cp's OEIS Frontend

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

Showing 1-9 of 9 results.

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

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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)

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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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

A028230 Bisection of A001353. Indices of square numbers which are also octagonal.

Original entry on oeis.org

1, 15, 209, 2911, 40545, 564719, 7865521, 109552575, 1525870529, 21252634831, 296011017105, 4122901604639, 57424611447841, 799821658665135, 11140078609864049, 155161278879431551, 2161117825702177665, 30100488280951055759, 419245718107612602961, 5839339565225625385695
Offset: 1

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Comments

Chebyshev S-sequence with Diophantine property.
4*b(n)^2 - 3*a(n)^2 = 1 with b(n) = A001570(n), n>=0.
y satisfying the Pellian x^2 - 3*y^2 = 1, for even x given by A094347(n). - Lekraj Beedassy, Jun 03 2004
a(n) = L(n,-14)*(-1)^n, where L is defined as in A108299; see also A001570 for L(n,+14). - Reinhard Zumkeller, Jun 01 2005
Product x*y, where the pair (x, y) solves for x^2 - 3y^2 = -2, i.e., a(n) = A001834(n)*A001835(n). - Lekraj Beedassy, Jul 13 2006
Numbers n such that RootMeanSquare(1,3,...,2*A001570(k)-1) = n. - Ctibor O. Zizka, Sep 04 2008
As n increases, this sequence is approximately geometric with common ratio r = lim(n -> oo, a(n)/a(n-1)) = (2 + sqrt(3))^2 = 7 + 4 * sqrt(3). - Ant King, Nov 15 2011

References

  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.
  • J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.

Crossrefs

Programs

  • GAP
    a:=[1,15];; for n in [3..30] do a[n]:=14*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019
  • Magma
    I:=[1,15]; [n le 2 select I[n] else 14*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 06 2019
    
  • Maple
    seq(coeff(series((1+x)/(1-14*x+x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Dec 06 2019
  • Mathematica
    LinearRecurrence[{14, - 1}, {1, 15}, 17] (* Ant King, Nov 15 2011 *)
    CoefficientList[Series[(1+x)/(1-14x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 17 2014 *)
  • PARI
    Vec((1+x)/(1-14*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Jun 16 2014
    
  • PARI
    isok(n) = ispolygonal(n^2, 8); \\ Michel Marcus, Jul 09 2017
    
  • Sage
    [(lucas_number2(n,14,1)-lucas_number2(n-1,14,1))/12 for n in range(1, 18)] # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) = 2*A001921(n)+1.
a(n) = 14*a(n-1) - a(n-2) for n>1.
a(n) = S(n, 14) + S(n-1, 14) = S(2*n, 4) with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x) = 0, S(n, 14) = A007655(n+1) and S(n, 4) = A001353(n+1).
G.f.: x*(1+x)/(1-14*x+x^2).
a(n) = (ap^(2*n+1) - am^(2*n+1))/(ap - am) with ap := 2+sqrt(3) and am := 2-sqrt(3).
a(n+1) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*16^(n-k), n>=0.
a(n) = sqrt((4*A001570(n-1)^2 - 1)/3).
a(n) ~ 1/6*sqrt(3)*(2 + sqrt(3))^(2*n-1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
4*a(n+1) = (A001834(n))^2 + 4*(A001835(n+1))^2 - (A001835(n))^2. E.g. 4*a(3) = 4*209 = 19^2 + 4*11^2 - 3^2 = (A001834(2))^2 + 4*(A001835(3))^2 - (A001835(2))^2. Generating floretion: 'i + 2'j + 3'k + i' + 2j' + 3k' + 4'ii' + 3'jj' + 4'kk' + 3'ij' + 3'ji' + 'jk' + 'kj' + 4e. - Creighton Dement, Dec 04 2004
a(n) = f(a(n-1),7) + f(a(n-2),7), where f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. - Marcos Carreira, Dec 27 2006
From Ant King, Nov 15 2011: (Start)
a(n) = 1/6 * sqrt(3) * ( (tan(5*Pi/12)) ^ (2n-1) - (tan(Pi/12)) ^ (2n-1) ).
a(n) = floor (1/6 * sqrt(3) * (tan(5*Pi/12)) ^ (2n-1)). (End)
a(n) = A001353(n)^2-A001353(n-1)^2. - Antonio Alberto Olivares, Apr 06 2020
E.g.f.: 1 - exp(7*x)*(3*cosh(4*sqrt(3)*x) - 2*sqrt(3)*sinh(4*sqrt(3)*x))/3. - Stefano Spezia, Dec 12 2022
a(n) = sqrt(A036428(n)). - Bernard Schott, Dec 19 2022

Extensions

Additional comments from Wolfdieter Lang, Nov 29 2002
Incorrect recurrence relation deleted by Ant King, Nov 15 2011
Minor edits by Vaclav Kotesovec, Jan 28 2015

A045899 Numbers k such that k+1 and 3*k+1 are perfect squares.

Original entry on oeis.org

0, 8, 120, 1680, 23408, 326040, 4541160, 63250208, 880961760, 12270214440, 170902040408, 2380358351280, 33154114877520, 461777249934008, 6431727384198600, 89582406128846400, 1247721958419651008, 17378525011746267720, 242051628206028097080, 3371344269872647091408
Offset: 1

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Author

Andrej Dujella (duje(AT)math.hr)

Keywords

Comments

Essentially the same as A051047.
It appears that a(n) = A046175(n)-A046174(n), that is, the triangular index of the n-th pentagonal triangular number minus its pentagonal index. - Jonathan Vos Post, Feb 28 2011
Sequence lists the nonnegative x solutions when (x + 1)*(3*x + 1) is a square. Positive x solutions when (x - 1)*(3*x - 1) is a square are in A011922. - Bruno Berselli, Feb 20 2018

Crossrefs

Programs

  • Mathematica
    f[n_] := FullSimplify[((Sqrt[3] + 2)*(7 + 4*Sqrt[3])^n - (Sqrt[3] - 2) (7 - 4 Sqrt[3])^n - 4)/6]; Array[f, 18, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006 *)
    Rest[CoefficientList[Series[-8*x^2/((x - 1)*(x^2 - 14*x + 1)), {x,0,50}], x]] (* G. C. Greubel, Jun 07 2017 *)
    LinearRecurrence[{15,-15,1},{0,8,120},20] (* Harvey P. Dale, Jul 14 2024 *)
  • PARI
    x='x+O('x^50); concat([0], Vec(-8*x^2/((x - 1)*(x^2 - 14*x + 1)))) \\ G. C. Greubel, Jun 07 2017

Formula

a(n) = A046184(n+1) - 1.
a(n) = 14*a(n-1) - a(n-2) + 8.
a(n) = ((2 + sqrt(3))*(7 + 4*sqrt(3))^n + (2 - sqrt(3))*(7 - 4*sqrt(3))^n - 4)/6. - Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006
a(n) = 8*A076139(n-1) = 4*A217855(n-1) = 2*A123480(n-1) = 8/3*A076140(n-1). - Peter Bala, Dec 31 2012
From Colin Barker, Jul 30 2013: (Start)
G.f.: -8*x^2 / ((x - 1)*(x^2 - 14*x + 1)).
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). (End)
E.g.f.: (-4*exp(x) + (2 + sqrt(3))*exp((7-4*sqrt(3))*x) + (2 - sqrt(3))*exp((7+4*sqrt(3))*x))/6. - Ilya Gutkovskiy, Apr 28 2016

A101386 Expansion of g.f.: (5 - 3*x)/(1 - 6*x + x^2).

Original entry on oeis.org

5, 27, 157, 915, 5333, 31083, 181165, 1055907, 6154277, 35869755, 209064253, 1218515763, 7102030325, 41393666187, 241259966797, 1406166134595, 8195736840773, 47768254910043, 278413792619485, 1622714500806867, 9457873212221717, 55124524772523435, 321289275422918893
Offset: 0

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Author

Creighton Dement, Jan 23 2005

Keywords

Comments

A floretion-generated sequence relating to NSW numbers and numbers n such that (n^2 - 8)/2 is a square. It is also possible to label this sequence as the "tesfor-transform of the zero-sequence" under the floretion given in the program code, below. This is because the sequence "vesseq" would normally have been A046184 (indices of octagonal numbers which are also a square) using the floretion given. This floretion, however, was purposely "altered" in such a way that the sequence "vesseq" would turn into A000004. As (a(n)) would not have occurred under "natural" circumstances, one could speak of it as the transform of A000004.
Floretion Algebra Multiplication Program FAMP code: - tesforseq[ + 3'i - 2'j + 'k + 3i' - 2j' + k' - 4'ii' - 3'jj' + 4'kk' - 'ij' - 'ji' + 3'jk' + 3'kj' + 4e], Note: vesforseq = A000004, lesforseq = A002315, jesforseq = A077445
From Wolfdieter Lang, Feb 05 2015: (Start)
All positive solutions x = a(n) of the (generalized) Pell equation x^2 - 2*y^2 = +7 based on the fundamental solution (x2,y2) = (5,3) of the second class of (proper) solutions. The corresponding y solutions are given by y(n) = A253811(n).
All other positive solutions come from the first class of (proper) solutions based on the fundamental solution (x1,y1) = (3,1). These are given in A038762 and A038761.
All solutions of this Pell equation are found in A077443(n+1) and A077442(n), for n >= 0. See, e.g., the Nagell reference on how to find all solutions.
(End)

References

  • T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!((5 - 3*x)/(1-6*x+x^2))); // G. C. Greubel, Jul 26 2018
    
  • Maple
    A101386:= (n) -> simplify(5*ChebyshevU(n, 3) - 3*ChebyshevU(n-1, 3)); seq( A101386(n), n = 0..30); # G. C. Greubel, Mar 17 2020
  • Mathematica
    CoefficientList[ Series[(5-3x)/(1-6x+x^2), {x,0,30}], x] (* Robert G. Wilson v, Jan 29 2005 *)
    LinearRecurrence[{6,-1},{5,27},30] (* Harvey P. Dale, Apr 23 2016 *)
  • PARI
    Vec((5-3*x)/(1-6*x+x^2) + O(x^30)) \\ Colin Barker, Feb 05 2015
    
  • SageMath
    [5*chebyshev_U(n,3) -3*chebyshev_U(n-1,3) for n in (0..30)] # G. C. Greubel, Mar 17 2020

Formula

a(n) = A002315(n) + A077445(n+1). Note: the offset of A077445 is 1.
a(n+1) - a(n) = 2*A054490(n+1).
a(n) = 6*a(n-1) - a(n-2), a(0)=5, a(1)=27. - Philippe Deléham, Nov 17 2008
From Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009: (Start)
a(n) = ((5+sqrt(18))*(3 + sqrt(8))^n + (5-sqrt(18))*(3 - sqrt(8))^n)/2.
Third binomial transform of A164737. (End)
a(n) = rational part of z(n), with z(n) = (5+3*sqrt(2))*(3+2*sqrt(2))^n, n >= 0, the general positive solutions of the second class of proper solutions. See the preceding formula. - Wolfdieter Lang, Feb 05 2015
a(n) = 5*A001109(n+1) - 3*A001109(n). - G. C. Greubel, Mar 17 2020
a(n) = Pell(2*n+2) + 3*Pell(2*n+1), where Pell(n) = A000129(n). - G. C. Greubel, Apr 17 2020
E.g.f.: exp(3*x)*(5*cosh(2*sqrt(2)*x) + 3*sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Mar 16 2024

Extensions

More terms from Robert G. Wilson v, Jan 29 2005

A098301 Member r=16 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 16, 225, 3136, 43681, 608400, 8473921, 118026496, 1643897025, 22896531856, 318907548961, 4441809153600, 61866420601441, 861688079266576, 12001766689130625, 167163045568562176, 2328280871270739841, 32428769152221795600, 451674487259834398561
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

Also m such that (3*m^2 + m)/4 = m*(3*m + 1)/4 is a perfect square. - Ctibor O. Zizka, Oct 15 2010
Consequently A049451(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{# - 1, -# + 1, 1}, {0, 1, #}, 20] &[16] (* Michael De Vlieger, Feb 23 2021 *)
  • PARI
    concat(0, Vec(x*(1+x)/((1-x)*(1-14*x+x^2)) + O(x^50))) \\ Colin Barker, Jun 15 2015

Formula

a(n) = (T(n, 7)-1)/6 with Chebyshev's polynomials of the first kind evaluated at x=7: T(n, 7) = A011943(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n)/2; therefore: a(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n - 2)/12.
a(n) = A001353(n)^2 = S(n-1, 4)^2 with Chebyshev's polynomials of the second kind evaluated at x=4, S(n, 4):=U(n, 2).
a(n) = 14*a(n-1) - a(n-2) + 2, n >= 2, a(0)=0, a(1)=1.
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3), n >= 3.
G.f.: x*(1+x)/((1-x)*(1 - 14*x + x^2)) = x*(1+x)/(1 - 15*x + 15*x^2 - x^3) (from the Stephan link, see A092184).
Conjecture: 4*A007655(n+1) + A046184(n) = A055793(n+2) + a(n+1). - Creighton Dement, Nov 01 2004
a(n) = (A001075(n)^2-1)/3. - Parker Grootenhuis, Nov 28 2017

A036428 Square octagonal numbers.

Original entry on oeis.org

1, 225, 43681, 8473921, 1643897025, 318907548961, 61866420601441, 12001766689130625, 2328280871270739841, 451674487259834398561, 87622522247536602581025, 16998317641534841066320321, 3297585999935511630263561281, 639714685669847721430064568225
Offset: 1

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Author

Jean-Francois Chariot (jean-francois.chariot(AT)afoc.alcatel.fr)

Keywords

Comments

Also, numbers simultaneously octagonal and centered octagonal. - Steven Schlicker, Apr 24 2007

Crossrefs

Programs

  • Magma
    [Floor(1/12*(2+Sqrt(3))^(4*n-2)): n in [1..20]]; // Vincenzo Librandi, Dec 04 2015
  • Maple
    A036428 := proc(n)
            option remember;
            if n < 4 then
                    op(n,[1,225,43681]) ;
            else
                    195*(procname(n-1)-procname(n-2))+procname(n-3) ;
            end if;
    end proc: # R. J. Mathar, Nov 11 2011
  • Mathematica
    LinearRecurrence[{195,-195,1}, {1,225,43681}, 12] (* Ant King, Nov 15 2011 *)
  • PARI
    Vec(-x*(x^2+30*x+1)/((x-1)*(x^2-194*x+1)) + O(x^20)) \\ Colin Barker, Jun 24 2015
    
  • PARI
    vector(15, n, floor((2+sqrt(3))^(4*n-2)/12)) \\ Altug Alkan, Oct 19 2015
    

Formula

Let x(n) + y(n)*sqrt(48) = (8+sqrt(48))*(7+sqrt(48))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+8*(s(n)^2-s(n))). - Steven Schlicker, Apr 24 2007
a(n+2) = 194*a(n+1) - a(n) + 32 and also a(n+1) = 97*a(n) + 56*sqrt(3*a(n)^2 + a(n)). - Richard Choulet, Sep 26 2007
G.f.: x*(x^2+30x+1)/((1-x)*(1-194x+x^2)).
From Ant King, Nov 15 2011: (Start)
lim_{n->oo} a(n)/a(n-1) = (2 + sqrt(3))^4 = 97 + 56*sqrt(3).
a(n) = (1/12) * ((2 + sqrt(3))^(4n-2) + (2 - sqrt(3))^(4n-2) - 2).
a(n) = floor((1/12) * (2 + sqrt(3))^(4n-2)).
a(n) = (1/12) * ((tan(5*Pi/12))^(4n-2) + (tan(Pi/12))^(4n-2) - 2).
a(n) = floor((1/12) * tan(5*Pi/12)^(4n-2)).
(End)
a(n) = A028230(n)^2. - Bernard Schott, Dec 23 2022

Extensions

More terms from Eric W. Weisstein
Edited by N. J. A. Sloane, Oct 02 2007

A103772 Larger of two sides in a (k,k,k-1)-integer-sided triangle with integer area.

Original entry on oeis.org

1, 17, 241, 3361, 46817, 652081, 9082321, 126500417, 1761923521, 24540428881, 341804080817, 4760716702561, 66308229755041, 923554499868017, 12863454768397201, 179164812257692801, 2495443916839302017, 34757050023492535441, 484103256412056194161
Offset: 1

Views

Author

Zak Seidov, Feb 23 2005

Keywords

Comments

Corresponding areas are 0, 120, 25080, 4890480, 949077360, 184120982760, ...
Values of (x^2 + y^2)/2, where the pair (x, y) satisfies x^2 - 3*y^2 = -2, i.e., a(n) = {(A001834(n))^2 + (A001835(n))^2}/2 = {(A001834(n))^2 + A046184(n)}/2. - Lekraj Beedassy, Jul 13 2006
The heights of these triangles are given in A028230. (A028230(n), A045899(n), A103772(n)) forms a primitive Pythagorean triple.
Shortest side of (k,k+2,k+3) triangle such that median to longest side is integral. Sequence of such medians is A028230. - James R. Buddenhagen, Nov 22 2013
Numbers n such that (n+1)*(3n-1) is a square. - James R. Buddenhagen, Nov 22 2013

Crossrefs

Programs

  • Magma
    I:=[1,17]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+4: n in [1..20]]; // Vincenzo Librandi, Mar 05 2016
  • Mathematica
    a[1] = 1; a[2] = 17; a[3] = 241; a[n_] := a[n] = 15a[n - 1] - 15a[n - 2] + a[n - 3]; Table[ a[n] - 1, {n, 17}] (* Robert G. Wilson v, Mar 24 2005 *)
    LinearRecurrence[{15,-15,1},{1,17,241},20] (* Harvey P. Dale, Jan 02 2016 *)
    RecurrenceTable[{a[1] == 1, a[2] == 17, a[n] == 14 a[n-1] - a[n-2] + 4}, a, {n, 20}] (* Vincenzo Librandi, Mar 05 2016 *)
  • PARI
    Vec(x*(1+x)^2/((1-x)*(1-14*x+x^2)) + O(x^25)) \\ Colin Barker, Mar 05 2016
    

Formula

a(n) = (4*A001570(n+1) - 1)/3, n > 0. - Ralf Stephan, May 20 2007
a(n) = A052530(n-1)*A052530(n) + 1. - Johannes Boot, May 21 2011
G.f.: x*(1+x)^2/((1-x)*(1-14*x+x^2)). - Colin Barker, Apr 09 2012
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3); a(1)=1, a(2)=17, a(3)=241. - Harvey P. Dale, Jan 02 2016
a(n) = (-1+(7-4*sqrt(3))^n*(2+sqrt(3))-(-2+sqrt(3))*(7+4*sqrt(3))^n)/3. - Colin Barker, Mar 05 2016
a(n) = 14*a(n-1) - a(n-2) + 4. - Vincenzo Librandi, Mar 05 2016
a(n) = A001353(n)^2 + A001353(n-1)^2. - Antonio Alberto Olivares, Apr 06 2020

Extensions

More terms from Robert G. Wilson v, Mar 24 2005

A251963 Numbers n such that the sum of the triangular numbers T(n) and T(n+1) is equal to an octagonal number N(m) for some m.

Original entry on oeis.org

0, 14, 208, 2910, 40544, 564718, 7865520, 109552574, 1525870528, 21252634830, 296011017104, 4122901604638, 57424611447840, 799821658665134, 11140078609864048, 155161278879431550, 2161117825702177664, 30100488280951055758, 419245718107612602960
Offset: 1

Views

Author

Colin Barker, Dec 11 2014

Keywords

Comments

Also nonnegative integers x in the solutions to 2*x^2-6*y^2+4*x+4*y+2+2 = 0, the corresponding values of y being A046184.

Examples

			14 is in the sequence because T(14)+T(15) = 105+120 = 225 = N(9).
		

Crossrefs

Programs

  • Magma
    I:=[0,14]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+12: n in [1..20]]; // Vincenzo Librandi, Mar 05 2016
  • Mathematica
    RecurrenceTable[{a[1] == 0, a[2] == 14, a[n] == 14 a[n-1]- a[n-2] + 12}, a, {n, 20}] (* Vincenzo Librandi, Mar 05 2016 *)
  • PARI
    concat(0, Vec(2*x^2*(x-7) / ((x-1)*(x^2-14*x+1)) + O(x^100)))
    

Formula

a(n) = 15*a(n-1)-15*a(n-2)+a(n-3).
G.f.: 2*x^2*(x-7) / ((x-1)*(x^2-14*x+1)).
a(n) = (-6-(7-4*sqrt(3))^n*(3+2*sqrt(3))+(-3+2*sqrt(3))*(7+4*sqrt(3))^n)/6. - Colin Barker, Mar 05 2016
a(n) = 14*a(n-1) - a(n-2) + 12. - Vincenzo Librandi, Mar 05 2016
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