cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

A001835 a(n) = 4*a(n-1) - a(n-2), with a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 3, 11, 41, 153, 571, 2131, 7953, 29681, 110771, 413403, 1542841, 5757961, 21489003, 80198051, 299303201, 1117014753, 4168755811, 15558008491, 58063278153, 216695104121, 808717138331, 3018173449203, 11263976658481, 42037733184721, 156886956080403, 585510091136891
Offset: 0

Views

Author

Keywords

Comments

See A079935 for another version.
Number of ways of packing a 3 X 2*(n-1) rectangle with dominoes. - David Singmaster.
Equivalently, number of perfect matchings of the P_3 X P_{2(n-1)} lattice graph. - Emeric Deutsch, Dec 28 2004
The terms of this sequence are the positive square roots of the indices of the octagonal numbers (A046184) - Nicholas S. Horne (nairon(AT)loa.com), Dec 13 1999
Terms are the solutions to: 3*x^2 - 2 is a square. - Benoit Cloitre, Apr 07 2002
Gives solutions x > 0 of the equation floor(x*r*floor(x/r)) == floor(x/r*floor(x*r)) where r = 1 + sqrt(3). - Benoit Cloitre, Feb 19 2004
a(n) = L(n-1,4), where L is defined as in A108299; see also A001834 for L(n,-4). - Reinhard Zumkeller, Jun 01 2005
Values x + y, where (x, y) solves for x^2 - 3*y^2 = 1, i.e., a(n) = A001075(n) + A001353(n). - Lekraj Beedassy, Jul 21 2006
Number of 01-avoiding words of length n on alphabet {0,1,2,3} which do not end in 0. (E.g., for n = 2 we have 02, 03, 11, 12, 13, 21, 22, 23, 31, 32, 33.) - Tanya Khovanova, Jan 10 2007
sqrt(3) = 2/2 + 2/3 + 2/(3*11) + 2/(11*41) + 2/(41*153) + 2/(153*571) + ... - Gary W. Adamson, Dec 18 2007
The lower principal convergents to 3^(1/2), beginning with 1/1, 5/3, 19/11, 71/41, comprise a strictly increasing sequence; numerators = A001834, denominators = A001835. - Clark Kimberling, Aug 27 2008
From Gary W. Adamson, Jun 21 2009: (Start)
A001835 and A001353 = bisection of denominators of continued fraction [1, 2, 1, 2, 1, 2, ...]; i.e., bisection of A002530.
a(n) = determinant of an n*n tridiagonal matrix with 1's in the super- and subdiagonals and (3, 4, 4, 4, ...) as the main diagonal.
Also, the product of the eigenvalues of such matrices: a(n) = Product_{k=1..(n-1)/2)} (4 + 2*cos(2*k*Pi/n).
(End)
Let M = a triangle with the even-indexed Fibonacci numbers (1, 3, 8, 21, ...) in every column, and the leftmost column shifted up one row. a(n) starting (1, 3, 11, ...) = lim_{n->oo} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
a(n+1) is the number of compositions of n when there are 3 types of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
For n >= 2, a(n) equals the permanent of the (2*n-2) X (2*n-2) tridiagonal matrix with sqrt(2)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Primes in the sequence are apparently those in A096147. - R. J. Mathar, May 09 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 4xy + y^2 + 2 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14xy + y^2 + 32 = 0. - Colin Barker, Feb 10 2014
The (1,1) element of A^n where A = (1, 1, 1; 1, 2, 1; 1, 1, 2). - David Neil McGrath, Jul 23 2014
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001075. - René Gy, Feb 25 2018
a(n+1) is the number of spanning trees of the graph T_n, where T_n is a 2 X n grid with an additional vertex v adjacent to (1,1) and (2,1). - Kevin Long, May 04 2018
a(n)/A001353(n) is the resistance of an n-ladder graph whose edges are replaced by one-ohm resistors. The resistance in ohms is measured at two nodes at one end of the ladder. It approaches sqrt(3) - 1 for n -> oo. See A342568, A357113, and A357115 for related information. - Hugo Pfoertner, Sep 17 2022
a(n) is the number of ways to tile a 1 X (n-1) strip with three types of tiles: small isosceles right triangles (with small side length 1), 1 X 1 squares formed by joining two of those right triangles along the hypotenuse, and large isosceles right triangles (with large side length 2) formed by joining two of those right triangles along a short leg. As an example, here is one of the a(6)=571 ways to tile a 1 X 5 strip with these kinds of tiles:
| / \ |\ /| |
|/_\|\/_||. - Greg Dresden and Arjun Datta, Jun 30 2023
From Klaus Purath, May 11 2024: (Start)
For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 4xy + y^2 = -2 = A028872(-1). In general, the following applies to all sequences (t) satisfying t(i) = 4t(i-1) - t(i-2) with t(0) = 1 and two consecutive terms (x,y): x^2 - 4xy + y^2 = A028872(t(1)-2). This includes and interprets the Feb 04 2014 comments here and on A001075 by Colin Barker and the Dec 12 2012 comment on A001353 by Max Alekseyev. By analogy to this, for three consecutive terms (x,y,z) y^2 - xz = A028872(t(1)-2). This includes and interprets the Jul 10 2021 comment on A001353 by Bernd Mulansky.
If (t) is a sequence satisfying t(k) = 3t(k-1) + 3t(k-2) - t(k-3) or t(k) = 4t(k-1) - t(k-2) without regard to initial values and including this sequence itself, then a(n) = (t(k+2n+1) + t(k))/(t(k+n+1) + t(k+n)) always applies, as long as t(k+n+1) + t(k+n) != 0 for integer k and n >= 1. (End)
Binomial transform of 1, 0, 2, 4, 12, ... (A028860 without the initial -1) and reverse binomial transform of 1, 2, 6, 24, 108, ... (A094433 without the initial 1). - Klaus Purath, Sep 09 2024

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).
  • Leonhard Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 375.
  • F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.
  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • 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).
  • R. P. Stanley, Enumerative Combinatorics I, p. 292.

Crossrefs

Row 3 of array A099390.
Essentially the same as A079935.
First differences of A001353.
Partial sums of A052530.
Pairwise sums of A006253.
Bisection of A002530, A005246 and A048788.
First column of array A103997.
Cf. A001519, A003699, A082841, A101265, A125077, A001353, A001542, A096147 (subsequence of primes).

Programs

  • GAP
    a:=[1,1];; for n in [3..20] do a[n]:=4*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
  • Haskell
    a001835 n = a001835_list !! n
    a001835_list =
       1 : 1 : zipWith (-) (map (4 *) $ tail a001835_list) a001835_list
    -- Reinhard Zumkeller, Aug 14 2011
    
  • Magma
    [n le 2 select 1 else 4*Self(n-1)-Self(n-2): n in [1..25]]; // Vincenzo Librandi, Sep 16 2016
    
  • Maple
    f:=n->((3+sqrt(3))^(2*n-1)+(3-sqrt(3))^(2*n-1))/6^n; [seq(simplify(expand(f(n))),n=0..20)]; # N. J. A. Sloane, Nov 10 2009
  • Mathematica
    CoefficientList[Series[(1-3x)/(1-4x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Jul 25 2011, after g.f. *)
    LinearRecurrence[{4,-1},{1,1},30] (* Harvey P. Dale, Jun 08 2013 *)
    Table[Round@Fibonacci[2n-1, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
    Table[(3*ChebyshevT[n, 2] - ChebyshevU[n, 2])/2, {n, 0, 20}] (* G. C. Greubel, Dec 23 2019 *)
  • PARI
    {a(n) = real( (2 + quadgen(12))^n * (1 - 1 / quadgen(12)) )} /* Michael Somos, Sep 19 2008 */
    
  • PARI
    {a(n) = subst( (polchebyshev(n) + polchebyshev(n-1)) / 3, x, 2)} /* Michael Somos, Sep 19 2008 */
    
  • Sage
    [lucas_number1(n,4,1)-lucas_number1(n-1,4,1) for n in range(25)] # Zerinvary Lajos, Apr 29 2009
    
  • Sage
    [(3*chebyshev_T(n,2) - chebyshev_U(n,2))/2 for n in (0..20)] # G. C. Greubel, Dec 23 2019
    

Formula

G.f.: (1 - 3*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
a(1-n) = a(n).
a(n) = ((3 + sqrt(3))^(2*n - 1) + (3 - sqrt(3))^(2*n - 1))/6^n. - Dean Hickerson, Dec 01 2002
a(n) = (8 + a(n-1)*a(n-2))/a(n-3). - Michael Somos, Aug 01 2001
a(n+1) = Sum_{k=0..n} 2^k * binomial(n + k, n - k), n >= 0. - Len Smiley, Dec 09 2001
Limit_{n->oo} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 10 2002
a(n) = 2*A061278(n-1) + 1 for n > 0. - Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n - i, i); then q(n, 2) = a(n+1). - Benoit Cloitre, Nov 10 2002
a(n+1) = Sum_{k=0..n} ((-1)^k)*((2*n+1)/(2*n + 1 - k))*binomial(2*n + 1 - k, k)*6^(n - k) (from standard T(n,x)/x, n >= 1, Chebyshev sum formula). The Smiley and Cloitre sum representation is that of the S(2*n, i*sqrt(2))*(-1)^n Chebyshev polynomial. - Wolfdieter Lang, Nov 29 2002
a(n) = S(n-1, 4) - S(n-2, 4) = T(2*n-1, sqrt(3/2))/sqrt(3/2) = S(2*(n-1), i*sqrt(2))*(-1)^(n - 1), 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(-2, x) = -1, S(n, 4) = A001353(n+1), T(-1, x) = x.
a(n+1) = sqrt((A001834(n)^2 + 2)/3), n >= 0 (see Cloitre comment).
Sequence satisfies -2 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = (1/6)*(3*(2 - sqrt(3))^n + sqrt(3)*(2 - sqrt(3))^n + 3*(2 + sqrt(3))^n - sqrt(3)*(2 + sqrt(3))^n) (Mathematica's solution to the recurrence relation). - Sarah-Marie Belcastro, Jul 04 2009
If p[1] = 3, p[i] = 2, (i > 1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n+1) = det A. - Milan Janjic, Apr 29 2010
a(n) = (a(n-1)^2 + 2)/a(n-2). - Irene Sermon, Oct 28 2013
a(n) = A001353(n+1) - 3*A001353(n). - R. J. Mathar, Oct 30 2015
a(n) = a(n-1) + 2*A001353(n-1). - Kevin Long, May 04 2018
From Franck Maminirina Ramaharo, Nov 11 2018: (Start)
a(n) = (-1)^n*(A125905(n) + 3*A125905(n-1)), n > 0.
E.g.f.: exp^(2*x)*(3*cosh(sqrt(3)*x) - sqrt(3)*sinh(sqrt(3)*x))/3. (End)
From Peter Bala, Feb 12 2024: (Start)
For n in Z, a(n) = A001353(n) + A001353(1-n).
For n, j, k in Z, a(n)*a(n+j+k) - a(n+j)*a(n+k) = 2*A001353(j)*A001353(k). The case j = 1, k = 2 is given above. (End)

A001834 a(0) = 1, a(1) = 5, a(n) = 4*a(n-1) - a(n-2).

Original entry on oeis.org

1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085
Offset: 0

Views

Author

Keywords

Comments

Sequence also gives values of x satisfying 3*y^2 - x^2 = 2, the corresponding y being given by A001835(n+1). Moreover, quadruples(p, q, r, s) satisfying p^2 + q^2 + r^2 = s^2, where p = q and r is either p+1 or p-1, are termed nearly isosceles Pythagorean and are given by p = {x + (-1)^n}/3, r = p-(-1)^n, s = y for n > 1. - Lekraj Beedassy, Jul 19 2002
a(n)= A002531(1+2*n). - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007
361 written in base A001835(n+1) - 1 is the square of a(n). E.g., a(12) = 2672279, A001835(13) - 1 = 1542840. We have 361_(1542840) = 3*1542840 + 6*1542840 + 1 = 2672279^2. - Richard Choulet, Oct 04 2007
The lower principal convergents to 3^(1/2), beginning with 1/1, 5/3, 19/11, 71/41, comprise a strictly increasing sequence; numerators=A001834, denominators=A001835. - Clark Kimberling, Aug 27 2008
General recurrence is a(n) = (a(1) - 1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x + 1)^n, k = (a(1) - 3), x = (1 + sqrt((a(1) + 1)/(a(1) - 3)))/2. Examples in OEIS: a(1) = 4 gives A002878, primes in it A121534. a(1) = 5 gives A001834, primes in it A086386. a(1) = 6 gives A030221, primes in it A299109. a(1) = 7 gives A002315, primes in it A088165. a(1) = 8 gives A033890, primes in it not in OEIS (do there exist any?). a(1) = 9 gives A057080, primes in {71, 34649, 16908641, ...}. a(1) = 10 gives A057081, primes in it {389806471, 192097408520951, ...}. - Ctibor O. Zizka, Sep 02 2008
Inverse binomial transform of A030192. - Philippe Deléham, Nov 19 2009
For positive n, a(n) equals the permanent of the (2*n) X (2*n) tridiagonal matrix with sqrt(6)'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
x-values in the solution to 3x^2 + 6 = y^2 (see A082841 for the y-values). - Sture Sjöstedt, Nov 25 2011
Pisano period lengths: 1, 1, 2, 4, 3, 2, 8, 4, 6, 3, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
The aerated sequence (b(n))A100047%20for%20a%20connection%20with%20Chebyshev%20polynomials.%20-%20_Peter%20Bala">{n>=1} = [1, 0, 5, 0, 19, 0, 71, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -2, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for a connection with Chebyshev polynomials. - _Peter Bala, Mar 22 2015
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001835 and with any member of A001075. - René Gy, Feb 26 2018
From Wolfdieter Lang, Oct 15 2020: (Start)
((-1)^n)*a(n) = X(n) = (-1)^n*(S(n, 4) + S(n-1, 4)) and Y(n) = X(n-1) gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 4*X*Y = +6, for n = -oo..+oo, with Chebyshev S polynomials (see A049310), with S(-1, x) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.
This binary indefinite quadratic form of discriminant 12, representing 6, has only this family of proper solutions (modulo sign flip), and no improper ones.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)
Floretion Algebra Multiplication Program, FAMP Code: A001834 = (4/3)vesseq[ - .25'i + 1.25'j - .25'k - .25i' + 1.25j' - .25k' + 1.25'ii' + .25'jj' - .75'kk' + .75'ij' + .25'ik' + .75'ji' - .25'jk' + .25'ki' - .25'kj' + .25e], apart from initial term

Examples

			G.f. = 1 + 5*x + 19*x^2 + 71*x^3 + 265*x^4 + 989*x^5 + 3691*x^6 + ...
		

References

  • Bastida, Julio R. 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)
  • Leonhard Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 375.
  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • 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).
  • P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.

Crossrefs

A bisection of sequence A002531.
Cf. A001352, A001835, A086386 (prime members).
Cf. A026150.
a(n)^2+1 = A094347(n+1).

Programs

  • Haskell
    a001834 n = a001834_list !! (n-1)
    a001834_list = 1 : 5 : zipWith (-) (map (* 4) $ tail a001834_list) a001834_list
    -- Reinhard Zumkeller, Jan 23 2012
    
  • Magma
    I:=[1,5]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015
  • Maple
    f:=n->((1+sqrt(3))^(2*n+1)+(1-sqrt(3))^(2*n+1))/2^(n+1); # N. J. A. Sloane, Nov 10 2009
  • Mathematica
    a[0] = 1; a[1] = 5; a[n_] := a[n] = 4a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 25}] (* Robert G. Wilson v, Apr 24 2004 *)
    Table[Expand[((1+Sqrt[3])^(2*n+1)+(1+Sqrt[3])^(2*n+1))/2^(n+1)],{n, 0, 20}] (* Anton Vrba, Feb 14 2007 *)
    LinearRecurrence[{4, -1}, {1, 5}, 50] (* Sture Sjöstedt, Nov 27 2011 *)
    a[c_, n_] := Module[{},
       p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
       d := Numerator[Convergents[Sqrt[c], n p]];
       t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
       Return[t];
    ] (* Complement of A002531 *)
    a[3, 20] (* Gerry Martens, Jun 07 2015 *)
    Round@Table[LucasL[2n+1, Sqrt[2]]/Sqrt[2], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
  • PARI
    {a(n) = real( (2 + quadgen(12))^n * (1 + quadgen(12)) )}; /* Michael Somos, Sep 19 2008 */
    
  • PARI
    {a(n) = subst( polchebyshev(n-1, 2) + polchebyshev(n, 2), x, 2)}; /* Michael Somos, Sep 19 2008 */
    
  • SageMath
    [(lucas_number2(n,4,1)-lucas_number2(n-1,4,1))/2 for n in range(1, 27)] # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) = ((1 + sqrt(3))^(2*n + 1) + (1 - sqrt(3))^(2*n + 1))/2^(n + 1). - N. J. A. Sloane, Nov 10 2009
a(n) = (1/2) * ((1 + sqrt(3))*(2 + sqrt(3))^n + (1 - sqrt(3))*(2 - sqrt(3))^n). - Dean Hickerson, Dec 01 2002
From Mario Catalani, Apr 11 2003: (Start)
With a = 2 + sqrt(3), b = 2 - sqrt(3): a(n) = (1/sqrt(2))(a^(n + 1/2) - b^(n + 1/2)).
a(n) - a(n-1) = A003500(n).
a(n) = sqrt(1 + 12*A061278(n) + 12*A061278(n)^2). (End)
a(n) = ((1 + sqrt(3))^(2*n + 1) + (1 - sqrt(3))^(2*n + 1))/2^(n + 1). - Anton Vrba, Feb 14 2007
G.f.: (1 + x)/((1 - 4*x + x^2)). Simon Plouffe in his 1992 dissertation.
a(n) = S(2*n, sqrt(6)) = S(n, 4) + S(n-1, 4); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 4) = A001353(n).
For all members x of the sequence, 3*x^2 + 6 is a square. Limit_{n->infinity} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 10 2002
a(n) = 2*A001571(n) + 1. - Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
Let q(n, x) = Sum_{i=0..n} x^(n - i)*binomial(2*n - i, i); then (-1)^n*q(n, -6) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n + 1, 2*k)*3^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = floor(sqrt(3)*A001835(n+1)). - Philippe Deléham, Mar 03 2004
a(n+1) - 2*a(n) = 3*A001835(n+1). Using the known relation A001835(n+1) = sqrt((a(n)^2 + 2)/3) it follows that a(n+1) - 2*a(n) = sqrt(3*(a(n)^2 + 2)). Therefore a(n+1)^2 + a(n)^2 - 4*a(n+1)*a(n) - 6 = 0. - Creighton Dement, Apr 18 2005
a(n) = L(n,-4)*(-1)^n, where L is defined as in A108299; see also A001835 for L(n,+4). - Reinhard Zumkeller, Jun 01 2005
a(n) = Jacobi_P(n, 1/2, -1/2, 2)/Jacobi_P(n, -1/2, 1/2, 1). - Paul Barry, Feb 03 2006
Equals binomial transform of A026150 starting (1, 4, 10, 28, 76, ...) and double binomial transform of (1, 3, 3, 9, 9, 27, 27, 81, 81, ...). - Gary W. Adamson, Nov 30 2007
Sequence satisfies 6 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(-1-n) = -a(n). - Michael Somos, Sep 19 2008
From Franck Maminirina Ramaharo, Nov 11 2018: (Start)
a(n) = (-1)^n*(5*A125905(n) + A125905(n+1)).
E.g.f.: exp(2*x)*(cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)). (End)
a(n) = A061278(n+1) - A061278(n-1) for n>=2. - John P. McSorley, Jun 20 2020
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 2), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n) - 2*a(n-1) = 3 * A001835(n) for n >= 1.
For arbitrary x, a(n+x)^2 - 4*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 6 with a(n) := (1/2) * ((1 + sqrt(3))*(2 + sqrt(3))^n + (1 - sqrt(3))*(2 - sqrt(3))^n) as above. The particular case x = 0 is noted above,
a(n+1/2) = sqrt(6) * A001353(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6*sqrt(6) + 12) * A001353(n+1).
a(n+3/4) - a(n+1/4) = sqrt(2*sqrt(6) - 4) * A001075(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/6 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A001352(n) + 1/A001352(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3*(1 - 2/A102206(k))). (End)

A005320 a(n) = 4*a(n-1) - a(n-2), with a(0) = 0, a(1) = 3.

Original entry on oeis.org

0, 3, 12, 45, 168, 627, 2340, 8733, 32592, 121635, 453948, 1694157, 6322680, 23596563, 88063572, 328657725, 1226567328, 4577611587, 17083879020, 63757904493, 237947738952, 888033051315, 3314184466308, 12368704813917, 46160634789360, 172273834343523, 642934702584732, 2399464975995405, 8954925201396888, 33420235829592147
Offset: 0

Views

Author

Keywords

Comments

For n > 1, a(n-1) is the determinant of the n X n band matrix which has {2,4,4,...,4,4,2} on the diagonal and a 1 on the entire super- and subdiagonal. This matrix appears when constructing a natural cubic spline interpolating n equally spaced data points. - g.degroot(AT)phys.uu.nl, Feb 14 2007
Integer values of x that make 9+3*x^2 a perfect square. - Lorenz H. Menke, Jr., Mar 26 2008
The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence whose numerators are the terms of this sequence and denominators are A001075. - Clark Kimberling, Aug 27 2008
a(n) also give the altitude to the middle side of a Super-Heronian Triangle. - Johannes Boot, Oct 14 2010
a(n) gives values of y satisfying 3*x^2 - 4*y^2 = 12; corresponding x values are given by A003500. - Sture Sjöstedt, Dec 19 2017

References

  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [3*Evaluate(ChebyshevSecond(n), 2): n in [0..40]]; // G. C. Greubel, Oct 10 2022
    
  • Maple
    A005320:=3*z/(1-4*z+z**2); # Simon Plouffe in his 1992 dissertation
    a:= n-> (Matrix([[3,0]]). Matrix([[4,1],[ -1,0]])^n)[1,2]: seq(a(n), n=0..50); # Alois P. Heinz, Aug 14 2008
  • Mathematica
    LinearRecurrence[{4,-1},{0,3},40] (* Harvey P. Dale, Mar 04 2012 *)
  • PARI
    Vec(3/(x^2-4*x+1)+O(x^99)) \\ Charles R Greathouse IV, Mar 05 2012
    
  • SageMath
    [3*chebyshev_U(n-1,2) for n in range(41)] # G. C. Greubel, Oct 10 2022

Formula

a(n) = (sqrt(3)/2)*( (2+sqrt(3))^n - (2-sqrt(3))^n ). - Antonio Alberto Olivares, Jan 17 2004
G.f.: 3*x/(1-4*x+x^2). - Harvey P. Dale, Mar 04 2012
a(n) = 3*A001353(n). - R. J. Mathar, Mar 14 2016

Extensions

Typo in definition corrected by Johannes Boot, Feb 05 2009

A084917 Positive numbers of the form 3*y^2 - x^2.

Original entry on oeis.org

2, 3, 8, 11, 12, 18, 23, 26, 27, 32, 39, 44, 47, 48, 50, 59, 66, 71, 72, 74, 75, 83, 92, 98, 99, 104, 107, 108, 111, 122, 128, 131, 138, 143, 146, 147, 156, 162, 167, 176, 179, 183, 188, 191, 192, 194, 200, 207, 218, 219, 227, 234, 236, 239, 242, 243, 251, 263, 264, 275, 282, 284
Offset: 1

Views

Author

Roger Cuculière, Jul 14 2003

Keywords

Comments

Positive integers k such that x^2 - 4xy + y^2 + k = 0 has integer solutions. (See the CROSSREFS section for sequences relating to solutions for particular k.)
Comments on method used, from Colin Barker, Jun 06 2014: (Start)
In general, we want to find the values of f, from 1 to 400 say, for which x^2 + bxy + y^2 + f = 0 has integer solutions for a given b.
In order to solve x^2 + bxy + y^2 + f = 0 we can solve the Pellian equation x^2 - Dy^2 = N, where D = b*b - 4 and N = 4*(b*b - 4)*f.
But since sqrt(D) < N, the classical method of solving x^2 - Dy^2 = N does not work. So I implemented the method described in the 1998 sci.math reference, which says:
"There are several methods for solving the Pellian equation when |N| > sqrt(d). One is to use a brute-force search. If N < 0 then search on y = sqrt(abs(n/d)) to sqrt((abs(n)(x1 + 1))/(2d)) and if N > 0 search on y = 0 to sqrt((n(x1 - 1))/(2d)) where (x1, y1) is the minimum positive solution (x, y) to x^2 - dy^2 = 1. If N < 0, for each positive (x, y) found by the search, also take (-x, y). If N > 0, also take (x, -y). In either case, all positive solutions are generated from these using (x1, y1) in the standard way."
Incidentally all my Pell code is written in B-Prolog, and is somewhat voluminous. (End)
Also, positive integers of the form -x^2 + 2xy + 2y^2 of discriminant 12. - N. J. A. Sloane, May 31 2014 [Corrected by Klaus Purath, May 07 2023]
The equivalent sequence for x^2 - 3xy + y^2 + k = 0 is A031363.
The equivalent sequence for x^2 - 5xy + y^2 + k = 0 is A237351.
A positive k does not appear in this sequence if and only if there is no integer solution of x^2 - 3*y^2 = -k with (i) 0 < y^2 <= k/2 and (ii) 0 <= x^2 <= k/2. See the Nagell reference Theorems 108 a and 109, pp. 206-7, with D = 3, N = k and (x_1,y_1) = (2,1). - Wolfdieter Lang, Jan 09 2015
From Klaus Purath, May 07 2023: (Start)
There are no squares in this sequence. Products of an odd number of terms as well as products of an odd number of terms and any terms of A014209 belong to the sequence.
Products of an even number of terms are terms of A014209. The union of this sequence and A014209 is closed under multiplication.
A positive number belongs to this sequence if and only if it contains an odd number of prime factors congruent to {2, 3, 11} modulo 12. If it contains prime factors congruent to {5, 7} modulo 12, these occur only with even exponents. (End)
From Klaus Purath, Jul 09 2023: (Start)
Any term of the sequence raised to an odd power also belongs to the sequence. Proof: t^(2n+1) = t*t^2n = (3*x^2 - y^2)*t^2n = 3*(x*t^n)^2 - (y*t^n)^2. It seems that t^(2n+1) occurs only if t also is in the sequence.
Joerg Arndt has proved that there are no squares in this sequence: Assume s^2 = 3*y^2 - x^2, then s^2 + x^2 = 3 * y^2, but the sum of two squares cannot be 3 * y^2, qed. (End)
That products of any 3 terms belong to the sequence can be proved by the following identity: (na^2 - b^2) (nc^2 - d^2) (ne^2 - f^2) = n[a(nce + df) + b(cf + de)]^2 - [na(cf + de) + b(nce + df)]^2. This can be verified by expanding both sides of the equation. - Klaus Purath, Jul 14 2023

Examples

			11 is in the sequence because 3 * 3^2 - 4^2 = 27 - 16 = 11.
12 is in the sequence because 3 * 4^2 - 6^2 = 48 - 36 = 12.
13 is not in the sequence because there is no solution in integers to 3y^2 - x^2 = 13.
From _Wolfdieter Lang_, Jan 09 2015: (Start)
Referring to the Jan 09 2015 comment above.
k = 1 is out because there is no integer solution of (i) 0 < y^2 <= 1/2.
For k = 4, 5, 6, and 7 one has y = 1, x = 0, 1 (and the negative of this). But x^2 - 3 is not -k for these k and x values. Therefore, these k values are missing.
For k = 8 .. 16 one has y = 1, 2 and x = 0, 1, 2. Only y = 2 has a chance and only for k = 8, 11 and 12 the x value 2, 1 and 0, respectively, solves x^2 - 12 = -k. Therefore 9, 10, 13, 14, 15, 16 are missing.
... (End)
		

References

  • T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

Crossrefs

With respect to solutions of the equation in the early comment, see comments etc. in: A001835 (k = 2), A001075 (k = 3), A237250 (k = 11), A003500 (k = 12), A082841 (k = 18), A077238 (k = 39).
A141123 gives the primes.
For a list of sequences giving numbers and/or primes represented by binary quadratic forms, see the "Binary Quadratic Forms and OEIS" link.

Programs

  • Mathematica
    r[n_] := Reduce[n == 3*y^2 - x^2 && x > 0 && y > 0, {x, y}, Integers]; Reap[For[n = 1, n <= 1000, n++, rn = r[n]; If[rn =!= False, Print["n = ", n, ", ", rn /. C[1] -> 1 // Simplify]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jan 21 2016 *)
    Select[Range[300],Length[FindInstance[3y^2-x^2==#,{x,y},Integers]]>0&] (* Harvey P. Dale, Apr 23 2023 *)

Extensions

Terms 26 and beyond from Colin Barker, Feb 06 2014

A102871 a(n) = a(n-3) - 5*a(n-2) + 5*a(n-1), a(0) = 1, a(1) = 3, a(2) = 10.

Original entry on oeis.org

1, 3, 10, 36, 133, 495, 1846, 6888, 25705, 95931, 358018, 1336140, 4986541, 18610023, 69453550, 259204176, 967363153, 3610248435, 13473630586, 50284273908, 187663465045, 700369586271, 2613814880038, 9754889933880, 36405744855481, 135868089488043, 507066613096690
Offset: 0

Views

Author

Creighton Dement, Mar 01 2005

Keywords

Comments

A floretion-generated sequence resulting from a particular transform of the periodic sequence (-1,1).
Floretion Algebra Multiplication Program, FAMP Code: .5em[J* ]forseq[ .25( 'i + 'j + 'k + i' + j' + k' + 'ii' + 'jj' + 'kk' + 'ij' + 'ik' + 'ji' + 'jk' + 'ki' + 'kj' + e ) ], em[J]forseq = A001834, vesforseq = (1,-1,1,-1). ForType 1A. Identity used: em[J]forseq + em[J* ]forseq = vesforseq.
Also indices of the centered triangular numbers which are triangular numbers - Richard Choulet, Oct 09 2007
Place a(n) red and b(n) blue balls in an urn; draw 2 balls without replacement. Probability(2 red balls) = 3*Probability(2 blue balls); b(n)=A101265(n). - Paul Weisenhorn, Aug 02 2010

Examples

			For n=5, a(5)=495; b(5)=286; binomial(495,2) = 122265 = 3*binomial(286,2). - _Paul Weisenhorn_, Aug 02 2010
		

Crossrefs

Cf. A001075 (first differences), A001834, A082841, A101265.

Programs

  • Maple
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=4*a[n-1]-a[n-2]-1 od: seq(a[n], n=1..23); # Zerinvary Lajos, Mar 08 2008
  • Mathematica
    LinearRecurrence[{5,-5,1},{1,3,10},30] (* Harvey P. Dale, Oct 04 2011 *)

Formula

2*a(n) - A001834(n) = (-1)^(n+1); a(n) = 4*a(n-1) - a(n) - 1;
G.f.: (2*x-1)/((x-1)*(x^2-4*x+1)).
Superseeker results: a(n+2) - 2a(n+1) + a(n) = A001834(n+1) (from this and the first relation involving A001834 it follows that 4a(n+1) - a(n+2) - a(n) = (-1)^n as well as the recurrence relation given for A001834 ); a(n+1) - a(n) = A001075(n+1); a(n+2) - a(n) = A082841(n+1).
a(j+3) - 3*a(j+2) - 3*a(j+1) + a(j) = -2 for all j.
a(n+1) = 2*a(n) - 1/2 + (1/2)*(12*a(n)^2 - 12*a(n) + 9)^(1/2). - Richard Choulet, Oct 09 2007
a(n) = (sqrt(12*b(n)*(b(n)-1) + 1) + 1)/2; b(n) = A101265(n). - Paul Weisenhorn, Aug 02 2010
a(n) = A001571(n) + 1. - Johannes Boot, Jun 17 2011
E.g.f.: (exp(2*x)*(cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) + cosh(x) + sinh(x))/2. - Stefano Spezia, Sep 19 2023

A227418 Array A(n,k) with all numbers m such that 3*m^2 +- 3^k is a square and their corresponding square roots, read by downward antidiagonals.

Original entry on oeis.org

0, 1, 1, 0, 2, 4, 3, 3, 7, 15, 0, 6, 12, 26, 56, 9, 9, 21, 45, 97, 209, 0, 18, 36, 78, 168, 362, 780, 27, 27, 63, 135, 291, 627, 1351, 2911, 0, 54, 108, 234, 504, 1086, 2340, 5042, 10864, 81, 81, 189, 405, 873, 1881, 4053, 8733, 18817, 40545
Offset: 0

Views

Author

Richard R. Forberg, Sep 02 2013

Keywords

Comments

Array is analogous to A228405 in goal and structure, with key differences.
Left column is A001353. Top row (not in OEIS) interleaves 0 with the powers of 3, as: 0, 1, 0, 3, 0, 9, 0, 27, 0, 81.
Either or both may be used as initializing values. See Formula section.
The left column is the second binomial transform of the top row. The intermediate transform sequence is A002605, not present in this array.
The columns of the array hold all values, in sequential order, of numbers m such that 3*m^2 + 3^k or 3*m^2 - 3^k are squares, and their corresponding square roots in the next column, which then form the "next round" of m values for column k+1.
For example: A(n,0) are numbers such that 3*m^2 + 1 are squares, the integer square roots of each are in A(n,1), which are then numbers m such that 3*m^2 - 3 are squares, with those square roots in A(n,2), etc. The sign alternates for each increment of k, etc. No integer square roots exist for the opposite sign in a given column, regardless of n.
Also, A(n,1) are values of m such that floor(m^2/3) is square, with the corresponding square roots given by A(n,0).
A(n,k)/A(n,k-2) = 3; A(n,k)/A(n,k-1) converges to sqrt(3) for large n.
A(n,k)/A(n-1,k) converges to 2 + sqrt(3) for large n.
Several ways of combining the first few columns give OEIS sequences:
A(n,0) + A(n,1) = A001835; A(n,1) + A(n,2)= A001834; A(n,2) + A(n,3) = A082841;
A(n,0)*A(n,1)/2 = A007655(n); A(n+2,0)*A(n+1,1) = A001922(n);
A(n,0)*A(n+1,1) = A001921(n); A(n,0)^2 + A(n,1)^2 = A103974(n);
A(n,1)^2 - A(n,0)^2 = A011922(n); (A(n+2,0)^2 + A(n+1,1)^2)/2 = A122770(n) = 2*A011916(n).
The main diagonal (without initial 0) = 2*A090018. The first subdiagonal = abs(A099842). First superdiagonal = A141041.
A001353 (in left column) are the only initializing set of numbers where the recursive square root equation (see below) produces exclusively integer values, for all iterations of k. For any other initial values only even iterations (at k = 2, 4, ...) produce integers.

Examples

			The array, A(n, k), begins as:
    0,    1,    0,    3,    0,     9,     0,    27, ... see A000244;
    1,    2,    3,    6,    9,    18,    27,    54, ... A038754;
    4,    7,   12,   21,   36,    63,   108,   189, ... A228879;
   15,   26,   45,   78,  135,   234,   405,   702, ...
   56,   97,  168,  291,  504,   873,  1512,  2619, ...
  209,  362,  627, 1086, 1881,  3258,  5643,  9774, ...
  780, 1351, 2340, 4053, 7020, 12159, 21060, 36477, ...
Antidiagonal triangle, T(n, k), begins as:
   0;
   1,  1;
   0,  2,   4;
   3,  3,   7,  15;
   0,  6,  12,  26,  56;
   9,  9,  21,  45,  97,  209;
   0, 18,  36,  78, 168,  362,  780;
  27, 27,  63, 135, 291,  627, 1351, 2911;
   0, 54, 108, 234, 504, 1086, 2340, 5042, 10864;
  81, 81, 189, 405, 873, 1881, 4053, 8733, 18817, 40545;
		

Crossrefs

Programs

  • Magma
    function A(n,k)
      if k lt 0 then return 0;
      elif n eq 0 then return Round((1/2)*(1-(-1)^k)*3^((k-1)/2));
      elif k eq 0 then return Evaluate(ChebyshevSecond(n), 2);
      else return 2*A(n, k-1) - A(n-1, k-1);
      end if; return A;
    end function;
    A227418:= func< n,k | A(k, n-k) >;
    [A227418(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Oct 09 2022
    
  • Mathematica
    A[n_, k_]:= If[k<0, 0, If[k==0, ChebyshevU[n-1, 2], 2*A[n, k-1] - A[n-1, k-1]]];
    T[n_, k_]:= A[k, n-k];
    Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 09 2022 *)
  • SageMath
    def A(n,k):
        if (k<0): return 0
        elif (k==0): return chebyshev_U(n-1,2)
        else: return 2*A(n, k-1) - A(n-1, k-1)
    def A227418(n, k): return A(k, n-k)
    flatten([[A227418(n,k) for k in range(n+1)] for n in range(15)]) # G. C. Greubel, Oct 09 2022

Formula

If using the left column and top row to initialize, then: A(n,k) = 2*A(n, k-1) - A(n-1, k-1).
If using only the top row to initialize, then: A(n,k) = 4*A(n-1,k) - A(n-2,k).
If using the left column to initialize, then: A(n,k) = sqrt(3*A(n,k-1) + (-3)^(k-1)), for all n, k > 0.
Other internal relationships that apply are: A(2*n-1, 2*k) = A(n,k)^2 - A(n-1,k)^2;
A(n+1,k) * A(n,k+1) - A(n+1, k+1) * A(n,k) = (-3)^k, for all n, k > 0.
A(n, 0) = A001353(n).
A(n, 1) = A001075(n).
A(n, 2) = A005320(n).
A(n, 3) = A151961(n).
A(1, k) = A038754(k).
A(n, n) = 2*A090018(n), for n > 0 (main diagonal).
A(n, n+1) = A141041(n-1) (superdiagonal).
A(n+1, n) = abs(A099842(n)) (subdiagonal).
From G. C. Greubel, Oct 09 2022: (Start)
T(n, 0) = (1/2)*(1-(-1)^n)*3^((n-1)/2).
T(n, 1) = A038754(n-1).
T(n, 2) = A228879(n-2).
T(2*n-1, n-1) = A141041(n-1).
T(2*n, n) = 2*A090018(n-1), n > 0.
T(n, n-4) = 3*A005320(n-4).
T(n, n-3) = 3*A001075(n-3).
T(n, n-2) = 3*A001353(n-2).
T(n, n-1) = A001075(n-1).
T(n, n) = A001353(n).
Sum_{k=0..n-1} T(n, k) = A084156(n).
Sum_{k=0..n} T(n, k) = A084156(n) + A001353(n). (End)

Extensions

Offset corrected by G. C. Greubel, Oct 09 2022

A236330 Positive integers n such that x^2 - 14xy + y^2 + n = 0 has integer solutions.

Original entry on oeis.org

32, 48, 128, 176, 192, 288, 368, 416, 432, 512, 624, 704, 752, 768, 800, 944, 1056, 1136, 1152, 1184, 1200, 1328, 1472, 1568, 1584, 1664, 1712, 1728, 1776, 1952, 2048, 2096, 2208, 2288, 2336, 2352, 2496, 2592, 2672, 2816, 2864, 2928, 3008, 3056, 3072, 3104
Offset: 1

Views

Author

Colin Barker, Feb 16 2014

Keywords

Examples

			48 is in the sequence because x^2 - 14xy + y^2 + 48 = 0 has integer solutions, for example (x, y) = (2, 26).
		

Crossrefs

Cf. A001835 (n = 32), A001075 (n = 48), A237250 (n = 176), A003500 (n = 192), A082841 (n = 288), A151961 (n = 432), A077238 (n = 624).

A001352 a(0) = 1, a(1) = 6, a(2) = 24; for n>=3, a(n) = 4a(n-1) - a(n-2).

Original entry on oeis.org

1, 6, 24, 90, 336, 1254, 4680, 17466, 65184, 243270, 907896, 3388314, 12645360, 47193126, 176127144, 657315450, 2453134656, 9155223174, 34167758040, 127515808986, 475895477904, 1776066102630, 6628368932616, 24737409627834, 92321269578720, 344547668687046
Offset: 0

Views

Author

Keywords

Comments

Also the coordination sequence of a {4,6} tiling of the hyperbolic plane, where there are 6 squares (with vertex angles Pi/3) around every vertex. - toen (tca110(AT)rsphysse.anu.edu.au), May 16 2005
a(n) is related to the almost-equilateral Heronian triangles because it is the area of the Heronian triangle with edge lengths A003500(n)-1, A003500(n)+1 and 4. - Herbert Kociemba, Mar 19 2021

References

  • Bastida, Julio R. 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

First differences of A082841. Pairwise sums of A001834.

Programs

  • Maple
    A001352 := proc(n) coeftayl(1+6*x/(1-4*x+x^2),x=0,n) ; end: for n from 0 to 30 do printf("%d,",A001352(n)) ; od ; # R. J. Mathar, Jun 06 2007
    A001352:=(z+1)**2/(1-4*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    Join[{1},LinearRecurrence[{4,-1},{6,24},30]] (* Harvey P. Dale, Jul 20 2011 *)
  • PARI
    Vec((x+1)^2/(x^2-4*x+1) + O(x^40)) \\ Colin Barker, Oct 12 2015

Formula

G.f.: 1+6x/(1-4x+x^2). - R. J. Mathar, Jun 06 2007
a(n) = sqrt(3)*(-(2-sqrt(3))^n+(2+sqrt(3))^n) for n>0. - Colin Barker, Oct 12 2015

Extensions

More terms from R. J. Mathar, Jun 06 2007

A237250 Values of x in the solutions to x^2 - 4xy + y^2 + 11 = 0, where 0 < x < y.

Original entry on oeis.org

2, 3, 5, 10, 18, 37, 67, 138, 250, 515, 933, 1922, 3482, 7173, 12995, 26770, 48498, 99907, 180997, 372858, 675490, 1391525, 2520963, 5193242, 9408362, 19381443, 35112485, 72332530, 131041578, 269948677, 489053827, 1007462178, 1825173730, 3759900035
Offset: 1

Views

Author

Colin Barker, Feb 05 2014

Keywords

Comments

The corresponding values of y are given by a(n+2).
Positive values of x (or y) satisfying x^2 - 14xy + y^2 + 176 = 0.

Examples

			10 is in the sequence because (x, y) = (10, 37) is a solution to x^2 - 4xy + y^2 + 11 = 0.
		

Crossrefs

Programs

  • PARI
    Vec(-x*(x-1)*(x+2)*(2*x+1)/(x^4-4*x^2+1) + O(x^100))

Formula

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

A083336 a(n) = 4*a(n-2) - a(n-4).

Original entry on oeis.org

3, 3, 9, 12, 33, 45, 123, 168, 459, 627, 1713, 2340, 6393, 8733, 23859, 32592, 89043, 121635, 332313, 453948, 1240209, 1694157, 4628523, 6322680, 17273883, 23596563, 64467009, 88063572, 240594153, 328657725, 897909603, 1226567328, 3351044259, 4577611587, 12506267433
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Apr 26 2003

Keywords

Comments

a(n)/A002531(n+1) converges to sqrt(3).

Programs

  • Mathematica
    CoefficientList[Series[(3+3x-3x^2)/(1-4x^2+x^4), {x, 0, 30}], x]
    Transpose[NestList[Flatten[{Rest[#],4#[[3]]-First[#]}]&, {3,3,9,12}, 50]][[1]]  (* Harvey P. Dale, Mar 26 2011 *)
    LinearRecurrence[{0, 4, 0, -1}, {3, 3, 9, 12}, 30] (* T. D. Noe, Mar 26 2011 *)

Formula

a(2*n) = A082841(n) = a(2*n-1) + 3*A002531(2*n).
a(2*n+1) = (a(2*n) + 3*A002531(2*n+1)) / 2.
G.f.: (3 + 3*x - 3*x^2) / (1 - 4*x^2 + x^4).
Showing 1-10 of 10 results.