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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A001541 a(0) = 1, a(1) = 3; for n > 1, a(n) = 6*a(n-1) - a(n-2).

Original entry on oeis.org

1, 3, 17, 99, 577, 3363, 19601, 114243, 665857, 3880899, 22619537, 131836323, 768398401, 4478554083, 26102926097, 152139002499, 886731088897, 5168247530883, 30122754096401, 175568277047523, 1023286908188737, 5964153172084899, 34761632124320657
Offset: 0

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Author

Keywords

Comments

Chebyshev polynomials of the first kind evaluated at 3.
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 8*y^2 = 1, the corresponding values of y are in A001109. For n > 0, the ratios a(n)/A001090(n) may be obtained as convergents to sqrt(8): either successive convergents of [3; -6] or odd convergents of [2; 1, 4]. - Lekraj Beedassy, Sep 09 2003 [edited by Jon E. Schoenfield, May 04 2014]
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r = sqrt(8). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions greater than 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r = sqrt(2). - Benoit Cloitre, Feb 24 2004
This sequence give numbers n such that (n-1)*(n+1)/2 is a perfect square. Remark: (i-1)*(i+1)/2 = (i^2-1)/2 = -1 = i^2 with i = sqrt(-1) so i is also in the sequence. - Pierre CAMI, Apr 20 2005
a(n) is prime for n = {1, 2, 4, 8}. Prime a(n) are {3, 17, 577, 665857}, which belong to A001601(n). a(2k-1) is divisible by a(1) = 3. a(4k-2) is divisible by a(2) = 17. a(8k-4) is divisible by a(4) = 577. a(16k-8) is divisible by a(8) = 665857. - Alexander Adamchuk, Nov 24 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=A001542. - Clark Kimberling, Aug 26 2008
Also index of sequence A082532 for which A082532(n) = 1. - Carmine Suriano, Sep 07 2010
Numbers n such that sigma(n-1) and sigma(n+1) are both odd numbers. - Juri-Stepan Gerasimov, Mar 28 2011
Also, numbers such that floor(a(n)^2/2) is a square: base 2 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001075. - M. F. Hasler, Jan 15 2012
Numbers such that 2n^2 - 2 is a square. Also integer square roots of the expression 2*n^2 + 1, at values of n given by A001542. Also see A228405 regarding 2n^2 -+ 2^k generally for k >= 0. - Richard R. Forberg, Aug 20 2013
Values of x (or y) in the solutions to x^2 - 6xy + y^2 + 8 = 0. - Colin Barker, Feb 04 2014
Panda and Ray call the numbers in this sequence the Lucas-balancing numbers C_n (see references and links).
Partial sums of X or X+1 of Pythagorean triples (X,X+1,Z). - Peter M. Chema, Feb 03 2017
a(n)/A001542(n) is the closest rational approximation to sqrt(2) with a numerator not larger than a(n), and 2*A001542(n)/a(n) is the closest rational approximation to sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001653 and A002315 give a complete set of closest rational approximations to sqrt(2) with restricted numerator or denominator. a(n)/A001542(n) > sqrt(2) > 2*A001542(n)/a(n). - A.H.M. Smeets, May 28 2017
x = a(n), y = A001542(n) are solutions of the Diophantine equation x^2 - 2y^2 = 1 (Pell equation). x = 2*A001542(n), y = a(n) are solutions of the Diophantine equation x^2 - 2y^2 = -2. Both together give the set of fractional approximations for sqrt(2) obtained from limited fractions obtained from continued fraction representation to sqrt(2). - A.H.M. Smeets, Jun 22 2017
a(n) is the radius of the n-th circle among the sequence of circles generated as follows: Starting with a unit circle centered at the origin, every subsequent circle touches the previous circle as well as the two limbs of hyperbola x^2 - y^2 = 1, and lies in the region y > 0. - Kaushal Agrawal, Nov 10 2018
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, 0A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0Michael Somos, Jun 26 2022

Examples

			99^2 + 99^2 = 140^2 + 2. - _Carmine Suriano_, Jan 05 2015
G.f. = 1 + 3*x + 17*x^2 + 99*x^3 + 577*x^4 + 3363*x^5 + 19601*x^6 + 114243*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)
  • J. W. L. Glaisher, On Eulerian numbers (formulas, residues, end-figures), with the values of the first twenty-seven, Quarterly Journal of Mathematics, vol. 45, 1914, pp. 1-51.
  • G. K. Panda, Some fascinating properties of balancing numbers, In Proc. of Eleventh Internat. Conference on Fibonacci Numbers and Their Applications, Cong. Numerantium 194 (2009), 185-189.
  • 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 A001333. A003499(n) = 2a(n).
Cf. A055997 = numbers n such that n(n-1)/2 is a square.
Row 1 of array A188645.
Cf. A055792 (terms squared), A132592.

Programs

  • Haskell
    a001541 n = a001541_list !! (n-1)
    a001541_list =
    1 : 3 : zipWith (-) (map (* 6) $ tail a001541_list) a001541_list
    -- Reinhard Zumkeller, Oct 06 2011
    (Scheme, with memoization-macro definec)
    (definec (A001541 n) (cond ((zero? n) 1) ((= 1 n) 3) (else (- (* 6 (A001541 (- n 1))) (A001541 (- n 2))))))
    ;; Antti Karttunen, Oct 04 2016
  • Magma
    [n: n in [1..10000000] |IsSquare(8*(n^2-1))]; // Vincenzo Librandi, Nov 18 2010
    
  • Maple
    a[0]:=1: a[1]:=3: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006
    A001541:=-(-1+3*z)/(1-6*z+z**2); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    Table[Simplify[(1/2) (3 + 2 Sqrt[2])^n + (1/2) (3 - 2 Sqrt[2])^n], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
    a[ n_] := If[n == 0, 1, With[{m = Abs @ n}, m Sum[4^i Binomial[m + i, 2 i]/(m + i), {i, 0, m}]]]; (* Michael Somos, Jul 11 2011 *)
    a[ n_] := ChebyshevT[ n, 3]; (* Michael Somos, Jul 11 2011 *)
    LinearRecurrence[{6, -1}, {1, 3}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
  • PARI
    {a(n) = real((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */
    
  • PARI
    {a(n) = subst( poltchebi( abs(n)), x, 3)}; /* Michael Somos, Apr 07 2003 */
    
  • PARI
    {a(n) = if( n<0, a(-n), polsym(1 - 6*x + x^2, n) [n+1] / 2)}; /* Michael Somos, Apr 07 2003 */
    
  • PARI
    {a(n) = polchebyshev( n, 1, 3)}; /* Michael Somos, Jul 11 2011 */
    
  • PARI
    a(n)=([1,2,2;2,1,2;2,2,3]^n)[3,3] \\ Vim Wenders, Mar 28 2007
    

Formula

G.f.: (1-3*x)/(1-6*x+x^2). - Barry E. Williams and Wolfdieter Lang, May 05 2000
E.g.f.: exp(3*x)*cosh(2*sqrt(2)*x). Binomial transform of A084128. - Paul Barry, May 16 2003
From N. J. A. Sloane, May 16 2003: (Start)
a(n) = sqrt(8*((A001109(n))^2) + 1).
a(n) = T(n, 3), with Chebyshev's T-polynomials A053120. (End)
a(n) = ((3+2*sqrt(2))^n + (3-2*sqrt(2))^n)/2.
a(n) = cosh(2*n*arcsinh(1)). - Herbert Kociemba, Apr 24 2008
a(n) ~ (1/2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all elements x of the sequence, 2*x^2 - 2 is a square. Limit_{n -> infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 10 2002 [corrected by Peter Pein, Mar 09 2009]
a(n) = 3*A001109(n) - A001109(n-1), n >= 1. - Barry E. Williams and Wolfdieter Lang, May 05 2000
For n >= 1, a(n) = A001652(n) - A001652(n-1). - Charlie Marion, Jul 01 2003
From Paul Barry, Sep 18 2003: (Start)
a(n) = ((-1+sqrt(2))^n + (1+sqrt(2))^n + (1-sqrt(2))^n + (-1-sqrt(2))^n)/4 (with interpolated zeros).
E.g.f.: cosh(x)*cosh(sqrt(2)x) (with interpolated zeros). (End)
For n > 0, a(n)^2 + 1 = 2*A001653(n-1)*A001653(n). - Charlie Marion, Dec 21 2003
a(n)^2 + a(n+1)^2 = 2*(A001653(2*n+1) - A001652(2*n)). - Charlie Marion, Mar 17 2003
a(n) = Sum_{k >= 0} binomial(2*n, 2*k)*2^k = Sum_{k >= 0} A086645(n, k)*2^k. - Philippe Deléham, Feb 29 2004
a(n)*A002315(n+k) = A001652(2*n+k) + A001652(k) + 1; for k > 0, a(n+k)*A002315(n) = A001652(2*n+k) - A001652(k-1). - Charlie Marion, Mar 17 2003
For n > k, a(n)*A001653(k) = A011900(n+k) + A053141(n-k-1). For n <= k, a(n)*A001653(k) = A011900(n+k) + A053141(k-n). - Charlie Marion, Oct 18 2004
A053141(n+1) + A055997(n+1) = a(n+1) + A001109(n+1). - Creighton Dement, Sep 16 2004
a(n+1) - A001542(n+1) = A090390(n+1) - A046729(n) = A001653(n); a(n+1) - 4*A079291(n+1) = (-1)^(n+1). Formula generated by the floretion - .5'i + .5'j - .5i' + .5j' - 'ii' + 'jj' - 2'kk' + 'ij' + .5'ik' + 'ji' + .5'jk' + .5'ki' + .5'kj' + e. - Creighton Dement, Nov 16 2004
a(n) = sqrt( A055997(2*n) ). - Alexander Adamchuk, Nov 24 2006
a(2n) = A056771(n). a(2*n+1) = 3*A077420(n). - Alexander Adamchuk, Feb 01 2007
a(n) = (A000129(n)^2)*4 + (-1)^n. - Vim Wenders, Mar 28 2007
2*a(k)*A001653(n)*A001653(n+k) = A001653(n)^2 + A001653(n+k)^2 + A001542(k)^2. - Charlie Marion, Oct 12 2007
a(n) = A001333(2*n). - Ctibor O. Zizka, Aug 13 2008
A028982(a(n)-1) + 2 = A028982(a(n)+1). - Juri-Stepan Gerasimov, Mar 28 2011
a(n) = 2*A001108(n) + 1. - Paul Weisenhorn, Dec 17 2011
a(n) = sqrt(2*x^2 + 1) with x being A001542(n). - Zak Seidov, Jan 30 2013
a(2n) = 2*a(n)^2 - 1 = a(n)^2 + 2*A001542(n)^2. a(2*n+1) = 1 + 2*A002315(n)^2. - Steven J. Haker, Dec 04 2013
a(n) = 3*a(n-1) + 4*A001542(n-1); e.g., a(4) = 99 = 3*17 + 4*12. - Zak Seidov, Dec 19 2013
a(n) = cos(n * arccos(3)) = cosh(n * log(3 + 2*sqrt(2))). - Daniel Suteu, Jul 28 2016
From Ilya Gutkovskiy, Jul 28 2016: (Start)
Inverse binomial transform of A084130.
Exponential convolution of A000079 and A084058.
Sum_{n>=0} (-1)^n*a(n)/n! = cosh(2*sqrt(2))/exp(3) = 0.4226407909842764637... (End)
a(2*n+1) = 2*a(n)*a(n+1) - 3. - Timothy L. Tiffin, Oct 12 2016
a(n) = a(-n) for all n in Z. - Michael Somos, Jan 20 2017
a(2^n) = A001601(n+1). - A.H.M. Smeets, May 28 2017
a(A298210(n)) = A002350(2*n^2). - A.H.M. Smeets, Jan 25 2018
a(n) = S(n, 6) - 3*S(n-1, 6), for n >= 0, with S(n, 6) = A001109(n+1), (Chebyshev S of A049310). See the first comment and the formula a(n) = T(n, 3). - Wolfdieter Lang, Nov 22 2020
From Peter Bala, Dec 31 2021: (Start)
a(n) = [x^n] (3*x + sqrt(1 + 8*x^2))^n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) hold for all prime p and positive integers n and k.
O.g.f. A(x) = 1 + x*d/dx(log(B(x))), where B(x) = 1/sqrt(1 - 6*x + x^2) is the o.g.f. of A001850. (End)
From Peter Bala, Aug 17 2022: (Start)
Sum_{n >= 1} 1/(a(n) - 2/a(n)) = 1/2.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 1/a(n)) = 1/4.
Sum_{n >= 1} 1/(a(n)^2 - 2) = 1/2 - 1/sqrt(8). (End)
From Peter Bala, Jun 23 2025: (Start)
Product_{n >= 0} (1 + 1/a(2^n)) = sqrt(2).
Product_{n >= 0} (1 - 1/(2*a(2^n))) = (4/7)*sqrt(2). See A002812. (End)

A058331 a(n) = 2*n^2 + 1.

Original entry on oeis.org

1, 3, 9, 19, 33, 51, 73, 99, 129, 163, 201, 243, 289, 339, 393, 451, 513, 579, 649, 723, 801, 883, 969, 1059, 1153, 1251, 1353, 1459, 1569, 1683, 1801, 1923, 2049, 2179, 2313, 2451, 2593, 2739, 2889, 3043, 3201, 3363, 3529, 3699, 3873, 4051
Offset: 0

Views

Author

Erich Friedman, Dec 12 2000

Keywords

Comments

Maximal number of regions in the plane that can be formed with n hyperbolas.
Also the number of different 2 X 2 determinants with integer entries from 0 to n.
Number of lattice points in an n-dimensional ball of radius sqrt(2). - David W. Wilson, May 03 2001
Equals A112295(unsigned) * [1, 2, 3, ...]. - Gary W. Adamson, Oct 07 2007
Binomial transform of A166926. - Gary W. Adamson, May 03 2008
a(n) = longest side a of all integer-sided triangles with sides a <= b <= c and inradius n >= 1. Triangle has sides (2n^2 + 1, 2n^2 + 2, 4n^2 + 1).
{a(k): 0 <= k < 3} = divisors of 9. - Reinhard Zumkeller, Jun 17 2009
Number of ways to partition a 3*n X 2 grid into 3 connected equal-area regions. - R. H. Hardin, Oct 31 2009
Let A be the Hessenberg matrix of order n defined by: A[1, j] = 1, A[i, i] := 2, (i > 1), A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 3, a(n - 1) = coeff(charpoly(A, x), x^(n - 2)). - Milan Janjic, Jan 26 2010
Except for the first term of [A002522] and [A058331] if X = [A058331], Y = [A087113], A = [A002522], we have, for all other terms, Pell's equation: [A058331]^2 - [A002522]*[A087113]^2 = 1; (X^2 - A*Y^2 = 1); e.g., 3^2 -2*2^2 = 1; 9^2 - 5*4^2 = 1; 129^2 - 65*16^2 = 1, and so on. - Vincenzo Librandi, Aug 07 2010
Niven (1961) gives this formula as an example of a formula that does not contain all odd integers, in contrast to 2n + 1 and 2n - 1. - Alonso del Arte, Dec 05 2012
Numbers m such that 2*m-2 is a square. - Vincenzo Librandi, Apr 10 2015
Number of n-tuples from the set {1,0,-1} where at most two elements are nonzero. - Michael Somos, Oct 19 2022
a(n) gives the x-value of the integral solution (x,y) of the Pellian equation x^2 - (n^2 + 1)*y^2 = 1. The y-value is given by 2*n (see Tattersall). - Stefano Spezia, Jul 23 2025

Examples

			a(1) = 3 since (0 0 / 0 0), (1 0 / 0 1) and (0 1 / 1 0) have different determinants.
G.f. = 1 + 3*x + 9*x^2 + 19*x^3 + 33*x^4 + 51*x^5 + 73*x^6 + ... - _Michael Somos_, Oct 19 2022
		

References

  • Ivan Niven, Numbers: Rational and Irrational, New York: Random House for Yale University (1961): 17.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 256.

Crossrefs

Cf. A000124.
Second row of array A099597.
See A120062 for sequences related to integer-sided triangles with integer inradius n.
Cf. A112295.
Column 2 of array A188645.
Cf. A001105 and A247375. - Bruno Berselli, Sep 16 2014

Programs

  • Haskell
    a058331 = (+ 1) . a001105  -- Reinhard Zumkeller, Dec 13 2014
    
  • Magma
    [2*n^2 + 1 : n in [0..100]]; // Wesley Ivan Hurt, Feb 02 2017
  • Mathematica
    b[g_] := Length[Union[Map[Det, Flatten[ Table[{{i, j}, {k, l}}, {i, 0, g}, {j, 0, g}, {k, 0, g}, {l, 0, g}], 3]]]] Table[b[g], {g, 0, 20}]
    2*Range[0, 49]^2 + 1 (* Alonso del Arte, Dec 05 2012 *)
  • PARI
    a(n)=2*n^2+1 \\ Charles R Greathouse IV, Jun 16 2011
    

Formula

G.f.: (1 + 3x^2)/(1 - x)^3. - Paul Barry, Apr 06 2003
a(n) = M^n * [1 1 1], leftmost term, where M = the 3 X 3 matrix [1 1 1 / 0 1 4 / 0 0 1]. a(0) = 1, a(1) = 3; a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). E.g., a(4) = 33 since M^4 *[1 1 1] = [33 17 1]. - Gary W. Adamson, Nov 11 2004
a(n) = cosh(2*arccosh(n)). - Artur Jasinski, Feb 10 2010
a(n) = 4*n + a(n-1) - 2 for n > 0, a(0) = 1. - Vincenzo Librandi, Aug 07 2010
a(n) = (((n-1)^2 + n^2))/2 + (n^2 + (n+1)^2)/2. - J. M. Bergot, May 31 2012
a(n) = A251599(3*n) for n > 0. - Reinhard Zumkeller, Dec 13 2014
a(n) = sqrt(8*(A000217(n-1)^2 + A000217(n)^2) + 1). - J. M. Bergot, Sep 03 2015
E.g.f.: (2*x^2 + 2*x + 1)*exp(x). - G. C. Greubel, Jul 14 2017
a(n) = A002378(n) + A002061(n). - Bruce J. Nicholson, Aug 06 2017
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(2))*coth(Pi/sqrt(2)))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(2))*csch(Pi/sqrt(2)))/2. (End)
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(2))*sinh(Pi).
Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(2))*csch(Pi/sqrt(2)). (End)
From Leo Tavares, May 23 2022: (Start)
a(n) = A000384(n+1) - 3*n.
a(n) = 3*A000217(n) + A000217(n-2). (End)
a(n) = a(-n) for all n in Z and A037235(n) = Sum_{k=0..n-1} a(k). - Michael Somos, Oct 19 2022

Extensions

Revised description from Noam Katz (noamkj(AT)hotmail.com), Jan 28 2001

A005899 Number of points on surface of octahedron; also coordination sequence for cubic lattice: a(0) = 1; for n > 0, a(n) = 4n^2 + 2.

Original entry on oeis.org

1, 6, 18, 38, 66, 102, 146, 198, 258, 326, 402, 486, 578, 678, 786, 902, 1026, 1158, 1298, 1446, 1602, 1766, 1938, 2118, 2306, 2502, 2706, 2918, 3138, 3366, 3602, 3846, 4098, 4358, 4626, 4902, 5186, 5478, 5778, 6086, 6402, 6726, 7058, 7398, 7746, 8102, 8466
Offset: 0

Views

Author

Keywords

Comments

Also, the number of regions the plane can be cut into by two overlapping concave (2n)-gons. - Joshua Zucker, Nov 05 2002
If X is an n-set and Y_i (i=1,2,3) are mutually disjoint 2-subsets of X then a(n-5) is equal to the number of 5-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007
Binomial transform of a(n) is A055580(n). - Wesley Ivan Hurt, Apr 15 2014
The identity (4*n^2+2)^2 - (n^2+1)*(4*n)^2 = 4 can be written as a(n)^2 - A002522(n)*A008586(n)^2 = 4. - Vincenzo Librandi, Jun 15 2014
Also the least number of unit cubes required, at the n-th iteration, to surround a 3D solid built from unit cubes, in order to hide all its visible faces, starting with a unit cube. - R. J. Cano, Sep 29 2015
Also, coordination sequence for "tfs" 3D uniform tiling. - N. J. A. Sloane, Feb 10 2018
Also, the number of n-th order specular reflections arriving at a receiver point from an emitter point inside a cuboid with reflective faces. - Michael Schutte, Sep 18 2018

References

  • H. S. M. Coxeter, "Polyhedral numbers," in R. S. Cohen et al., editors, For Dirk Struik. Reidel, Dordrecht, 1974, pp. 25-35.
  • Gmelin Handbook of Inorg. and Organomet. Chem., 8th Ed., 1994, TYPIX search code (225) cF8
  • B. Grünbaum, Uniform tilings of 3-space, Geombinatorics, 4 (1994), 49-56. See tilings #16 and #22.
  • R. W. Marks and R. B. Fuller, The Dymaxion World of Buckminster Fuller. Anchor, NY, 1973, p. 46.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Partial sums give A001845.
Column 2 * 2 of array A188645.
The 28 uniform 3D tilings: cab: A299266, A299267; crs: A299268, A299269; fcu: A005901, A005902; fee: A299259, A299265; flu-e: A299272, A299273; fst: A299258, A299264; hal: A299274, A299275; hcp: A007899, A007202; hex: A005897, A005898; kag: A299256, A299262; lta: A008137, A299276; pcu: A005899, A001845; pcu-i: A299277, A299278; reo: A299279, A299280; reo-e: A299281, A299282; rho: A008137, A299276; sod: A005893, A005894; sve: A299255, A299261; svh: A299283, A299284; svj: A299254, A299260; svk: A010001, A063489; tca: A299285, A299286; tcd: A299287, A299288; tfs: A005899, A001845; tsi: A299289, A299290; ttw: A299257, A299263; ubt: A299291, A299292; bnn: A007899, A007202. See the Proserpio link in A299266 for overview.
Row 3 of A035607, A266213, A343599.
Column 3 of A113413, A119800, A122542.

Programs

Formula

G.f.: ((1+x)/(1-x))^3. - Simon Plouffe in his 1992 dissertation
Binomial transform of [1, 5, 7, 1, -1, 1, -1, 1, ...]. - Gary W. Adamson, Nov 02 2007
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=1, a(1)=6, a(2)=18, a(3)=38. - Harvey P. Dale, Nov 08 2011
Recurrence: n*a(n) = (n-2)*a(n-2) + 6*a(n-1), a(0)=1, a(1)=6. - Fung Lam, Apr 15 2014
For n > 0, a(n) = A001844(n-1) + A001844(n) = (n-1)^2 + 2n^2 + (n+1)^2. - Doug Bell, Aug 18 2015
For n > 0, a(n) = A010014(n) - A195322(n). - R. J. Cano, Sep 29 2015
For n > 0, a(n) = A000384(n+1) + A014105(n-1). - Bruce J. Nicholson, Oct 08 2017
a(n) = A008574(n) + A008574(n-1) + a(n-1). - Bruce J. Nicholson, Dec 18 2017
a(n) = 2*d*Hypergeometric2F1(1-d, 1-n, 2, 2) where d=3, n>0. - Shel Kaphan, Feb 16 2023
a(n) = A035597(n)*3/n, for n>0. - Shel Kaphan, Feb 26 2023
E.g.f.: exp(x)*(2 + 4*x + 4*x^2) - 1. - Stefano Spezia, Mar 08 2023
Sum_{n>=0} 1/a(n) = 3/4 + Pi *sqrt(2)*coth( Pi/sqrt 2)/8 = 1.31858... - R. J. Mathar, Apr 27 2024

A003499 a(n) = 6*a(n-1) - a(n-2), with a(0) = 2, a(1) = 6.

Original entry on oeis.org

2, 6, 34, 198, 1154, 6726, 39202, 228486, 1331714, 7761798, 45239074, 263672646, 1536796802, 8957108166, 52205852194, 304278004998, 1773462177794, 10336495061766, 60245508192802, 351136554095046, 2046573816377474, 11928306344169798, 69523264248641314
Offset: 0

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Author

Keywords

Comments

Two times Chebyshev polynomials of the first kind evaluated at 3.
Also 2(a(2*n)-2) and a(2*n+1)-2 are perfect squares. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
Chebyshev polynomials of the first kind evaluated at 3, then multiplied by 2. - Michael Somos, Apr 07 2003
Also gives solutions > 2 to the equation x^2 - 3 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004
Output of Lu and Wu's formula for the number of perfect matchings of an m X n Klein bottle where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009
It appears that for prime P = 8*n +- 3, that a((P-1)/2) == -6 (mod P) and for all composites C = 8*n +- 3, there is at least one i < (C-1)/2 such that a(i) == -6 (mod P). Only a few of the primes P of the form 8*n +-3, e.g., 29, had such an i less than (P-1)/2. As for primes P = 8*n +- 1, it seems that the sum of the two adjacent terms, a((P-1)/2) and a((P+1)/2), is congruent to 8 (mod P). - Kenneth J Ramsey, Feb 14 2012 and Mar 05 2012
For n >= 1, a(n) is also the curvature of circles (rounded to the nearest integer) successively inscribed toward angle 90 degree of tangent lines, starting with a unit circle. The expansion factor is 5.828427... or 1/(3 - 2*sqrt(2)), which is also 3 + 2*sqrt(2) or A156035. See illustration in links. - Kival Ngaokrajang, Sep 04 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 6*x*y + y^2 + 32 = 0. - Colin Barker, Feb 08 2014

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 198.
  • 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 and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

A081555(n) = 1 + a(n).
Bisection of A002203.
First row of array A103999.
Row 1 * 2 of array A188645. A174501.

Programs

  • GAP
    a:=[2,6];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 16 2020
  • Magma
    I:=[2,6]; [n le 2 select I[n] else 6*Self(n-1) -Self(n-2): n in [1..25]]; // G. C. Greubel, Jan 16 2020
    
  • Magma
    R:=PowerSeriesRing(Integers(), 25); Coefficients(R!( (2-6*x)/(1 - 6*x + x^2) )); // Marius A. Burtea, Jan 16 2020
    
  • Maple
    A003499:=-2*(-1+3*z)/(1-6*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    a[0]=2; a[1]=6; a[n_]:= 6a[n-1] -a[n-2]; Table[a[n], {n,0,25}] (* Robert G. Wilson v, Jan 30 2004 *)
    Table[Tr[MatrixPower[{{6, -1}, {1, 0}}, n]], {n, 25}] (* Artur Jasinski, Apr 22 2008 *)
    LinearRecurrence[{6, -1}, {2, 6}, 25] (* Vladimir Joseph Stephan Orlovsky, Feb 26 2012 *)
    CoefficientList[Series[(2-6x)/(1-6x+x^2), {x,0,25}], x] (* Vincenzo Librandi, Jun 07 2013 *)
    (* From Eric W. Weisstein, Apr 17 2018 *)
    Table[(3-2Sqrt[2])^n + (3+2Sqrt[2])^n, {n,0,25}]//Expand
    Table[(1+Sqrt[2])^(2n) + (1-Sqrt[2])^(2n), {n,0,25}]//FullSimplify
    Join[{2}, Table[Fibonacci[4n, 2]/Fibonacci[2n, 2], {n, 25}]]
    2*ChebyshevT[Range[0, 25], 3] (* End *)
  • PARI
    a(n)=2*real((3+quadgen(32))^n)
    
  • PARI
    a(n)=2*subst(poltchebi(abs(n)),x,3)
    
  • PARI
    a(n)=if(n<0,a(-n),polsym(1-6*x+x^2,n)[n+1])
    
  • Sage
    [lucas_number2(n,6,1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008
    

Formula

G.f.: (2-6*x)/(1 - 6*x + x^2).
a(n) = (3+2*sqrt(2))^n + (3-2*sqrt(2))^n = 2*A001541(n).
For all sequence elements n, 2*n^2 - 8 is a perfect square. Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 06 2002
a(2*n)+2 is a perfect square, 2(a(2*n+1)+2) is a perfect square. a(n), a(n-1) and A077445(n), n > 0, satisfy the Diophantine equation x^2 + y^2 - 3*z^2 = -8. - Mario Catalani (mario.catalani(AT)unito.it), Mar 24 2003
a(n+1) is the trace of n-th power of matrix {{6, -1}, {1, 0}}. - Artur Jasinski, Apr 22 2008
a(n) = Product_{r=1..n} (4*sin^2((4*r-1)*Pi/(4*n)) + 4). [Lu/Wu] - Sarah-Marie Belcastro, Jul 04 2009
a(n) = (1 + sqrt(2))^(2*n) + (1 + sqrt(2))^(-2*n). - Gerson Washiski Barbosa, Sep 19 2010
For n > 0, a(n) = A001653(n) + A001653(n+1). - Charlie Marion, Dec 27 2011
For n > 0, a(n) = b(4*n)/b(2*n) where b(n) is the Pell sequence, A000129. - Kenneth J Ramsey, Feb 14 2012
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 3 - 2*sqrt(2). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.16585 37786 96882 80543 ... = 2 + 1/(6 + 1/(34 + 1/(198 + ...))). Cf. A174501.
Also F(-alpha) = 0.83251219269380007634 ... has the continued fraction representation 1 - 1/(6 - 1/(34 - 1/(198 - ...))) and the simple continued fraction expansion 1/(1 + 1/((6-2) + 1/(1 + 1/((34-2) + 1/(1 + 1/((198-2) + 1/(1 + ...))))))). Cf. A174501 and A003500.
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((6^2-4) + 1/(1 + 1/((34^2-4) + 1/(1 + 1/((198^2-4) + 1/(1 + ...))))))).
(End)
G.f.: G(0), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013
Inverse binomial transform of A228568 [Bhadouria]. - R. J. Mathar, Nov 10 2013
From Peter Bala, Oct 16 2019: (Start)
4*Sum_{n >= 1} 1/(a(n) - 8/a(n)) = 1.
8*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 4/a(n)) = 1.
Series acceleration formulas for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/4 - 8*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 8)) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 4*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 4)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(3-2*sqrt(2)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(2*sqrt(2)-3))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A067902.
(End)
E.g.f.: 2*exp(3*x)*cosh(2*sqrt(2)*x). - Stefano Spezia, Oct 18 2019
a(2*n)+2 = a(n)^2. - Greg Dresden and Shraya Pal, Jun 29 2021

A023039 a(n) = 18*a(n-1) - a(n-2).

Original entry on oeis.org

1, 9, 161, 2889, 51841, 930249, 16692641, 299537289, 5374978561, 96450076809, 1730726404001, 31056625195209, 557288527109761, 10000136862780489, 179445175002939041, 3220013013190122249, 57780789062419261441
Offset: 0

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Author

Keywords

Comments

The primitive Heronian triangle 3*a(n) +- 2, 4*a(n) has the latter side cut into 2*a(n) +- 3 by the corresponding altitude and has area 10*a(n)*A060645(n). - Lekraj Beedassy, Jun 25 2002
Chebyshev polynomials T(n,x) evaluated at x=9.
{a(n)} gives all (unsigned, integer) solutions of Pell equation a(n)^2 - 80*b(n)^2 = +1 with b(n) = A049660(n), n >= 0.
{a(n)} gives all possible solutions for x in Pell equation x^2 - D*y^2 = 1 for D=5, D=20 and D=80. The corresponding values for y are A060645 (D=5), A207832 (D=20) and A049660 (D=80). - Herbert Kociemba, Jun 05 2022
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r=sqrt(5). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions > 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r=sqrt(5). - Benoit Cloitre, Feb 24 2004
For all terms x of the sequence, 5*x^2 - 5 is a square, A004292(n)^2.
The a(n) are the x-values in the nonnegative integer solutions of x^2 - 5y^2 = 1, see A060645(n) for the corresponding y-values. - Sture Sjöstedt, Nov 29 2011
Rightmost digits alternate repeatedly: 1 and 9 in fact, a(2) = 18*9 - 1 == 1 (mod 10); a(3) = 18*1 - 9 == 9 (mod 10) therefore a(2n) == 1 (mod 10), a(2n+1) == 9 (mod 10). - Carmine Suriano, Oct 03 2013

Examples

			G.f. = 1 + 9*x + 161*x^2 + 2889*x^3 + 51841*x4 + 930249*x^5 + 16692641*x^6 + ...
		

Crossrefs

Row 2 of array A188645.
Row 4 of A322790.

Programs

  • Magma
    I:=[1, 9]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 13 2012
    
  • Maple
    a := n -> hypergeom([n, -n], [1/2], -4):
    seq(simplify(a(n)), n=0..16); # Peter Luschny, Jul 26 2020
  • Mathematica
    LinearRecurrence[{18, -1}, {1, 9}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
    CoefficientList[Series[(1-9*x)/(1-18*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)
  • PARI
    {a(n) = fibonacci(6*n) / 2 + fibonacci(6*n - 1)}; /* Michael Somos, Aug 11 2009 */
    
  • PARI
    x='x+O('x^30); Vec((1-9*x)/(1-18*x+x^2)) \\ G. C. Greubel, Dec 19 2017

Formula

a(n) ~ (1/2)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->infinity} a(n)/a(n-1) = phi^6 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 9) = (S(n, 18) - S(n-2, 18))/2, with S(n, x) := U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 18)=A049660(n+1).
a(n) = sqrt(80*A049660(n)^2 + 1) (cf. Richardson comment).
a(n) = ((9 + 4*sqrt(5))^n + (9 - 4*sqrt(5))^n)/2.
G.f.: (1 - 9*x)/(1 - 18*x + x^2).
a(n) = cosh(2*n*arcsinh(2)). - Herbert Kociemba, Apr 24 2008
a(n) = A001077(2*n). - Michael Somos, Aug 11 2009
From Johannes W. Meijer, Jul 01 2010: (Start)
a(n) = 2*A167808(6*n+1) - A167808(6*n+3).
Limit_{k->infinity} a(n+k)/a(k) = a(n) + A060645(n)*sqrt(5).
Limit_{n->infinity} a(n)/A060645(n) = sqrt(5).
(End)
a(n) = (1/2)*A087215(n) = (1/2)*(sqrt(5) + 2)^(2*n) + (1/2)*(sqrt(5) - 2)^(2*n).
Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1/8. Compare with A005248, A002878 and A075796. - Peter Bala, Nov 29 2013
a(n) = 2*A115032(n-1) - 1 = S(n, 18) - 9*S(n-1, 18), with A115032(-1) = 1, and see the above formula with S(n, 18) using its recurrence. - Wolfdieter Lang, Aug 22 2014
a(n) = A128052(3n). - A.H.M. Smeets, Oct 02 2017
a(n) = A049660(n+1) - 9*A049660(n). - R. J. Mathar, May 24 2018
a(n) = hypergeom([n, -n], [1/2], -4). - Peter Luschny, Jul 26 2020
a(n) = L(6*n)/2 for L(n) the Lucas sequence A000032(n). - Greg Dresden, Dec 07 2021
a(n) = cosh(6*n*arccsch(2)). - Peter Luschny, May 25 2022

Extensions

Chebyshev and Pell comments from Wolfdieter Lang, Nov 08 2002
Sture Sjöstedt's comment corrected and reformulated by Wolfdieter Lang, Aug 24 2014

A078986 Chebyshev T(n,19) polynomial.

Original entry on oeis.org

1, 19, 721, 27379, 1039681, 39480499, 1499219281, 56930852179, 2161873163521, 82094249361619, 3117419602578001, 118379850648602419, 4495316905044313921, 170703662541035326579, 6482243859654298096081, 246154563004322292324499, 9347391150304592810234881, 354954709148570204496600979, 13478931556495363178060602321
Offset: 0

Views

Author

Wolfdieter Lang, Jan 10 2003

Keywords

Comments

a(n+1)^2 - 10*(6*A078987(n))^2 = 1, n >= 0 (Pell equation +1, see A033313 and A033317).
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r=sqrt(10). - Benoit Cloitre, Feb 14 2004
Numbers n such that 10*(n^2 - 1) is a square. - Vincenzo Librandi, Aug 08 2010

Crossrefs

Row 3 of array A188645.

Programs

  • Mathematica
    LinearRecurrence[{38, -1},{1, 19},15] (* Ray Chandler, Aug 11 2015 *)
  • PARI
    a(n) = polchebyshev(n, 1, 19); \\ Michel Marcus, Jan 14 2018
  • Sage
    [lucas_number2(n,38,1)/2 for n in range(0, 16)] # Zerinvary Lajos, Nov 07 2009
    

Formula

a(n) = 38*a(n-1) - a(n-2), a(-1) := 19, a(0)=1.
G.f.: (1-19*x)/(1-38*x+x^2).
a(n) = T(n, 19) = (S(n, 38)-S(n-2, 38))/2 = S(n, 38)-19*S(n-1, 38) with T(n, x), resp. S(n, x), Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(n, 38) = A078987(n).
a(n) = (ap^n + am^n)/2 with ap := 19+6*sqrt(10) and am := 19-6*sqrt(10).
a(n) = Sum_{k=0..floor(n/2)} ((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*19)^(n-2*k), n >= 1.
a(n) = cosh(2*arcsinh(3)*n). - Herbert Kociemba, Apr 24 2008

Extensions

More terms from Indranil Ghosh, Feb 04 2017

A188647 Array read by antidiagonals of a(n) = a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2+1)+x)^2 for integers x>=1.

Original entry on oeis.org

1, 5, 1, 29, 17, 1, 169, 305, 37, 1, 985, 5473, 1405, 65, 1, 5741, 98209, 53353, 4289, 101, 1, 33461, 1762289, 2026009, 283009, 10301, 145, 1, 195025, 31622993, 76934989, 18674305, 1050601, 21169, 197, 1, 1136689, 567451585, 2921503573, 1232221121, 107151001, 3090529, 39005, 257, 1
Offset: 0

Views

Author

Charles L. Hohn, Apr 06 2011

Keywords

Comments

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=1 results shown here; y=-1 results are A188646.
Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is (1/sqrt(n^2+1)) * T_{2*k+1}(sqrt(n^2+1)), with T the Chebyshev polynomial of the first kind. - Seiichi Manyama, Jan 02 2019

Examples

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   5,     29,       169,          985, ...
   2 | 1,  17,    305,      5473,        98209, ...
   3 | 1,  37,   1405,     53353,      2026009, ...
   4 | 1,  65,   4289,    283009,     18674305, ...
   5 | 1, 101,  10301,   1050601,    107151001, ...
   6 | 1, 145,  21169,   3090529,    451196065, ...
   7 | 1, 197,  39005,   7722793,   1529074009, ...
   8 | 1, 257,  66305,  17106433,   4413393409, ...
   9 | 1, 325, 105949,  34539049,  11259624025, ...
  10 | 1, 401, 161201,  64802401,  26050404001, ...
  11 | 1, 485, 235709, 114554089,  55673051545, ...
  12 | 1, 577, 333505, 192765313, 111418017409, ...
  13 | 1, 677, 459005, 311204713, 210996336409, ...
  14 | 1, 785, 617009, 484968289, 381184458145, ...
  15 | 1, 901, 812701, 733055401, 661215159001, ...
  ...
		

Crossrefs

Row 1 is A001653, row 2 is A007805, row 3 is A097315, row 4 is A078988, row 5 is A097727, row 6 is A097730, row 7 is A097733, row 8 is A097736, row 9 is A097739, row 10 is A097742, row 11 is A097767, row 12 is A097770, row 13 is A097773.
Column 1 is A053755.
A(n,n) gives A323012.
Cf. A188645, A188646 (f(x, y) as above with y=-1).

Formula

A(n,k) = 2 * A188645(n,k) - A(n,k-1).
A(n,k) = Sum_{j=0..k} binomial(2*k+1,2*j)*(n^2+1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 02 2019

Extensions

Edited and extended by Seiichi Manyama, Jan 02 2019

A087215 Lucas(6*n): a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

Original entry on oeis.org

2, 18, 322, 5778, 103682, 1860498, 33385282, 599074578, 10749957122, 192900153618, 3461452808002, 62113250390418, 1114577054219522, 20000273725560978, 358890350005878082, 6440026026380244498, 115561578124838522882, 2073668380220713167378
Offset: 0

Views

Author

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 19 2003

Keywords

Comments

a(n+1)/a(n) converges to 9 + sqrt(80) = 17.9442719... a(0)/a(1) = 2/18; a(1)/a(2) = 18/322; a(2)/a(3) = 322/5778; a(3)/a(4) = 5778/103682; etc.
Lim_{n -> oo} a(n)/a(n+1) = 0.05572809000084... = 1/(9 + sqrt(80)) = 9 - sqrt(80).
From Peter Bala, Oct 13 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = (1/2)*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^6) = 1.0555459720... = 1 + 1/(18 + 1/(322 + 1/(5778 + ...))).
Also F(-Phi^6) = 0.9444348576... has the continued fraction representation 1 - 1/(18 - 1/(322 - 1/(5788 - ...))) and the simple continued fraction expansion 1/(1 + 1/((18 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/((5788 - 2) + 1/(1 + ...))))))).
F(Phi^6)*F(-Phi^6) = 0.9968944099... has the simple continued fraction expansion 1/(1 + 1/((18^2 - 4) + 1/(1 + 1/((322^2 - 4) + 1/(1 + 1/((5788^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^6)/F(-Phi^6) = 1.0588241282... has the simple continued fraction expansion 1 + 1/((18 - 2) + 1/(1 + 1/((5778 - 2) + 1/(1 + 1/(1860498 - 2) + 1/(1 + ...))))). (End)

Examples

			a(4) = 103682 = 18*a(3) - a(2) = 18*5778 - 322 = (9 + sqrt(80))^4 + (9 - sqrt(80))^4 = 103681.99999035512... + 0.00000964487... = 103682.
		

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
  • R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Crossrefs

Cf. A074919.
Row 2 * 2 of array A188645.
Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

Programs

  • Magma
    [ Lucas(6*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011
    
  • Mathematica
    a[0] = 2; a[1] = 18; a[n_] := 18a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
    Table[LucasL[6n], {n, 0, 18}]  (* or *) CoefficientList[Series[2*(1 - 9*x)/(1 - 18*x + x^2), {x, 0, 17}], x] (* Indranil Ghosh, Mar 15 2017 *)
  • PARI
    Vec(2*(1-9*x)/(1-18*x+x^2) + O(x^20)) \\ Colin Barker, Jan 24 2016
    
  • PARI
    a(n) = if(n<2, 17^n + 1, 18*a(n - 1) - a(n - 2));
    for(n=0, 17, print1(a(n),", ")) \\ Indranil Ghosh, Mar 15 2017

Formula

a(n) = A000032(6*n).
a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.
a(n) = (9 + sqrt(80))^n + (9 - sqrt(80))^n.
G.f.: 2*(1-9*x)/(1-18*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = 2*A023039(n). - R. J. Mathar, Oct 22 2010
From Peter Bala, Oct 13 2019: (Start)
a(n) = F(6*n+6)/F(6) - F(6*n-6)/F(6) = A049660(n+1) - A049660(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^6 = [5, 8; 8, 13].
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).
16*Sum_{n >= 1} 1/(a(n) - 20/a(n)) = 1: (20 = Lucas(6) + 2 and 16 = Lucas(6) - 2)
20*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 16/a(n)) = 1.
Series acceleration formulas for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/16 - 20*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 20)).
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/20 + 16*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 16)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(9-4*sqrt(5)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(4*sqrt(5)-9))^2 )/4,
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 18*x^2 + 323*x^3 + ... is the o.g.f. for A049660. (End)
E.g.f.: 2*exp(9*x)*cosh(4*sqrt(5)*x). - Stefano Spezia, Oct 18 2019
a(n) = L(2n-1)^2 * F(2n+1) + L(2n+1)^2 * F(2n-1), where F(n) = A000045(n) and L(n) = A000032(n). - Diego Rattaggi, Nov 12 2020
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^3 - 3*Lucas(2*n) = 2*T(3, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind; more generally, for k >= 0, Lucas(2*k*n) = 2*T(k, Lucas(2*n)/2).
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3(9 - sqrt(80))^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3(sqrt(80) - 9)^2), where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

A188644 Array of (k^n + k^(-n))/2 where k = (sqrt(x^2-1) + x)^2 for integers x >= 1.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 97, 17, 1, 1, 1351, 577, 31, 1, 1, 18817, 19601, 1921, 49, 1, 1, 262087, 665857, 119071, 4801, 71, 1, 1, 3650401, 22619537, 7380481, 470449, 10081, 97, 1, 1, 50843527, 768398401, 457470751, 46099201, 1431431, 18817, 127, 1
Offset: 0

Views

Author

Charles L. Hohn, Apr 06 2011

Keywords

Comments

Conjecture: Given the function f(x,y) = (sqrt(x^2+y) + x)^2 and constant k=f(x,y), then for all integers x >= 1 and y=[+-]1, k may be irrational, but (k^n + k^(-n))/2 always produces integer sequences; y=-1 results shown here; y=1 results are A188645.
Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{2*k}(x), evaluated at x=n. - Seiichi Manyama, Dec 30 2018

Examples

			Row 2 gives {( (2+sqrt(3))^(2*n) + (2-sqrt(3))^(2*n) )/2}.
Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   1,      1,         1,            1, ...
   2 | 1,   7,     97,      1351,        18817, ...
   3 | 1,  17,    577,     19601,       665857, ...
   4 | 1,  31,   1921,    119071,      7380481, ...
   5 | 1,  49,   4801,    470449,     46099201, ...
   6 | 1,  71,  10081,   1431431,    203253121, ...
   7 | 1,  97,  18817,   3650401,    708158977, ...
   8 | 1, 127,  32257,   8193151,   2081028097, ...
   9 | 1, 161,  51841,  16692641,   5374978561, ...
  10 | 1, 199,  79201,  31521799,  12545596801, ...
  11 | 1, 241, 116161,  55989361,  26986755841, ...
  12 | 1, 287, 164737,  94558751,  54276558337, ...
  13 | 1, 337, 227137, 153090001, 103182433537, ...
  14 | 1, 391, 305761, 239104711, 186979578241, ...
  15 | 1, 449, 403201, 362074049, 325142092801, ...
  ...
		

Crossrefs

Row 2 is A011943, row 3 is A056771, row 8 is A175633, (row 2)*2 is A067902, (row 9)*2 is A089775.
(column 1)*2 is A060626.
Cf. A188645 (f(x, y) as above with y=1).
Diagonals give A173129, A322899.

Programs

  • Mathematica
    max = 9; y = -1; t = Table[k = ((x^2 + y)^(1/2) + x)^2; ((k^n) + (k^(-n)))/2 // FullSimplify, {n, 0, max - 1}, {x, 1, max}]; Table[ t[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)

Formula

A(n,k) = (A188646(n,k-1) + A188646(n,k))/2.
A(n,k) = Sum_{j=0..k} binomial(2*k,2*j)*(n^2-1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Extensions

Edited by Seiichi Manyama, Dec 30 2018
More terms from Seiichi Manyama, Jan 01 2019

A099370 Chebyshev polynomial of the first kind, T(n,x), evaluated at x=33.

Original entry on oeis.org

1, 33, 2177, 143649, 9478657, 625447713, 41270070401, 2723199198753, 179689877047297, 11856808685922849, 782369683393860737, 51624542295308885793, 3406437421806992601601, 224773245296966202819873
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

Used in A099369.
Solutions of the Pell equation x^2 - 17y^2 = 1 (x values). After initial term this sequence bisects A041024. See 8*A097316(n-1) with A097316(-1) = 0 for corresponding y values. a(n+1)/a(n) apparently converges to (4+sqrt(17))^2. (See related comments in A088317, which this sequence also bisects.). - Rick L. Shepherd, Jul 31 2006
From a(n) = T(n, 33) (see the formula section) and the de Moivre-Binet formula for T(n,x=33) follows a(n+1)/a(n) = 33 + 8*sqrt(17), which is the conjectured value (4+sqrt(17))^2 given in the previous comment by Rick L. Shepherd. - Wolfdieter Lang, Jun 28 2013
Also numbers k such that 17*(k-1)*(k+1) is a square. - Bruno Berselli, May 31 2025

Examples

			a(1)^2 - 17*A121470(1)^2 = 33^2 - 17*8^2 = 1089 - 1088 = 1.
		

Crossrefs

Row 4 of array A188645.

Programs

  • Mathematica
    LinearRecurrence[{66, -1},{1, 33},14] (* Ray Chandler, Aug 11 2015 *)
  • PARI
    \\ Program uses fact that continued fraction for sqrt(17) = [4,8,8,...].
    print1("1, "); forstep(n=2,40,2,v=vector(n,i,if(i>1,8,4)); print1(contfracpnqn(v)[1,1],", ")) \\ Rick L. Shepherd, Jul 31 2006
    
  • PARI
    vector(20,n,polchebyshev(n-1,1,33)) \\ Joerg Arndt, Jan 01 2021

Formula

a(n) = 66*a(n-1) - a(n-2), a(-1):= 33, a(0)=1.
a(n) = T(n, 33) = (S(n, 66)-S(n-2, 66))/2 = S(n, 66)-33*S(n-1, 66) with T(n, x), resp. S(n, x), Chebyshev polynomials of the first, resp.second, kind. See A053120 and A049310. S(n, 66) = A097316(n).
a(n) = ((33+8*sqrt(17))^n + (33-8*sqrt(17))^n)/2.
a(n) = Sum_{k=0..floor(n/2)} ((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*33)^(n-2*k), for n>=1, a(0)=1.
G.f.: (1-33*x)/(1-66*x+x^2).

Extensions

A-number for y values in Pell equation corrected by Wolfdieter Lang, Jun 28 2013
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