) = (a + Nb)/(a + b) gives fractions converging to N^(1/2). - Amarnath Murthy, Mar 22 2003
Sqrt(3) = 2/2 + 2/3 + 2/(3*11) + 2/(11*41) + 2/(41*153) + 2/(153*571), ...; the sum of the first 6 terms of this series = 1.7320490367..., while sqrt(3) = 1.7320508075... - Gary W. Adamson, Dec 15 2007
From Clark Kimberling, Aug 27 2008: (Start)
Related convergents (numerator/denominator):
lower principal convergents: A001834/A001835
upper principal convergents: A001075/A001353
intermediate convergents: A005320/A001075
principal and intermediate convergents: A143642/A140827
lower principal and intermediate convergents: A143643/A005246. (End)
Row sums of triangle A152063 = (1, 3, 4, 11, ...). - Gary W. Adamson, Nov 26 2008
From Alois P. Heinz, Apr 13 2011: (Start)
Also number of domino tilings of the 3 X (n-1) rectangle with upper left corner removed iff n is even. For n=4 the 4 domino tilings of the 3 X 3 rectangle with upper left corner removed are:
. ._. . ._. . ._. . ._.
.|__| .|__| .| | | .|___|
| |_| | | | | | ||| |_| |
||__| |||_| ||__| |_|_| (End)
This is the sequence of Lehmer numbers u_n(sqrt(R),Q) with the parameters R = 2 and Q = -1. It is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all natural numbers n and m. - Peter Bala, Apr 18 2014
2^(-floor(n/2))*(1 + sqrt(3))^n = A002531(n) + a(n)*sqrt(3); integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 11 2018
Let T(n) = 2^(n mod 2), U(n) = a(n), V(n) = A002531(n), x(n) = V(n)/U(n). Then T(n*m) * U(n+m) = U(n)*V(m) + U(m)*V(n), T(n*m) * V(n+m) = 3*U(n)*U(m) + V(m)*V(n), x(n+m) = (3 + x(n)*x(m))/(x(n) + x(m)). - Michael Somos, Nov 29 2022

Examples

			Convergents to sqrt(3) are: 1, 2, 5/3, 7/4, 19/11, 26/15, 71/41, 97/56, 265/153, 362/209, 989/571, 1351/780, 3691/2131, ... = A002531/A002530 for n >= 1.
1 + 1/(1 + 1/(2 + 1/(1 + 1/2))) = 19/11 so a(5) = 11.
G.f. = x + x^2 + 3*x^3 + 4*x^4 + 11*x^5 + 15*x^6 + 41*x^7 + ... - _Michael Somos_, Mar 18 2022
		

References

  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • Russell Lyons, A bird's-eye view of uniform spanning trees and forests, in Microsurveys in Discrete Probability, AMS, 1998.
  • I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 181.
  • Murat Sahin and Elif Tan, Conditional (strong) divisibility sequences, Fib. Q., 56 (No. 1, 2018), 18-31.
  • 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).
  • A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.

Crossrefs

Cf. A002531 (numerators of convergents to sqrt(3)), A048788, A003297.
Bisections: A001353 and A001835.
Cf. A152063.
Analog for sqrt(m): A000129 (m=2), A001076 (m=5), A041007 (m=6), A041009 (m=7), A041011 (m=8), A005668 (m=10), A041015 (m=11), A041017 (m=12), ..., A042935 (m=999), A042937 (m=1000).

Programs

  • Magma
    I:=[0,1,1,3]; [n le 4 select I[n] else 4*Self(n-2) - Self(n-4): n in [1..50]]; // G. C. Greubel, Feb 25 2019
    
  • Maple
    a := proc(n) option remember; if n=0 then 0 elif n=1 then 1 elif n=2 then 1 elif n=3 then 3 else 4*a(n-2)-a(n-4) fi end; [ seq(a(i),i=0..50) ];
    A002530:=-(-1-z+z**2)/(1-4*z**2+z**4); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
  • Mathematica
    Join[{0},Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[3],n]]], {n,1,50}]] (* Stefan Steinerberger, Apr 01 2006 *)
    Join[{0},Denominator[Convergents[Sqrt[3],50]]] (* or *) LinearRecurrence[ {0,4,0,-1},{0,1,1,3},50] (* Harvey P. Dale, Jan 29 2013 *)
    a[ n_] := If[n<0, -(-1)^n, 1] SeriesCoefficient[ x*(1+x-x^2)/(1-4*x^2+x^4), {x, 0, Abs@n}]; (* Michael Somos, Apr 18 2019 *)
    a[ n_] := ChebyshevU[n-1, Sqrt[-1/2]]*Sqrt[2]^(Mod[n, 2]-1)/I^(n-1) //Simplify; (* Michael Somos, Nov 29 2022 *)
  • PARI
    {a(n) = if( n<0, -(-1)^n * a(-n), contfracpnqn(vector(n, i, 1 + (i>1) * (i%2)))[2, 1])}; /* Michael Somos, Jun 05 2003 */
    
  • PARI
    { for (n=0, 50, a=contfracpnqn(vector(n, i, 1+(i>1)*(i%2)))[2, 1]; write("b002530.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 01 2009
    
  • PARI
    my(w=quadgen(12)); A002530(n)=real((2+w)^(n\/2)*if(bittest(n,0),1-w/3,w/3));
    apply(A002530, [0..30]) \\ M. F. Hasler, Nov 04 2019
    
  • Python
    from functools import cache
    @cache
    def a(n): return [0, 1, 1, 3][n] if n < 4 else 4*a(n-2) - a(n-4)
    print([a(n) for n in range(36)]) # Michael S. Branicky, Nov 13 2022
  • Sage
    (x*(1+x-x^2)/(1-4*x^2+x^4)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019
    

Formula

G.f.: x*(1 + x - x^2)/(1 - 4*x^2 + x^4).
a(n) = 4*a(n-2) - a(n-4). [Corrected by László Szalay, Feb 21 2014]
a(n) = -(-1)^n * a(-n) for all n in Z, would satisfy the same recurrence relation. - Michael Somos, Jun 05 2003
a(2*n) = a(2*n-1) + a(2*n-2), a(2*n+1) = 2*a(2*n) + a(2*n-1).
From Benoit Cloitre, Dec 15 2002: (Start)
a(2*n) = ((2 + sqrt(3))^n - (2 - sqrt(3))^n)/(2*sqrt(3)).
a(2*n) = A001353(n).
a(2*n-1) = ceiling((1 + 1/sqrt(3))/2*(2 + sqrt(3))^n) = ((3 + sqrt(3))^(2*n - 1) + (3 - sqrt(3))^(2*n - 1))/6^n.
a(2*n-1) = A001835(n). (End)
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n - k, k) * 2^floor((n - 2*k)/2). - Paul Barry, Jul 13 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(floor(n/2) + k, floor((n - 1)/2 - k))*2^k. - Paul Barry, Jun 22 2005
G.f.: (sqrt(6) + sqrt(3))/12*Q(0), where Q(k) = 1 - a/(1 + 1/(b^(2*k) - 1 - b^(2*k)/(c + 2*a*x/(2*x - g*m^(2*k)/(1 + a/(1 - 1/(b^(2*k + 1) + 1 - b^(2*k + 1)/(h - 2*a*x/(2*x + g*m^(2*k + 1)/Q(k + 1)))))))))). - Sergei N. Gladkovskii, Jun 21 2012
a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, and a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even, where alpha = 1/2*(sqrt(2) + sqrt(6)) and beta = (1/2)*(sqrt(2) - sqrt(6)). Cf. A108412. - Peter Bala, Apr 18 2014
a(n) = (-sqrt(2)*i)^n*S(n, sqrt(2)*i)*2^(-floor(n/2)) = A002605(n)*2^(-floor(n/2)), n >= 0, with i = sqrt(-1) and S the Chebyshev polynomials (A049310). - Wolfdieter Lang, Feb 10 2018
a(n+1)*a(n+2) - a(n+3)*a(n) = (-1)^n, n >= 0. - Kai Wang, Feb 06 2020
E.g.f.: sinh(sqrt(3/2)*x)*(sinh(x/sqrt(2)) + sqrt(2)*cosh(x/sqrt(2)))/sqrt(3). - Stefano Spezia, Feb 07 2020
a(n) = ((1 + sqrt(3))^n - (1 - sqrt(3))^n)/(2*2^floor(n/2))/sqrt(3) = A002605(n)/2^floor(n/2). - Robert FERREOL, Apr 13 2023

Extensions

Definition edited by M. F. Hasler, Nov 04 2019

A004187 a(n) = 7*a(n-1) - a(n-2) with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 7, 48, 329, 2255, 15456, 105937, 726103, 4976784, 34111385, 233802911, 1602508992, 10983760033, 75283811239, 516002918640, 3536736619241, 24241153416047, 166151337293088, 1138818207635569, 7805576116155895, 53500214605455696, 366695926122033977
Offset: 0

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Author

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Comments

Define the sequence T(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}= 0 . A004187 (with initial 0 omitted) is T(1,7).
This is a divisibility sequence.
For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 7's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
a(n) and b(n) := A056854(n) are the proper and improper nonnegative solutions of the Pell equation b(n)^2 - 5*(3*a(n))^2 = +4. see the cross-reference to A056854 below. - Wolfdieter Lang, Jun 26 2013
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2,3,4,5,6}. - Milan Janjic, Jan 25 2015
The digital root is A253298, which shares its digital root with A253368. - Peter M. Chema, Jul 04 2016
Lim_{n->oo} a(n+1)/a(n) = 2 + 3*phi = 1+ A090550 = 6.854101... - Wolfdieter Lang, Nov 16 2023

Examples

			a(2) = 7*a(1) - a(0) = 7*7 - 1 = 48. - _Michael B. Porter_, Jul 04 2016
		

Crossrefs

Cf. A000027, A001906, A001353, A004254, A001109, A049685, A033888. a(n)=sqrt((A056854(n)^2 - 4)/45).
Second column of array A028412.

Programs

  • Magma
    [Fibonacci(4*n)/3 : n in [0..30]]; // Vincenzo Librandi, Jun 07 2011
    
  • Magma
    /* By definition: */ [n le 2 select n-1 else 7*Self(n-1)-Self(n-2): n in [1..23]]; // Bruno Berselli, Dec 24 2012
  • Maple
    seq(combinat:-fibonacci(4*n)/3, n = 0 .. 30); # Robert Israel, Jan 26 2015
  • Mathematica
    LinearRecurrence[{7,-1},{0,1},30] (* Harvey P. Dale, Jul 13 2011 *)
    CoefficientList[Series[x/(1 - 7*x + x^2), {x, 0, 50}], x] (* Vincenzo Librandi, Dec 23 2012 *)
  • Maxima
    a[0]:0$ a[1]:1$ a[n]:=7*a[n-1] - a[n-2]$ A004187(n):=a[n]$ makelist(A004187(n),n,0,30); /* Martin Ettl, Nov 11 2012 */
    
  • MuPAD
    numlib::fibonacci(4*n)/3 $ n = 0..25; // Zerinvary Lajos, May 09 2008
    
  • PARI
    a(n)=fibonacci(4*n)/3 \\ Charles R Greathouse IV, Mar 09 2012
    
  • PARI
    concat(0, Vec(x/(1-7*x+x^2) + O(x^99))) \\ Altug Alkan, Jul 03 2016
    
  • Sage
    [lucas_number1(n,7,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [fibonacci(4*n)/3 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009
    

Formula

G.f.: x/(1-7*x+x^2).
a(n) = F(4*n)/3 = A033888(n)/3, where F=A000045 (the Fibonacci sequence).
a(n) = S(2*n-1, sqrt(9))/sqrt(9) = S(n-1, 7); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = Sum_{i = 0..n-1} C(2*n-1-i, i)*5^(n-i-1). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
[A049685(n-1), a(n)] = [1,5; 1,6]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167816(4*n). - Reinhard Zumkeller, Nov 13 2009
a(n) = (((7+sqrt(45))/2)^n-((7-sqrt(45))/2)^n)/sqrt(45). - Noureddine Chair, Aug 31 2011
a(n+1) = Sum_{k = 0..n} A101950(n,k)*6^k. - Philippe Deléham, Feb 10 2012
a(n) = (A081072(n)/3)-1. - Martin Ettl, Nov 11 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = (1/5)*(5 + 3*sqrt(5)).
Product {n >= 2} (1 - 1/a(n)) = (1/14)*(5 + 3*sqrt(5)). (End)
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -A(x)*A(-x), where A(x) = Sum_{n >= 1} Fibonacci(2*n)* x^n.
1 + 5*Sum_{n >= 1} a(n)*x^(2*n) = F(x)*F(-x) = G(x)*G(-x), where F(x) = 1 + A(x) and G(x) = 1 + 5*A(x).
1 + Sum_{n >= 1} a(n)*x^(2*n) = H(x)*H(-x) = I(x)*I(-x), where H(x) = 1 + Sum_{n >= 1} Fibonacci(2*n + 3)*x^n and I(x) = 1 + x + x*Sum_{n >= 1} Fibonacci(2*n - 1)*x^n. (End)
E.g.f.: 2*exp(7*x/2)*sinh(3*sqrt(5)*x/2)/(3*sqrt(5)). - Ilya Gutkovskiy, Jul 03 2016
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*9^k*binomial(n+k, 2*k+1). - Peter Bala, Jul 17 2023
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k*7^(n-2*k)*binomial(n-k, k). - Greg Dresden, Aug 03 2024
From Peter Bala, Jul 22 2025: (Start)
The following products telescope:
Product {n >= 2} (1 + (-1)^n/a(n)) = (3/14)*(3 + sqrt(5)).
Product {n >= 1} (1 - (-1)^n/a(n)) = (1/3)*(3 + sqrt(5)).
Product_{n >= 1} (a(2*n) + 1)/(a(2*n) - 1) = (3/5)*sqrt(5). (End)

Extensions

Entry improved by comments from Michael Somos and Wolfdieter Lang, Aug 02 2000

A290890 p-INVERT of the positive integers, where p(S) = 1 - S^2.

Original entry on oeis.org

0, 1, 4, 11, 28, 72, 188, 493, 1292, 3383, 8856, 23184, 60696, 158905, 416020, 1089155, 2851444, 7465176, 19544084, 51167077, 133957148, 350704367, 918155952, 2403763488, 6293134512, 16475640049, 43133785636, 112925716859, 295643364940, 774004377960
Offset: 0

Views

Author

Clark Kimberling, Aug 15 2017

Keywords

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
Note that in A290890, s = (1,2,3,4,...); i.e., A000027(n+1) for n>=0, whereas in A290990, s = (0,1,2,3,4,...); i.e., A000027(n) for n>=0.
Guide to p-INVERT sequences using s = (1,2,3,4,5,...) = A000027:
p(S) t(1,2,3,4,5,...)
1 - S A001906
1 - S^2 A290890; see A113067 for signed version
1 - S^3 A290891
1 - S^4 A290892
1 - S^5 A290893
1 - S^6 A290894
1 - S^7 A290895
1 - S^8 A290896
1 - S - S^2 A289780
1 - S - S^3 A290897
1 - S - S^4 A290898
1 - S^2 - S^4 A290899
1 - S^2 - S^3 A290900
1 - S^3 - S^4 A290901
1 - 2S A052530; (1/2)*A052530 = A001353
1 - 3S A290902; (1/3)*A290902 = A004254
1 - 4S A003319; (1/4)*A003319 = A001109
1 - 5S A290903; (1/5)*A290903 = A004187
1 - 2*S^2 A290904; (1/2)*A290904 = A290905
1 - 3*S^2 A290906; (1/3)*A290906 = A290907
1 - 4*S^2 A290908; (1/4)*A290908 = A099486
1 - 5*S^2 A290909; (1/5)*A290909 = A290910
1 - 6*S^2 A290911; (1/6)*A290911 = A290912
1 - 7*S^2 A290913; (1/7)*A290913 = A290914
1 - 8*S^2 A290915; (1/8)*A290915 = A290916
(1 - S)^2 A290917
(1 - S)^3 A290918
(1 - S)^4 A290919
(1 - S)^5 A290920
(1 - S)^6 A290921
1 - S - 2*S^2 A290922
1 - 2*S - 2*S^2 A290923; (1/2)*A290923 = A290924
1 - 3*S - 2*S^2 A290925
(1 - S^2)^2 A290926
(1 - S^2)^3 A290927
(1 - S^3)^2 A290928
(1 - S)(1 - S^2) A290929
(1 - S^2)(1 - S^4) A290930
1 - 3 S + S^2 A291025
1 - 4 S + S^2 A291026
1 - 5 S + S^2 A291027
1 - 6 S + S^2 A291028
1 - S - S^2 - S^3 A291029
1 - S - S^2 - S^3 - S^4 A201030
1 - 3 S + 2 S^3 A291031
1 - S - S^2 - S^3 + S^4 A291032
1 - 6 S A291033
1 - 7 S A291034
1 - 8 S A291181
1 - 3 S + 2 S^3 A291031
1 - 3 S + 2 S^2 A291182
1 - 4 S + 2 S^3 A291183
1 - 4 S + 3 S^3 A291184

Examples

			(See the examples at A289780.)
		

Crossrefs

Cf. A000027, A113067, A289780, A113067 (signed version of same sequence).

Programs

  • Mathematica
    z = 60; s = x/(1 - x)^2; p = 1 - s^2;
    Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
    Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290890 *)

Formula

G.f.: x/(1 - 4 x + 5 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) - 5*a(n-2) + 4*a(n-3) - a(n-4).

A004189 a(n) = 10*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 10, 99, 980, 9701, 96030, 950599, 9409960, 93149001, 922080050, 9127651499, 90354434940, 894416697901, 8853812544070, 87643708742799, 867583274883920, 8588189040096401, 85014307126080090, 841554882220704499, 8330534515080964900, 82463790268588944501, 816307368170808480110
Offset: 0

Views

Author

Keywords

Comments

Indices of square numbers which are also generalized pentagonal numbers.
If t(n) denotes the n-th triangular number, t(A105038(n))=a(n)*a(n+1). - Robert Phillips (bobanne(AT)bellsouth.net), May 25 2008
The n-th term is a(n) = ((5+sqrt(24))^n - (5-sqrt(24))^n)/(2*sqrt(24)). - Sture Sjöstedt, May 31 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 10's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
a(n) and b(n) (A001079) are the nonnegative proper solutions of the Pell equation b(n)^2 - 6*(2*a(n))^2 = +1. See the cross reference to A001079 below. - Wolfdieter Lang, Jun 26 2013
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,9}. - Milan Janjic, Jan 25 2015
For n > 1, this also gives the number of (n-1)-decimal-digit numbers which avoid a particular two-digit number with distinct digits. For example, there are a(5) = 9701 4-digit numbers which do not include "39" as a substring; see Wikipedia link. - Charles R Greathouse IV, Jan 14 2016
All possible solutions for y in Pell equation x^2 - 24*y^2 = 1. The values for x are given in A001079. - Herbert Kociemba, Jun 05 2022
Dickson on page 384 gives the Diophantine equation "(20) 24x^2 + 1 = y^2" and later states "... three consecutive sets (x_i, y_i) of solutions of (20) or 2x^2 + 1 = 3y^2 satisfy x_{n+1} = 10x_n - x_{n-1}, y_{n+1} = 10y_n - y_{n-1} with (x_1, y_1) = (0, 1) or (1, 1), (x_2, y_2) = (1, 5) or (11, 9), respectively." The first set of values (x_n, y_n) = (A001079(n-1), a(n-1)). - Michael Somos, Jun 19 2023

Examples

			a(2)=10 and (3(-8)^2-(-8))/2=10^2, a(3)=99 and (3(81)^2-(81))/2=99^2. - _Michael Somos_, Sep 05 2006
G.f. = x + 10*x^2 + 99*x^3 + 980*x^4 + 9701*x^5 + 96030*x^6 + ...
		

References

  • L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, p. 384.

Crossrefs

Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), this sequence (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

Programs

  • GAP
    m:=5;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
  • Magma
    [ n eq 1 select 0 else n eq 2 select 1 else 10*Self(n-1)-Self(n-2): n in [1..20] ]; // Vincenzo Librandi, Aug 19 2011
    
  • Maple
    A004189 := proc(n)
        option remember;
        if n <= 1 then
            n ;
        else
            10*procname(n-1)-procname(n-2) ;
        end if;
    end proc:
    seq(A004189(n),n=0..20) ; # R. J. Mathar, Apr 30 2017
    seq( simplify(ChebyshevU(n-1, 5)), n=0..20); # G. C. Greubel, Dec 23 2019
  • Mathematica
    Table[GegenbauerC[n-1,1,5], {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008; modified by G. C. Greubel, Jun 06 2019 *)
    LinearRecurrence[{10, -1}, {0, 1}, 20] (* Jean-François Alcover, Nov 15 2017 *)
    ChebyshevU[Range[21] -2, 5] (* G. C. Greubel, Dec 23 2019 *)
  • PARI
    {a(n) = subst(poltchebi(n+1) - 5*poltchebi(n), 'x, 5) / 24}; /* Michael Somos, Sep 05 2006 */
    
  • PARI
    a(n)=([9,1;8,1]^(n-1)*[1;1])[1,1] \\ Charles R Greathouse IV, Jan 14 2016
    
  • PARI
    vector(21, n, n--; polchebyshev(n-1, 2, 5) ) \\ G. C. Greubel, Dec 23 2019
    
  • Sage
    [lucas_number1(n,10,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [chebyshev_U(n-1,5) for n in (0..20)] # G. C. Greubel, Dec 23 2019
    

Formula

a(n) = S(2*n-1, sqrt(12))/sqrt(12) = S(n-1, 10); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(-1, x) := 0.
A001079(n) = sqrt(24*(a(n)^2)+1), that is a(n) = sqrt((A001079(n)^2-1)/24).
From Barry E. Williams, Aug 18 2000: (Start)
a(n) = ( (5+2*sqrt(6))^n - (5-2*sqrt(6))^n )/(4*sqrt(6)).
G.f.: x/(1-10*x+x^2). (End)
a(-n) = -a(n). - Michael Somos, Sep 05 2006
From Mohamed Bouhamida, May 26 2007: (Start)
a(n) = 9*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 10*a(n-1) - a(n-2). (End)
a(n+1) = Sum_{k=0..n} A101950(n,k)*9^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = 1/2*(2 + sqrt(6)).
Product {n >= 2} (1 - 1/a(n)) = 1/5*(2 + sqrt(6)). (End)
a(n) = (A054320(n-1) + A072256(n))/2. - Richard R. Forberg, Nov 21 2013
a(2*n - 1) = A046173(n).
E.g.f.: exp(5*x)*sinh(2*sqrt(6)*x)/(2*sqrt(6)). - Stefano Spezia, Dec 12 2022
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*8^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*12^k. - Peter Bala, Jul 18 2023

A001090 a(n) = 8*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 8, 63, 496, 3905, 30744, 242047, 1905632, 15003009, 118118440, 929944511, 7321437648, 57641556673, 453811015736, 3572846569215, 28128961537984, 221458845734657, 1743541804339272, 13726875588979519, 108071462907496880, 850844827670995521, 6698687158460467288
Offset: 0

Views

Author

Keywords

Comments

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 15*y^2 = 1; the corresponding values of x are in A001091. - Vincenzo Librandi, Nov 12 2010 [edited by Jon E. Schoenfield, May 02 2014]
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 8's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,7}. - Milan Janjic, Jan 25 2015
From Klaus Purath, Jul 25 2024: (Start)
For any three consecutive terms (x, y, z) y^2 - x*z = 1 always applies.
a(n) = (t(i+2n) - t(i))/(t(i+n+1) - t(i+n-1)) where (t) is any recurrence t(k) = 9t(k-1) - 9t(k-2) + t(k-3) or t(k) = 8t(k-1) - t(k-2) without regard to initial values.
In particular, if the recurrence (t) of the form (9,-9,1) has the initial values t(0) = 1, t(1) = 2, t(2) = 9, a(n) = t(n) - 1 applies. (End)

Examples

			G.f. = x + 8*x^2 + 63*x^3 + 496*x^4 + 3905*x^5 + 30744*x^6 + 242047*x^7 + ...
		

References

  • Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
  • 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

Equals one-third A136325.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), this sequence (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

Programs

  • GAP
    m:=4;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
  • Magma
    I:=[0,1]; [n le 2 select I[n] else 8*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 20 2017
    
  • Maple
    A001090:=1/(1-8*z+z**2); # Simon Plouffe in his 1992 dissertation
    seq( simplify(ChebyshevU(n-1, 4)), n=0..20); # G. C. Greubel, Dec 23 2019
  • Mathematica
    Table[GegenbauerC[n-1, 1, 4], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
    LinearRecurrence[{8,-1},{0,1},30] (* Harvey P. Dale, Aug 29 2012 *)
    a[n_]:= ChebyshevU[n-1, 4]; (* Michael Somos, May 28 2014 *)
    CoefficientList[Series[x/(1-8*x+x^2), {x,0,20}], x] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    {a(n) = subst(poltchebi(n+1) - 4 * poltchebi(n), x, 4) / 15}; /* Michael Somos, Apr 05 2008 */
    
  • PARI
    {a(n) = polchebyshev(n-1, 2, 4)}; /* Michael Somos, May 28 2014 */
    
  • PARI
    my(x='x+O('x^30)); concat([0], Vec(x/(1-8*x-x^2))) \\ G. C. Greubel, Dec 20 2017
    
  • SageMath
    [lucas_number1(n,8,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008
    
  • SageMath
    [chebyshev_U(n-1,4) for n in (0..20)] # G. C. Greubel, Dec 23 2019
    

Formula

15*a(n)^2 - A001091(n)^2 = -1.
a(n) = sqrt((A001091(n)^2 - 1)/15).
a(n) = S(2*n-1, sqrt(10))/sqrt(10) = S(n-1, 8); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310, with S(-1, x) := 0.
From Barry E. Williams, Aug 18 2000: (Start)
a(n) = ((4+sqrt(15))^n - (4-sqrt(15))^n)/(2*sqrt(15)).
G.f.: x/(1-8*x+x^2). (End)
Limit_{n->infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 13 2002
[A070997(n-1), a(n)] = [1,6; 1,7]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(-n) = -a(n). - Michael Somos, Apr 05 2008
a(n+1) = Sum_{k=0..n} A101950(n,k)*7^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n >= 1} (1 + 1/a(n)) = (1/3)*(3 + sqrt(15)).
Product_{n >= 2} (1 - 1/a(n)) = (1/8)*(3 + sqrt(15)).
(End)
a(n) = A136325(n)/3. - Greg Dresden, Sep 12 2019
E.g.f.: exp(4*x)*sinh(sqrt(15)*x)/sqrt(15). - Stefano Spezia, Dec 12 2022
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*6^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*10^k. - Peter Bala, Jul 17 2023

Extensions

More terms from Wolfdieter Lang, Aug 02 2000

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

Original entry on oeis.org

2, 4, 14, 52, 194, 724, 2702, 10084, 37634, 140452, 524174, 1956244, 7300802, 27246964, 101687054, 379501252, 1416317954, 5285770564, 19726764302, 73621286644, 274758382274, 1025412242452, 3826890587534, 14282150107684, 53301709843202, 198924689265124
Offset: 0

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Author

Keywords

Comments

a(n) gives values of x satisfying x^2 - 3*y^2 = 4; corresponding y values are given by 2*A001353(n).
If M is any given term of the sequence, then the next one is 2*M + sqrt(3*M^2 - 12). - Lekraj Beedassy, Feb 18 2002
For n > 0, the three numbers a(n) - 1, a(n), and a(n) + 1 form a Fleenor-Heronian triangle, i.e., a Heronian triangle with consecutive sides, whose area A(n) may be obtained from the relation [4*A(n)]^2 = 3([a(2n)]^2 - 4); or A(n) = 3*A001353(2*n)/2 and whose semiperimeter is 3*a[n]/2. The sequence is symmetrical about a[0], i.e., a[-n] = a[n].
For n > 0, a(n) + 2 is the number of dimer tilings of a 2*n X 2 Klein bottle (cf. A103999).
Tsumura shows that, for prime p, a(p) is composite (contrary to a conjecture of Juricevic). - Charles R Greathouse IV, Apr 13 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 12 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 192 = 0. - Colin Barker, Feb 16 2014
A268281(n) - 1 is a member of this sequence iff A268281(n) is prime. - Frank M Jackson, Feb 27 2016
a(n) gives values of x satisfying 3*x^2 - 4*y^2 = 12; corresponding y values are given by A005320. - Sture Sjöstedt, Dec 19 2017
Middle side lengths of almost-equilateral Heronian triangles. - Wesley Ivan Hurt, May 20 2020
For all elements k of the sequence, 3*(k-2)*(k+2) is a square. - Davide Rotondo, Oct 25 2020

References

  • B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 82.
  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p.91.
  • Michael P. Cohen, Generating Heronian Triangles With Consecutive Integer Sides. Journal of Recreational Mathematics, vol. 30 no. 2 1999-2000 p. 123.
  • L. E. Dickson, History of The Theory of Numbers, Vol. 2 pp. 197;198;200;201. Chelsea NY.
  • Charles R. Fleenor, Heronian Triangles with Consecutive Integer Sides, Journal of Recreational Mathematics, Volume 28, no. 2 (1996-7) 113-115.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
  • V. D. To, "Finding All Fleenor-Heronian Triangles", Journal of Recreational Mathematics vol. 32 no.4 2003-4 pp. 298-301 Baywood NY.

Crossrefs

Cf. A011945 (areas), A334277 (perimeters).
Cf. this sequence (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).

Programs

  • Haskell
    a003500 n = a003500_list !! n
    a003500_list = 2 : 4 : zipWith (-)
       (map (* 4) $ tail a003500_list) a003500_list
    -- Reinhard Zumkeller, Dec 17 2011
    
  • Magma
    I:=[2,4]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2018
  • Maple
    A003500 := proc(n) option remember; if n <= 1 then 2*n+2 else 4*procname(n-1)-procname(n-2); fi;
    end proc;
  • Mathematica
    a[0]=2; a[1]=4; a[n_]:= a[n]= 4a[n-1] -a[n-2]; Table[a[n], {n, 0, 23}]
    LinearRecurrence[{4,-1},{2,4},30] (* Harvey P. Dale, Aug 20 2011 *)
    Table[Round@LucasL[2n, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
  • PARI
    x='x+O('x^99); Vec(-2*(-1+2*x)/(1-4*x+x^2)) \\ Altug Alkan, Apr 04 2016
    
  • Sage
    [lucas_number2(n,4,1) for n in range(0, 24)] # Zerinvary Lajos, May 14 2009
    

Formula

a(n) = ( 2 + sqrt(3) )^n + ( 2 - sqrt(3) )^n.
a(n) = 2*A001075(n).
G.f.: 2*(1 - 2*x)/(1 - 4*x + x^2). Simon Plouffe in his 1992 dissertation.
a(n) = A001835(n) + A001835(n+1).
a(n) = trace of n-th power of the 2 X 2 matrix [1 2 / 1 3]. - Gary W. Adamson, Jun 30 2003 [corrected by Joerg Arndt, Jun 18 2020]
From the addition formula, a(n+m) = a(n)*a(m) - a(m-n), it is easy to derive multiplication formulas, such as: a(2*n) = (a(n))^2 - 2, a(3*n) = (a(n))^3 - 3*(a(n)), a(4*n) = (a(n))^4 - 4*(a(n))^2 + 2, a(5*n) = (a(n))^5 - 5*(a(n))^3 + 5*(a(n)), a(6*n) = (a(n))^6 - 6*(a(n))^4 + 9*(a(n))^2 - 2, etc. The absolute values of the coefficients in the expansions are given by the triangle A034807. - John Blythe Dobson, Nov 04 2007
a(n) = 2*A001353(n+1) - 4*A001353(n). - R. J. Mathar, Nov 16 2007
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n=0..infinity} (1 + x^(4*n + 1))/(1 + x^(4*n + 3)). Let alpha = 2 - sqrt(3). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.24561 99455 06551 88869 ... = 2 + 1/(4 + 1/(14 + 1/(52 + ...))). Cf. A174500.
Also F(-alpha) = 0.74544 81786 39692 68884 ... has the continued fraction representation 1 - 1/(4 - 1/(14 - 1/(52 - ...))) and the simple continued fraction expansion 1/(1 + 1/((4 - 2) + 1/(1 + 1/((14 - 2) + 1/(1 + 1/((52 - 2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((4^2 - 4) + 1/(1 + 1/((14^2 - 4) + 1/(1 + 1/((52^2 - 4) + 1/(1 + ...))))))).
(End)
a(2^n) = A003010(n). - John Blythe Dobson, Mar 10 2014
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 12*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
E.g.f.: 2*exp(2*x)*cosh(sqrt(3)*x). - Ilya Gutkovskiy, Apr 27 2016
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*n*(n - k - 1)!/(k!*(n - 2*k)!)*4^(n - 2*k) for n >= 1. - Peter Luschny, May 10 2016
From Peter Bala, Oct 15 2019: (Start)
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; -1, 4].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
2*Sum_{n >= 1} 1/( a(n) - 6/a(n) ) = 1.
6*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 2/a(n) ) = 1.
8*Sum_{n >= 1} 1/( a(n) + 24/(a(n) - 12/(a(n))) ) = 1.
8*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 8/(a(n) + 4/(a(n))) ) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/2 - 6*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 6)),
Sum_{n >= 1} 1/a(n) = 1/8 + 24*Sum_{n >= 1} 1/(a(n)*(a(n)^2 + 12)),
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/6 + 2*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 2)) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 12)).
Sum_{n >= 1} 1/a(n) = ( theta_3(2-sqrt(3))^2 - 1 )/4 = 0.34770 07561 66992 06261 .... See Borwein and Borwein, Proposition 3.5 (i), p.91.
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - theta_3(sqrt(3)-2)^2 )/4. Cf. A003499 and A153415. (End)
a(n) = tan(Pi/12)^n + tan(5*Pi/12)^n. - Greg Dresden, Oct 01 2020
From Wolfdieter Lang, Sep 06 2021: (Start)
a(n) = S(n, 4) - S(n-2, 4) = 2*T(n, 2), for n >= 0, with S and T Chebyshev polynomials, with S(-1, x) = 0 and S(-2, x) = -1. S(n, 4) = A001353(n+1), for n >= -1, and T(n, 2) = A001075(n).
a(2*k) = A067902(k), a(2*k+1) = 4*A001570(k+1), for k >= 0. (End)
a(n) = sqrt(2 + 2*A011943(n+1)) = sqrt(2 + 2*A102344(n+1)), n>0. - Ralf Steiner, Sep 23 2021
Sum_{n>=1} arctan(3/a(n)^2) = Pi/6 - arctan(1/3) = A019673 - A105531 (Ohtskua, 2024). - Amiram Eldar, Aug 29 2024

Extensions

More terms from James Sellers, May 03 2000
Additional comments from Lekraj Beedassy, Feb 14 2002

A001570 Numbers k such that k^2 is centered hexagonal.

Original entry on oeis.org

1, 13, 181, 2521, 35113, 489061, 6811741, 94875313, 1321442641, 18405321661, 256353060613, 3570537526921, 49731172316281, 692665874901013, 9647591076297901, 134373609193269601, 1871582937629476513, 26067787517619401581, 363077442309042145621
Offset: 1

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Keywords

Comments

Chebyshev T-sequence with Diophantine property. - Wolfdieter Lang, Nov 29 2002
a(n) = L(n,14), where L is defined as in A108299; see also A028230 for L(n,-14). - Reinhard Zumkeller, Jun 01 2005
Numbers x satisfying x^2 + y^3 = (y+1)^3. Corresponding y given by A001921(n)={A028230(n)-1}/2. - Lekraj Beedassy, Jul 21 2006
Mod[ a(n), 12 ] = 1. (a(n) - 1)/12 = A076139(n) = Triangular numbers that are one-third of another triangular number. (a(n) - 1)/4 = A076140(n) = Triangular numbers T(k) that are three times another triangular number. - Alexander Adamchuk, Apr 06 2007
Also numbers n such that RootMeanSquare(1,3,...,2*n-1) is an integer. - Ctibor O. Zizka, Sep 04 2008
a(n), with n>1, is the length of the cevian of equilateral triangle whose side length is the term b(n) of the sequence A028230. This cevian divides the side (2*x+1) of the triangle in two integer segments x and x+1. - Giacomo Fecondo, Oct 09 2010
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(12)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Beal's conjecture would imply that set intersection of this sequence with the perfect powers (A001597) equals {1}. In other words, existence of a nontrivial perfect power in this sequence would disprove Beal's conjecture. - Max Alekseyev, Mar 15 2015
Numbers n such that there exists positive x with x^2 + x + 1 = 3n^2. - Jeffrey Shallit, Dec 11 2017
Given by the denominators of the continued fractions [1,(1,2)^i,3,(1,2)^{i-1},1]. - Jeffrey Shallit, Dec 11 2017
A near-isosceles integer-sided triangle with an angle of 2*Pi/3 is a triangle whose sides (a, a+1, c) satisfy Diophantine equation (a+1)^3 - a^3 = c^2. For n >= 2, the largest side c is given by a(n) while smallest and middle sides (a, a+1) = (A001921(n-1), A001922(n-1)) (see Julia link). - Bernard Schott, Nov 20 2022

Examples

			G.f. = x + 13*x^2 + 181*x^3 + 2521*x^4 + 35113*x^5 + 489061*x^6 + 6811741*x^7 + ...
		

References

  • E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - N. J. A. Sloane, Mar 03 2022
  • 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

Bisection of A003500/4. Cf. A006051, A001921, A001922.
One half of odd part of bisection of A001075. First differences of A007655.
Cf. A077417 with companion A077416.
Row 14 of array A094954.
A122571 is another version of the same sequence.
Row 2 of array A188646.
Cf. similar sequences listed in A238379.
Cf. A028231, which gives the corresponding values of x in 3n^2 = x^2 + x + 1.
Similar sequences of the type cosh((2*m+1)*arccosh(k))/k are listed in A302329. This is the case k=2.

Programs

  • Magma
    [((2 + Sqrt(3))^(2*n - 1) + (2 - Sqrt(3))^(2*n - 1))/4: n in [1..50]]; // G. C. Greubel, Nov 04 2017
  • Maple
    A001570:=-(-1+z)/(1-14*z+z**2); # Simon Plouffe in his 1992 dissertation.
  • Mathematica
    NestList[3 + 7*#1 + 4*Sqrt[1 + 3*#1 + 3*#1^2] &, 0, 24] (* Zak Seidov, May 06 2007 *)
    f[n_] := Simplify[(2 + Sqrt@3)^(2 n - 1) + (2 - Sqrt@3)^(2 n - 1)]/4; Array[f, 19] (* Robert G. Wilson v, Oct 28 2010 *)
    a[c_, n_] := Module[{},
       p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
       d := Denominator[Convergents[Sqrt[c], n p]];
       t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
       Return[t];
      ] (* Complement of A041017 *)
    a[12, 20] (* Gerry Martens, Jun 07 2015 *)
    LinearRecurrence[{14, -1}, {1, 13}, 19] (* Jean-François Alcover, Sep 26 2017 *)
    CoefficientList[Series[x (1-x)/(1-14x+x^2),{x,0,20}],x] (* Harvey P. Dale, Sep 18 2024 *)
  • PARI
    {a(n) = real( (2 + quadgen( 12)) ^ (2*n - 1)) / 2}; /* Michael Somos, Feb 15 2011 */
    

Formula

a(n) = ((2 + sqrt(3))^(2*n - 1) + (2 - sqrt(3))^(2*n - 1)) / 4. - Michael Somos, Feb 15 2011
G.f.: x * (1 - x) / (1 -14*x + x^2). - Michael Somos, Feb 15 2011
Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i) then a(n) = q(n, 12). - Benoit Cloitre, Dec 10 2002
a(n) = S(n, 14) - S(n-1, 14) = T(2*n+1, 2)/2 with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 14)=A007655(n+1) and T(n, 2)=A001075(n). - Wolfdieter Lang, Nov 29 2002
a(n) = A001075(n)*A001075(n+1) - 1 and thus (a(n)+1)^6 has divisors A001075(n)^6 and A001075(n+1)^6 congruent to -1 modulo a(n) (cf. A350916). - Max Alekseyev, Jan 23 2022
4*a(n)^2 - 3*b(n)^2 = 1 with b(n)=A028230(n+1), n>=0.
a(n)*a(n+3) = 168 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = 14*a(n-1) - a(n-2), a(0) = a(1) = 1. a(1 - n) = a(n) (compare A122571).
a(n) = 12*A076139(n) + 1 = 4*A076140(n) + 1. - Alexander Adamchuk, Apr 06 2007
a(n) = (1/12)*((7-4*sqrt(3))^n*(3-2*sqrt(3))+(3+2*sqrt(3))*(7+4*sqrt(3))^n -6). - Zak Seidov, May 06 2007
a(n) = A102871(n)^2+(A102871(n)-1)^2; sum of consecutive squares. E.g. a(4)=36^2+35^2. - Mason Withers (mwithers(AT)semprautilities.com), Jan 26 2008
a(n) = sqrt((3*A028230(n+1)^2 + 1)/4).
a(n) = A098301(n+1) - A001353(n)*A001835(n).
a(n) = A000217(A001571(n-1)) + A000217(A133161(n)), n>=1. - Ivan N. Ianakiev, Sep 24 2013
a(n)^2 = A001922(n-1)^3 - A001921(n-1)^3, for n >= 1. - Bernard Schott, Nov 20 2022
a(n) = 2^(2*n-3)*Product_{k=1..2*n-1} (2 - sin(2*Pi*k/(2*n-1))). Michael Somos, Dec 18 2022
a(n) = A003154(A101265(n)). - Andrea Pinos, Dec 19 2022

A092184 Sequence S_6 of the S_r family.

Original entry on oeis.org

0, 1, 6, 25, 96, 361, 1350, 5041, 18816, 70225, 262086, 978121, 3650400, 13623481, 50843526, 189750625, 708158976, 2642885281, 9863382150, 36810643321, 137379191136, 512706121225, 1913445293766, 7141075053841, 26650854921600, 99462344632561, 371198523608646
Offset: 0

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Author

Rainer Rosenthal, Apr 03 2004

Keywords

Comments

The r-family of sequences is S_r(n) = 2*(T(n,(r-2)/2) - 1)/(r-4) provided r is not equal to 4 and S_4(n) = n^2 = A000290(n). Here T(n,x) are Chebyshev's polynomials of the first kind. See their coefficient triangle A053120. See also the R. Stephan link for the explicit formula for s_k(n) for k not equal to 4 (Stephan's s_k(n) is identical with S_r(n)).
An integer n is in this sequence iff mutually externally tangent circles with radii n, n+1, n+2 have Soddy circles (i.e., circles tangent to all three) of rational radius. - James R. Buddenhagen, Nov 16 2005
This sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. It is the case P1 = 6, P2 = 8, Q = 1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 25 2014
a(n) is the block size of the (n-1)-th design in a sequence of multi-set designs with 2 blocks, see A335649. - John P. McSorley, Jun 22 2020

Examples

			a(3)=25 because a(1)=1 and a(2)=6 and a(1)*a(3) = 1*25 = (6-1)^2 = (a(2)-1)^2.
		

Crossrefs

See A001110=S_36 for further references to S_r sequences.
Other members of this r-family are: A007877 (r=2), |A078070| (r=3), A004146 (r=5), A054493 (r=7). A098306, A100047. A001353, A001834. A001350, A052530.

Programs

  • Magma
    [Floor(1/2*(-2+(2+Sqrt(3))^n+(2-Sqrt(3))^n)): n in [0..30]]; // Vincenzo Librandi, Oct 14 2015
  • Maple
    A092184 := proc(n)
        option remember;
        if n <= 1 then
            n;
        else
            4*procname(n-1)-procname(n-2)+2 ;
        end if ;
    end proc:
    seq(A092184(n),n=0..10) ;# Zerinvary Lajos, Mar 09 2008
  • Mathematica
    Table[Simplify[ -((2 + Sqrt[3])^n - 1)*((2 - Sqrt[3])^n - 1)]/2, {n, 0, 26}] (* Stefan Steinerberger, May 15 2007 *)
    LinearRecurrence[{5,-5,1},{0,1,6},27] (* Ray Chandler, Jan 27 2014 *)
    CoefficientList[Series[x (1 + x)/(1 - 5 x + 5 x^2 - x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 14 2015 *)
  • PARI
    Vec(x*(1+x)/(1 - 5*x + 5*x^2 - x^3) + O(x^50)) \\ Michel Marcus, Oct 14 2015
    

Formula

S_r type sequences are defined by a(0)=0, a(1)=1, a(2)=r and a(n-1)*a(n+1) = (a(n)-1)^2. This sequence emanates from r=6.
a(n) = 1/2*(-2 + (2+sqrt(3))^n + (2-sqrt(3))^n). - Ralf Stephan, Apr 14 2004
G.f.: x*(1+x)/(1 - 5*x + 5*x^2 - x^3) = x*(1+x)/((1-x)*(1 - 4*x + x^2)). [from the Ralf Stephan link]
a(n) = T(n, 2)-1 = A001075(n)-1, with Chebyshev's polynomials T(n, 2) of the first kind.
a(n) = b(n) + b(n-1), n >= 1, with b(n):=A061278(n) the partial sums of S(n, 4) = U(n, 2) = A001353(n+1) Chebyshev's polynomials of the second kind.
An integer k is in this sequence iff k is nonnegative and (k^2 + 2*k)/3 is a square. - James R. Buddenhagen, Nov 16 2005
a(0)=0, a(1)=1, a(n+1) = 3 + floor(a(n)*(2+sqrt(3))). - Anton Vrba (antonvrba(AT)yahoo.com), Jan 16 2007
a(n) = 4*a(n-1) - a(n-2) + 2. - Zerinvary Lajos, Mar 09 2008
From Peter Bala, Mar 25 2014: (Start)
a(2*n) = 6*A001353(n)^2; a(2*n+1) = A001834(n)^2.
a(n) = u(n)^2, where {u(n)} is the Lucas sequence in the quadratic integer ring Z[sqrt(6)] defined by the recurrence u(0) = 0, u(1) = 1, u(n) = sqrt(6)*u(n-1) - u(n-2) for n >= 2.
Equivalently, a(n) = U(n-1,sqrt(6)/2)^2, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = (1/2)*( ((sqrt(6) + sqrt(2))/2)^n - ((sqrt(6) - sqrt(2))/2)^n )^2.
a(n) = bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -2; 1, 3] and T(n,x) denotes the Chebyshev polynomial of the first kind. Cf. A098306.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
exp( Sum_{n >= 1} 2*a(n)*x^n/n ) = 1 + Sum_{n >= 1} A052530(n)*x^n. Cf. A001350. - Peter Bala, Mar 19 2015
E.g.f.: exp(2*x)*cosh(sqrt(3)*x) - cosh(x) - sinh(x). - Stefano Spezia, Oct 13 2019

Extensions

Extension and Chebyshev comments from Wolfdieter Lang, Sep 10 2004

A007655 Standard deviation of A007654.

Original entry on oeis.org

0, 1, 14, 195, 2716, 37829, 526890, 7338631, 102213944, 1423656585, 19828978246, 276182038859, 3846719565780, 53577891882061, 746243766783074, 10393834843080975, 144767444036350576, 2016350381665827089, 28084137899285228670, 391161580208327374291, 5448177985017298011404
Offset: 1

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Author

Keywords

Comments

a(n) corresponds also to one-sixth the area of Fleenor-Heronian triangle with middle side A003500(n). - Lekraj Beedassy, Jul 15 2002
a(n) give all (nontrivial, integer) solutions of Pell equation b(n+1)^2 - 48*a(n+1)^2 = +1 with b(n+1)=A011943(n), n>=0.
For n>=3, a(n) equals the permanent of the (n-2) X (n-2) tridiagonal matrix with 14's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,13}. - Milan Janjic, Jan 25 2015
6*a(n)^2 = 6*S(n-1, 14)^2 is the triangular number Tri((T(n, 7) - 1)/2) with Tri = A000217 and T = A053120. This is instance k = 3 of the general k-identity given in a comment to A001109. - Wolfdieter Lang, Feb 01 2016

Examples

			G.f. = x^2 + 14*x^3 + 195*x^4 + 2716*x^5 + 37829*x^6 + 526890*x^7 + ...
		

References

  • D. A. Benaron, personal communication.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), this sequence (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

Programs

  • GAP
    m:=7;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
  • Magma
    [n le 2 select n-1 else 14*Self(n-1)-Self(n-2): n in [1..70]]; // Vincenzo Librandi, Feb 02 2016
    
  • Maple
    0,seq(orthopoly[U](n,7),n=0..30); # Robert Israel, Feb 04 2016
  • Mathematica
    Table[GegenbauerC[n, 1, 7], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
    LinearRecurrence[{14,-1}, {0,1}, 20] (* Vincenzo Librandi, Feb 02 2016 *)
    ChebyshevU[Range[21] -2, 7] (* G. C. Greubel, Dec 23 2019 *)
    Table[Sum[Binomial[n, 2 k - 1]*7^(n - 2 k + 1)*48^(k - 1), {k, 1, n}], {n, 0, 15}] (* Horst H. Manninger, Jan 16 2022 *)
  • PARI
    concat(0, Vec((x^2/(1-14*x+x^2) + O(x^30)))) \\ Michel Marcus, Feb 02 2016
    
  • PARI
    vector(21, n, polchebyshev(n-2, 2, 7) ) \\ G. C. Greubel, Dec 23 2019
    
  • Sage
    [lucas_number1(n,14,1) for n in range(0,20)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [chebyshev_U(n,7) for n in (-1..20)] # G. C. Greubel, Dec 23 2019
    

Formula

a(n) = 14*a(n-1) - a(n-2).
G.f.: x^2/(1-14*x+x^2).
a(n+1) ~ 1/24*sqrt(3)*(2 + sqrt(3))^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n+1) = S(n-1, 14), n>=0, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
a(n+1) = ( (7+4*sqrt(3))^n - (7-4*sqrt(3))^n )/(8*sqrt(3)).
a(n+1) = sqrt((A011943(n)^2 - 1)/48), n>=0.
Chebyshev's polynomials U(n-2, x) evaluated at x=7.
a(n) = A001353(2n)/4. - Lekraj Beedassy, Jul 15 2002
4*a(n+1) + A046184(n) = A055793(n+2) + A098301(n+1) 4*a(n+1) + A098301(n+1) + A055793(n+2) = A046184(n+1) (4*a(n+1))^2 = A098301(2n+1) (conjectures). - Creighton Dement, Nov 02 2004
(4*a(n))^2 = A103974(n)^2 - A011922(n-1)^2. - Paul D. Hanna, Mar 06 2005
From Mohamed Bouhamida, May 26 2007: (Start)
a(n) = 13*( a(n-1) + a(n-2) ) - a(n-3).
a(n) = 15*( a(n-1) - a(n-2) ) + a(n-3). (End)
a(n) = b such that (-1)^n/4*Integral_{x=-Pi/2..Pi/2} (sin((2*n-2)*x))/(2-sin(x)) dx = c+b*log(3). - Francesco Daddi, Aug 02 2011
a(n+2) = Sum_{k=0..n} A101950(n,k)*13^k. - Philippe Deléham, Feb 10 2012
Product {n >= 1} (1 + 1/a(n)) = 1/3*(3 + 2*sqrt(3)). - Peter Bala, Dec 23 2012
Product {n >= 2} (1 - 1/a(n)) = 1/7*(3 + 2*sqrt(3)). - Peter Bala, Dec 23 2012
a(n) = (A028230(n) - A001570(n))/2. - Richard R. Forberg, Nov 14 2013
E.g.f.: 1 - exp(7*x)*(12*cosh(4*sqrt(3)*x) - 7*sqrt(3)*sinh(4*sqrt(3)*x))/12. - Stefano Spezia, Dec 11 2022

Extensions

Chebyshev comments from Wolfdieter Lang, Nov 08 2002
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