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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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A127672 Monic integer version of Chebyshev T-polynomials (increasing powers).

Original entry on oeis.org

2, 0, 1, -2, 0, 1, 0, -3, 0, 1, 2, 0, -4, 0, 1, 0, 5, 0, -5, 0, 1, -2, 0, 9, 0, -6, 0, 1, 0, -7, 0, 14, 0, -7, 0, 1, 2, 0, -16, 0, 20, 0, -8, 0, 1, 0, 9, 0, -30, 0, 27, 0, -9, 0, 1, -2, 0, 25, 0, -50, 0, 35, 0, -10, 0, 1, 0, -11, 0, 55, 0, -77, 0, 44, 0, -11, 0, 1, 2, 0, -36, 0, 105, 0, -112, 0, 54, 0, -12, 0, 1, 0, 13, 0, -91
Offset: 0

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Author

Wolfdieter Lang, Mar 07 2007

Keywords

Comments

The row polynomials R(n,x) := Sum_{m=0..n} a(n,m)*x^m have been called Chebyshev C_n(x) polynomials in the Abramowitz-Stegun handbook, p. 778, 22.5.11 (see A049310 for the reference, and note that on p. 774 the S and C polynomials have been mixed up in older printings). - Wolfdieter Lang, Jun 03 2011
This is a signed version of triangle A114525.
The unsigned column sequences (without zeros) are, for m=1..11: A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.
The row polynomials R(n,x) := Sum_{m=0..n} a(n,m)*x*m, give for n=2,3,...,floor(N/2) the positive zeros of the Chebyshev S(N-1,x)-polynomial (see A049310) in terms of its largest zero rho(N):= 2*cos(Pi/N) by putting x=rho(N). The order of the positive zeros is falling: n=1 corresponds to the largest zero rho(N) and n=floor(N/2) to the smallest positive zero. Example N=5: rho(5)=phi (golden section), R(2,phi)= phi^2-2 = phi-1, the second largest (and smallest) positive zero of S(4,x). - Wolfdieter Lang, Dec 01 2010
The row polynomial R(n,x), for n >= 1, factorizes into minimal polynomials of 2*cos(Pi/k), called C(k,x), with coefficients given in A187360, as follows.
R(n,x) = Product_{d|oddpart(n)} C(2*n/d,x)
= Product_{d|oddpart(n)} C(2^(k+1)*d,x),
with oddpart(n)=A000265(n), and 2^k is the largest power of 2 dividing n, where k=0,1,2,...
(Proof: R and C are monic, the degree on both sides coincides, and the zeros of R(n,x) appear all on the r.h.s.) - Wolfdieter Lang, Jul 31 2011 [Theorem 1B, eq. (43) in the W. Lang link. - Wolfdieter Lang, Apr 13 2018]
The zeros of the row polynomials R(n,x) are 2*cos(Pi*(2*k+1)/(2*n)), k=0,1, ..., n-1; n>=1 (from those of the Chebyshev T-polynomials). - Wolfdieter Lang, Sep 17 2011
The discriminants of the row polynomials R(n,x) are found under A193678. - Wolfdieter Lang, Aug 27 2011
The determinant of the N X N matrix M(N) with entries M(N;n,m) = R(m-1,x[n]), 1 <= n,m <= N, N>=1, and any x[n], is identical with twice the Vandermondian Det(V(N)) with matrix entries V(N;n,m) = x[n]^(m-1). This is an instance of the general theorem given in the Vein-Dale reference on p. 59. Note that R(0,x) = 2 (not 1). See also the comments from Aug 26 2013 under A049310 and from Aug 27 2013 under A000178. - Wolfdieter Lang, Aug 27 2013
This triangle a(n,m) is also used to express in the regular (2*(n+1))-gon, inscribed in a circle of radius R, the length ratio side/R, called s(2*(n+1)), as a polynomial in rho(2*(n+1)), the length ratio (smallest diagonal)/side. See the bisections ((-1)^(k-s))*A111125(k,s) and A127677 for comments and examples. - Wolfdieter Lang, Oct 05 2013
From Tom Copeland, Nov 08 2015: (Start)
These are the characteristic polynomials a_n(x) = 2*T_n(x/2) for the adjacency matrix of the Coxeter simple Lie algebra B_n, related to the Cheybshev polynomials of the first kind, T_n(x) = cos(n*q) with x = cos(q) (see p. 20 of Damianou). Given the polynomial (x - t)*(x - 1/t) = 1 - (t + 1/t)*x + x^2 = e2 - e1*x + x^2, the symmetric power sums p_n(t,1/t) = t^n + t^(-n) of the zeros of this polynomial may be expressed in terms of the elementary symmetric polynomials e1 = t + 1/t = y and e2 = t*1/t = 1 as p_n(t,1/t) = a_n(y) = F(n,-y,1,0,0,...), where F(n,b1,b2,...,bn) are the Faber polynomials of A263916.
The partial sum of the first n+1 rows given t and y = t + 1/t is PS(n,t) = Sum_{k=0..n} a_n(y) = (t^(n/2) + t^(-n/2))*(t^((n+1)/2) - t^(-(n+1)/2)) / (t^(1/2) - t^(-1/2)). (For n prime, this is related simply to the cyclotomic polynomials.)
Then a_n(y) = PS(n,t) - PS(n-1,t), and for t = e^(iq), y = 2*cos(q), and, therefore, a_n(2*cos(q)) = PS(n,e^(iq)) - PS(n-1,e^(iq)) = 2*cos(nq) = 2*T_n(cos(q)) with PS(n,e^(iq)) = 2*cos(nq/2)*sin((n+1)q/2) / sin(q/2).
(End)
R(45, x) is the famous polynomial used by Adriaan van Roomen (Adrianus Romanus) in his Ideae mathematicae from 1593 to pose four problems, solved by Viète. See, e.g., the Havil reference, pp. 69-74. - Wolfdieter Lang, Apr 28 2018
From Wolfdieter Lang, May 05 2018: (Start)
Some identities for the row polynomials R(n, x) following from the known ones for Chebyshev T-polynomials (A053120) are:
(1) R(-n, x) = R(n, x).
(2) R(n*m, x) = R(n, R(m, x)) = R(m, R(n, x)).
(3) R(2*k+1, x) = (-1)^k*x*S(2*k, sqrt(4-x^2)), k >= 0, with the S row polynomials of A049310.
(4) R(2*k, x) = R(k, x^2-2), k >= 0.
(End)
For y = z^n + z^(-n) and x = z + z^(-1), Hirzebruch notes that y(z) = R(n,x) for the row polynomial of this entry. - Tom Copeland, Nov 09 2019

Examples

			Row n=4: [2,0,-4,0,1] stands for the polynomial 2*y^0 - 4*y^2 + 1*y^4. With y^m replaced by 2^(m-1)*x^m this becomes T(4,x) = 1 - 8*x^2 + 8*x^4.
Triangle begins:
n\m   0   1   2   3   4   5   6   7   8   9  10 ...
0:    2
1:    0   1
2:   -2   0   1
3:    0  -3   0   1
4:    2   0  -4   0   1
5:    0   5   0  -5   0   1
6:   -2   0   9   0  -6   0   1
7:    0  -7   0  14   0  -7   0   1
8:    2   0 -16   0  20   0  -8   0   1
9:    0   9   0 -30   0  27   0  -9   0   1
10:  -2   0  25   0 -50   0  35   0 -10   0   1 ...
Factorization into minimal C-polynomials:
R(12,x) = R((2^2)*3,x) = C(24,x)*C(8,x) = C((2^3)*1,x)*C((2^3)*3,x). - _Wolfdieter Lang_, Jul 31 2011
		

References

  • Julian Havil, The Irrationals, A Story of the Numbers You Can't Count On, Princeton University Press, Princeton and Oxford, 2012, pp. 69-74.
  • F. Hirzebruch et al., Manifolds and Modular Forms, Vieweg 1994 pp. 77, 105.
  • R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer, 1999.

Crossrefs

Row sums (signed): A057079(n-1). Row sums (unsigned): A000032(n) (Lucas numbers). Alternating row sums: A099837(n+3).
Bisection: A127677 (even n triangle, without zero entries), ((-1)^(n-m))*A111125(n, m) (odd n triangle, without zero entries).

Programs

  • Maple
    seq(seq(coeff(2*orthopoly[T](n,x/2),x,j),j=0..n),n=0..20); # Robert Israel, Aug 04 2015
  • Mathematica
    a[n_, k_] := SeriesCoefficient[(2 - t*x)/(1 - t*x + x^2), {x, 0, n}, {t, 0, k}]; Flatten[Table[a[n, k], {n, 0, 12}, {k, 0, n}]] (* L. Edson Jeffery, Nov 02 2017 *)

Formula

a(n,0) = 0 if n is odd, a(n,0) = 2*(-1)^(n/2) if n is even, else a(n,m) = t(n,m)/2^(m-1) with t(n,m):=A053120(n,m) (coefficients of Chebyshev T-polynomials).
G.f. for m-th column (signed triangle): 2/(1+x^2) if m=0 else (x^m)*(1-x^2)/(1+x^2)^(m+1).
Riordan type matrix ((1-x^2)/(1+x^2),x/(1+x^2)) if one puts a(0,0)=1 (instead of 2).
O.g.f. for row polynomials: R(x,z) := Sum_{n>=0} R(n,x)*z^n = (2-x*z)*S(x,z), with the o.g.f. S(x,z) = 1/(1 - x*z + z^2) for the S-polynomials (see A049310).
Note that R(n,x) = R(2*n,sqrt(2+x)), n>=0 (from the o.g.f.s of both sides). - Wolfdieter Lang, Jun 03 2011
a(n,m) := 0 if n < m or n+m odd; a(n,0) = 2*(-1)^(n/2) (n even); else a(n,m) = ((-1)^((n+m)/2 + m))*n*binomial((n+m)/2-1,m-1)/m.
Recursion for n >= 2 and m >= 2: a(n,m) = a(n-1,m-1) - a(n-2,m), a(n,m) = 0 if n < m, a(2*k,1) = 0, a(2*k+1,1) = (2*k+1)*(-1)^k. In addition, for column m=0: a(2*k,0) = 2*(-1)^k, a(2*k+1,0) = 0, k>=0.
Chebyshev T(n,x) = Sum{m=0..n} a(n,m)*2^(m-1)*x^m. - Wolfdieter Lang, Jun 03 2011
R(n,x) = 2*T(n,x/2) = S(n,x) - S(n-2,x), n>=0, with Chebyshev's T- and S-polynomials, showing that they are integer and monic polynomials. - Wolfdieter Lang, Nov 08 2011
From Tom Copeland, Nov 08 2015: (Start)
a(n,x) = sqrt(2 + a(2n,x)), or 2 + a(2n,x) = a(n,x)^2, is a reflection of the relation of the Chebyshev polynomials of the first kind to the cosine and the half-angle formula, cos(q/2)^2 = (1 + cos(q))/2.
Examples: For n = 2, -2 + x^2 = sqrt(2 + 2 - 4*x^2 + x^4).
For n = 3, -3*x + x^3 = sqrt(2 - 2 + 9*x^2 - 6*x^4 + x^6).
(End)
L(x,h1,h2) = -log(1 - h1*x + h2*x^2) = Sum_{n>0} F(n,-h1,h2,0,...,0) x^n/n = h1*x + (-2*h2 + h1^2) x^2/2 + (-3*h1*h2 + h1^3) x^3/3 + ... is a log series generator of the bivariate row polynomials where T(0,0) = 0 and F(n,b1,b2,...,bn) are the Faber polynomials of A263916. exp(L(x,h1,h2)) = 1 / (1 - h1*x + h2*x^2) is the o.g.f. of A049310. - Tom Copeland, Feb 15 2016

Extensions

Name changed and table rewritten by Wolfdieter Lang, Nov 08 2011

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)

A029635 The (1,2)-Pascal triangle (or Lucas triangle) read by rows.

Original entry on oeis.org

2, 1, 2, 1, 3, 2, 1, 4, 5, 2, 1, 5, 9, 7, 2, 1, 6, 14, 16, 9, 2, 1, 7, 20, 30, 25, 11, 2, 1, 8, 27, 50, 55, 36, 13, 2, 1, 9, 35, 77, 105, 91, 49, 15, 2, 1, 10, 44, 112, 182, 196, 140, 64, 17, 2, 1, 11, 54, 156, 294, 378, 336, 204, 81, 19, 2, 1, 12, 65, 210, 450, 672, 714, 540, 285, 100
Offset: 0

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Keywords

Comments

This is also called Vieta's array. - N. J. A. Sloane, Nov 22 2017
Dropping the first term and changing the boundary conditions to T(n,1)=n, T(n,n-1)=2 (n>=2), T(n,n)=1 yields the number of nonterminal symbols (which generate strings of length k) in a certain context-free grammar in Chomsky normal form that generates all permutations of n symbols. Summation over k (1<=k<=n) results in A003945. For the number of productions of this grammar: see A090327. Example: 1; 2, 1; 3, 2, 1; 4, 5, 2, 1; 5, 9, 7, 2, 1; 6, 14, 16, 9, 2, 1; In addition to the example of A090327 we have T(3,3)=#{S}=1, T(3,2)=#{D,E}=2 and T(3,1)=#{A,B,C}=3. - Peter R. J. Asveld, Jan 29 2004
Much as the original Pascal triangle gives the Fibonacci numbers as sums of its diagonals, this triangle gives the Lucas numbers (A000032) as sums of its diagonals; see Posamentier & Lehmann (2007). - Alonso del Arte, Apr 09 2012
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
It appears that for the infinite set of (1,N) Pascal's triangles, the binomial transform of the n-th row (n>0), followed by zeros, is equal to the n-th partial sum of (1, N, N, N, ...). Example: for the (1,2) Pascal's triangle, the binomial transform of the second row followed by zeros, i.e., of (1, 3, 2, 0, 0, 0, ...), is equal to the second partial sum of (1, 2, 2, 2, ...) = (1, 4, 9, 16, ...). - Gary W. Adamson, Aug 11 2015
Given any (1,N) Pascal triangle, let the binomial transform of the n-th row (n>1) followed by zeros be Q(x). It appears that the binomial transform of the (n-1)-th row prefaced by a zero is Q(n-1). Example: In the (1,2) Pascal triangle the binomial transform of row 3: (1, 4, 5, 2, 0, 0, 0, ...) is A000330 starting with 1: (1, 5, 14, 30, 55, 91, ...). The binomial transform of row 2 prefaced by a zero and followed by zeros, i.e., of (0, 1, 3, 2, 0, 0, 0, ...) is (0, 1, 5, 14, 30, 55, ...). - Gary W. Adamson, Sep 28 2015
It appears that in the array accompanying each (1,N) Pascal triangle (diagonals of the triangle), the binomial transform of (..., 1, N, 0, 0, 0, ...) preceded by (n-1) zeros generates the n-th row of the array (n>0). Then delete the zeros in the result. Example: in the (1,2) Pascal triangle, row 3 (1, 5, 14, 30, ...) is the binomial transform of (0, 0, 1, 2, 0, 0, 0, ...) with the resulting zeros deleted. - Gary W. Adamson, Oct 11 2015
Read as a square array (similar to the Example section Sq(m,j), but with Sq(0,0)=0 and Sq(m,j)=P(m+1,j) otherwise), P(n,k) are the multiplicities of the eigenvalues, lambda_n = n(n+k-1), of the Laplacians on the unit k-hypersphere, given by Teo (and Choi) as P(n,k) = (2n-k+1)(n+k-2)!/(n!(k-1)!). P(n,k) is also the numerator of a Dirichlet series for the Minakashisundarum-Pleijel zeta function for the sphere. Also P(n,k) is the dimension of the space of homogeneous, harmonic polynomials of degree k in n variables (Shubin, p. 169). For relations to Chebyshev polynomials and simple Lie algebras, see A034807. - Tom Copeland, Jan 10 2016
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Jan 15 2016

Examples

			Triangle begins:
  [0] [2]
  [1] [1, 2]
  [2] [1, 3,  2]
  [3] [1, 4,  5,  2]
  [4] [1, 5,  9,  7,   2]
  [5] [1, 6, 14, 16,   9,  2]
  [6] [1, 7, 20, 30,  25, 11,  2]
  [7] [1, 8, 27, 50,  55, 36, 13,  2]
  [8] [1, 9, 35, 77, 105, 91, 49, 15, 2]
.
Read as a square, the array begins:
  n\k| 0  1   2    3    4    5
  --------------------------------------
  0 |  2  2   2    2    2    2   A040000
  1 |  1  3   5    7    9   11   A005408
  2 |  1  4   9   16   25   36   A000290
  3 |  1  5  14   30   55   91   A000330
  4 |  1  6  20   50  105  196   A002415
  5 |  1  7  27   77  182  378   A005585
  6 |  1  8  35  112  294  672   A040977
		

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers. New York: Prometheus Books (2007): 97 - 105.
  • M. Shubin and S. Andersson, Pseudodifferential Operators and Spectral Theory, Springer Series in Soviet Mathematics, 1987.

Crossrefs

Cf. A003945 (row sums), A007318, A034807, A061896, A029653 (row-reversed), A157000.
Sums along ascending antidiagonals give Lucas numbers, n>0.

Programs

  • Haskell
    a029635 n k = a029635_tabl !! n !! k
    a029635_row n = a029635_tabl !! n
    a029635_tabl = [2] : iterate
       (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,2]
    -- Reinhard Zumkeller, Mar 12 2012, Feb 23 2012
    
  • Maple
    T := proc(n, k) option remember;
    if n = k then 2 elif k = 0 then 1 else T(n-1, k-1) + T(n-1, k) fi end:
    for n from 0 to 8 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Dec 22 2024
  • Mathematica
    t[0, 0] = 2; t[n_, k_] := If[k < 0 || k > n, 0, Binomial[n, k] + Binomial[n-1, k-1]]; Flatten[Table[t[n, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, May 03 2011 *)
    (* The next program cogenerates A029635 and A029638. *)
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := u[n - 1, x] + v[n - 1, x]
    v[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1
    Table[Factor[u[n, x]], {n, 1, z}]
    Table[Factor[v[n, x]], {n, 1, z}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]   (* A029638  *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]   (* A029635 *)
    (* Clark Kimberling, Feb 20 2012 *)
    Table[Binomial[n,k]+Binomial[n-1,k-1],{n,0,20},{k,0,n}]//Flatten (* Harvey P. Dale, Feb 08 2024 *)
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, (n==0) + binomial(n, k) + binomial(n-1, k-1))}; /* Michael Somos, Jul 15 2003 */
    
  • Sage
    # uses[riordan_array from A256893]
    riordan_array((2-x)/(1-x), x/(1-x), 8) # Peter Luschny, Nov 09 2019

Formula

From Henry Bottomley, Apr 26 2002; (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k).
T(n, k) = C(n, k) + C(n-1, k-1).
T(n, k) = C(n, k)*(n + k)/n.
T(n, k) = A007318(n, k) + A007318(n-1, k-1).
T(n, k) = A061896(n + k, k) but with T(0, 0) = 1 and T(1, 1) = 2.
Row sum is floor(3^2(n-1)) i.e., A003945. (End)
G.f.: 1 + (1 + x*y) / (1 - x - x*y). - Michael Somos, Jul 15 2003
G.f. for n-th row: (x+2*y)*(x+y)^(n-1).
O.g.f. for row n: (1+x)/(1-x)^(n+1). The entries in row n are the nonzero entries in column n of A053120 divided by 2^(n-1). - Peter Bala, Aug 14 2008
T(2n, n) - T(2n, n+1)= Catalan(n)= A000108(n). - Philippe Deléham, Mar 19 2009
With T(0, 0) = 1 : Triangle T(n, k), read by rows, given by [1,0,0,0,0,0,...] DELTA [2,-1,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 10 2011
With T(0, 0) = 1, as in the Example section below, this is known as Vieta's array. The LU factorization of the square array is given by Yang and Leida, equation 20. - Peter Bala, Feb 11 2012
For n > 0: T(n, k) = A097207(n-1, k), 0 <= k < n. - Reinhard Zumkeller, Mar 12 2012
For n > 0: T(n, k) = A029600(n, k) - A007318(n, k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Riordan array ((2-x)/(1-x), x/(1-x)). - Philippe Deléham, Mar 15 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 4*x + 5*x^2/2! + 2*x^3/3!) = 1 + 5*x + 14*x^2/2! + 30*x^3/3! + 55*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
For n>=1: T(n, 0) + T(n, 1) + T(n, 2) = A000217(n+1). T(n, n-2) = (n-1)^2. - Bob Selcoe, Mar 29 2016:

Extensions

More terms from David W. Wilson
a(0) changed to 2 (was 1) by Daniel Forgues, Jul 06 2010

A049660 a(n) = Fibonacci(6*n)/8.

Original entry on oeis.org

0, 1, 18, 323, 5796, 104005, 1866294, 33489287, 600940872, 10783446409, 193501094490, 3472236254411, 62306751484908, 1118049290473933, 20062580477045886, 360008399296352015, 6460088606857290384, 115921586524134874897, 2080128468827570457762
Offset: 0

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Comments

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 18'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 >= 2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,17}. - Milan Janjic, Jan 25 2015
10*a(n)^2 = Tri(4)*S(n-1, 18)^2 is the triangular number Tri((T(n, 9) - 1)/2), with Tri, S and T given in A000217, A049310 and A053120. This is instance k = 4 of the k-family of identities given in a comment on A001109. - Wolfdieter Lang, Feb 01 2016
Possible solutions for y in Pell equation x^2 - 80*y^2 = 1. The values for x are given in A023039. - Herbert Kociemba, Jun 05 2022

Examples

			a(3) = F(6 * 3) / 8 = F(18) / 8 = 2584 / 8 = 323. - _Indranil Ghosh_, Feb 06 2017
		

Crossrefs

Column m=6 of array A028412.
Partial sums of A007805.

Programs

Formula

G.f.: x/(1 - 18*x + x^2).
a(n) = A134492(n)/8.
a(n) ~ (1/40)*sqrt(5)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all terms k of the sequence, 80*k^2 + 1 is a square. Limit_{n->oo} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 14 2002
a(n) = S(n-1, 18) with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n))/(8*sqrt(5)).
a(n) = sqrt((A023039(n)^2 - 1)/80) (cf. Richardson comment).
a(n) = 18*a(n-1) - a(n-2). - Gregory V. Richardson, Oct 14 2002
a(n) = A001076(2n)/4.
a(n) = 17*(a(n-1) + a(n-2)) - a(n-3) = 19*(a(n-1) - a(n-2)) + a(n-3). - Mohamed Bouhamida, May 26 2007
a(n+1) = Sum_{k=0..n} A101950(n,k)*17^k. - Philippe Deléham, Feb 10 2012
Product_{n>=1} (1 + 1/a(n)) = (1/2)*(2 + sqrt(5)). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = (2/9)*(2 + sqrt(5)). - Peter Bala, Dec 23 2012
a(n) = (1/32)*(F(6*n + 3) - F(6*n - 3)).
Sum_{n>=1} 1/(4*a(n) + 1/(4*a(n))) = 1/4. Compare with A001906 and A049670. - Peter Bala, Nov 29 2013
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -G(x)*G(-x), where G(x) = Sum_{n >= 1} A001076(n)*x^n.
1 + 4*Sum_{n >= 1} a(n)*x^(2*n) = (1 + F(x))*(1 + F(-x)) = (1 + 2*x*G(x))*(1 - 2*x*G(-x)), where F(x) = Sum_{n >= 1} Fibonacci(3*n + 3)*x^n.
1 + 7*Sum_{n >= 1} a(n)*x^(2*n) = (1 + G(x))*(1 + G(-x)) = (1 + 7*G(x))*(1 + 7*G(-x)).
1 + 12*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*G(x))*(1 + 2*G(-x)) = (1 + 6*G(x))*(1 + 6*G(-x)) = (1 + A(x))*(1 + A(-x)), where A(x) = Sum_{n >= 1} Fibonacci(3*n)*x^n is the o.g.f for A014445.
1 + 15*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 5*G(x))*(1 + 5*G(-x)) = (1 + 3*G(x))*(1 + 3*G(-x)) = H(x)*H(-x), where H(x) = Sum_{n >= 0} A155179(n)*x^n.
1 + 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 4*G(x))*(1 + 4*G(-x)) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*(-x)^n) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*(-x)^n).
1 + 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} Lucas(3*n)*x^n)*(1 + Sum_{n >= 1} Lucas(3*n)*(-x)^n).
1 - 5*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} A001077(n+1)*x^n)*(1 + Sum_{n >= 1} A001077(n+1)*(-x)^n).
1 - 9*Sum_{n >= 1} a(n)*x^(2*n) = (1 - G(x))*(1 - G(-x)) = (1 + 9*G(x))*(1 + 9*G(-x)).
1 - 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*Sum_{n >= 1} A099843(n)*x^n)*(1 + 2*Sum_{n >= 1} A099843(n)*(-x)^n).
1 - 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 - 2*G(x))*(1 - 2*G(-x)) = (1 + 10*G(x))*(1 + 10*G(-x)).
(End)

Extensions

Chebyshev and other comments from Wolfdieter Lang, Nov 08 2002

A001793 a(n) = n*(n+3)*2^(n-3).

Original entry on oeis.org

1, 5, 18, 56, 160, 432, 1120, 2816, 6912, 16640, 39424, 92160, 212992, 487424, 1105920, 2490368, 5570560, 12386304, 27394048, 60293120, 132120576, 288358400, 627048448, 1358954496, 2936012800, 6325010432, 13589544960, 29125246976
Offset: 1

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Comments

Coefficients of Chebyshev T polynomials: the subdiagonal A053120(n+3, n-1), for n > = 1. [rewritten by Wolfdieter Lang, Nov 25 2019]
Number of 132-avoiding permutations of [n+3] containing exactly two 123 patterns. - Emeric Deutsch, Jul 13 2001
Number of Dyck paths of semilength n+2 having pyramid weight n+1 (for pyramid weight see Denise and Simion). Example: a(2)=5 because the Dyck paths of semilength 4 having pyramid weight 3 are: (ud)u(ud)(ud)d, u(ud)(ud)d(ud), u(ud)(ud)(ud)d, u(ud)(uudd)d and u(uudd)(ud)d [here u=(1,1), d=(1,-1) and the maximal pyramids, of total length 3, are shown between parentheses]. - Emeric Deutsch, Mar 10 2004
a(n) is the number of dissections of a regular (n+3)-gon using n-1 noncrossing diagonals such that every piece of the dissection contains at least one non-base side of the (n+3)-gon. (One side of the (n+3)-gon is designated the base.) - David Callan, Mar 23 2004
If X_1,X_2,...,X_n are 2-blocks of a (2n+1)-set X then a(n) is the number of (n+2)-subsets of X intersecting each X_i, (i=1..n). - Milan Janjic, Nov 18 2007
The second corrector line for transforming 2^n offset 0 with a leading 1 into the Fibonacci sequence. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
Sum of all nodes of all integer compositions of n, see example. - Olivier Gérard, Oct 22 2011
Number of compositions of 2n with exactly two odd summands (see example). - Mamuka Jibladze, Sep 04 2013
4*a(n) is the number of North-East paths from (0,0) to (n+2,n+2) with exactly two east steps below y = x-1 or above y = x+1. It is related to paired pattern P_1 and P_6 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
From Paul Weisenhorn, Oct 18 2019: (Start)
The polynomials S(n,x)= Sum_(k>=1) b(n,k)*x^k has the recurrence relation S(n+2,x)=2*S(n+1,x))-x*S(n) with S(1,x)=1, S(2,x)=2-x and are generated by the coefficients b(n,k). b(n,k) is defined by b(n,k)=Sum_(j=1..k) binomials(k+1,j)*b(n-j,k) or by b(n,k)=((n-2+k)!*(n-1+2k)*2^n)/(4*(n-1)!*k!). b(n,1)=A001792, b(n,2)=A001793, b(n,3)=A001794, b(n,4)=A006974, b(n,5)=A006975, b(n,6)=A006976, b(n,7)=A209404.
The general formula for the sequences with k>=1: a(n)=((n-2+k)!*(n-1+2k)*2^n)/(4*(n-1)!*k!) with n >= 1. (End) [See a comment in A053120 on subdiagonal sequences. - Wolfdieter Lang, Jan 03 2020]

Examples

			a(2)=5 since 32415, 32451, 34125, 42135 and 52134 are the only 132-avoiding permutations of 12345 containing exactly two increasing subsequences of length 3.
a(4)=56: the compositions of 4 are 4, 3+1, 1+3, 2+2, 2+1+1, 1+2+1, 1+1+2, 1+1+1+1, the corresponding nodes (partial sums) are {0, 4}, {0, 3, 4}, {0, 1, 4}, {0, 2, 4}, {0, 2, 3, 4}, {0, 1, 3, 4}, {0, 1, 2, 4}, {0, 1, 2, 3, 4}, with individual sums {4, 7, 5, 6, 9, 8, 7, 10} and total 56. - _Olivier Gérard_, Oct 22 2011
The a(3)=18 compositions of 2*3=6 with two odd summands are 5+1, 1+5, 3+3, 4+1+1, 1+4+1, 1+1+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2. - _Mamuka Jibladze_, Sep 04 2013
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
  • 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

a(n) = A039991(n+3, 4) = A055252(n, 1).
Cf. A053120.

Programs

Formula

G.f.: x*(1-x)/(1-2*x)^3. Binomial transform of squares [1, 4, 9, ...].
a(n) = Sum_{k=0..floor((n+4)/2)} C(n+4, 2k)*C(k, 2). - Paul Barry, May 15 2003
With two leading zeros, binomial transform of quarter-squares A002620. - Paul Barry, May 27 2003
a(n) = Sum_{k=0..n+2} C(n+2, k) * floor(k^2/4). - Paul Barry, May 27 2003
a(n) = Sum_{i=0..j} binomial(i+1, 2)*binomial(j, i). - Jon Perry, Feb 26 2004
With one leading zero, binomial transform of triangular numbers A000217. - Philippe Deléham, Aug 02 2005
a(n) = Sum_{k=0..n+1} (-1)^(n-k+1)*C(k, n-k+1)*k*C(2k, k)/2. - Paul Barry, Oct 07 2005
Left-shifted sequence is binomial transform of left-shifted squares (A000290). - Franklin T. Adams-Watters, Nov 29 2006
Binomial transform of a(n) = n^2 offset 1. a(3)=18. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
a(n) = (1/n) * Sum_{k=0..n} binomial(n,k)*k^3. - Gary Detlefs, Nov 26 2011
For n > 1, a(n) = Sum_{k=0..n-1} Sum_{i=0..n} (k+2) * C(n-2,i). - Wesley Ivan Hurt, Sep 20 2017
a(n) = a(-3-n)*2^(2*n+3), a(n)*(n+3) = -A058645(-3-n)*2^(2*n+3) for all n in Z. - Michael Somos, Apr 19 2019
E.g.f.: (1/2)*exp(2*x)*x*(2 + x). - Stefano Spezia, Aug 17 2019
From Amiram Eldar, Jan 05 2022: (Start)
Sum_{n>=1} 1/a(n) = 128/9 - 56*log(2)/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 24*log(3/2) - 80/9. (End)

A001077 Numerators of continued fraction convergents to sqrt(5).

Original entry on oeis.org

1, 2, 9, 38, 161, 682, 2889, 12238, 51841, 219602, 930249, 3940598, 16692641, 70711162, 299537289, 1268860318, 5374978561, 22768774562, 96450076809, 408569081798, 1730726404001, 7331474697802, 31056625195209
Offset: 0

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Comments

a(2*n+1) with b(2*n+1) := A001076(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 5*b^2 = -1.
a(2*n) with b(2*n) := A001076(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 5*b^2 = +1 (see Emerson reference).
Bisection: a(2*n) = T(n,9) = A023039(n), n >= 0 and a(2*n+1) = 2*S(2*n, 2*sqrt(5)) = A075796(n+1), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
From Greg Dresden, May 21 2023: (Start)
For n >= 2, 8*a(n) is the number of ways to tile this T-shaped figure of length n-1 with four colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 8*a(6) = 23112 different tilings.
_
|| _
|||_|||
|_|
(End)

Examples

			1  2  9  38  161  (A001077)
-, -, -, --, ---, ...
0  1  4  17   72  (A001076)
1 + 2*x + 9*x^2 + 38*x^3 + 161*x^4 + 682*x^5 + 2889*x^6 + 12238*x^7 + ... - _Michael Somos_, Aug 11 2009
		

References

  • 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).
  • V. Thébault, Les Récréations Mathématiques, Gauthier-Villars, Paris, 1952, p. 282.

Crossrefs

Programs

  • Magma
    I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017
  • Maple
    A001077:=(-1+2*z)/(-1+4*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation
    with(combinat): a:=n->fibonacci(n+1, 4)-2*fibonacci(n, 4): seq(a(n), n=0..30); # Zerinvary Lajos, Apr 04 2008
  • Mathematica
    LinearRecurrence[{4, 1}, {1, 2}, 30]
    Join[{1},Numerator[Convergents[Sqrt[5],30]]] (* Harvey P. Dale, Mar 23 2016 *)
    CoefficientList[Series[(1-2*x)/(1-4*x-x^2), {x, 0, 30}], x] (* G. C. Greubel, Dec 19 2017 *)
    LucasL[3*Range[0,30]]/2 (* Rigoberto Florez, Apr 03 2019 *)
    a[ n_] := LucasL[n, 4]/2; (* Michael Somos, Nov 02 2021 *)
  • PARI
    {a(n) = fibonacci(3*n) / 2 + fibonacci(3*n - 1)}; /* Michael Somos, Aug 11 2009 */
    
  • PARI
    a(n)=if(n<2,n+1,my(t=4);for(i=1,n-2,t=4+1/t);numerator(2+1/t)) \\ Charles R Greathouse IV, Dec 05 2011
    
  • PARI
    x='x+O('x^30); Vec((1-2*x)/(1-4*x-x^2)) \\ G. C. Greubel, Dec 19 2017
    
  • Sage
    [lucas_number2(n,4,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, May 14 2009
    

Formula

G.f.: (1-2*x)/(1-4*x-x^2).
a(n) = 4*a(n-1) + a(n-2), a(0)=1, a(1)=2.
a(n) = ((2 + sqrt(5))^n + (2 - sqrt(5))^n)/2.
a(n) = A014448(n)/2.
Limit_{n->infinity} a(n)/a(n-1) = phi^3 = 2 + sqrt(5). - Gregory V. Richardson, Oct 13 2002
a(n) = ((-i)^n)*T(n, 2*i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1.
Binomial transform of A084057. - Paul Barry, May 10 2003
E.g.f.: exp(2x)cosh(sqrt(5)x). - Paul Barry, May 10 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)*5^k*2^(n-2k). - Paul Barry, Nov 15 2003
a(n) = 4*a(n-1) + a(n-2) when n > 2; a(1) = 1, a(2) = 2. - Alex Vinokur (alexvn(AT)barak-online.net), Oct 25 2004
a(n) = A001076(n+1) - 2*A001076(n) = A097924(n) - A015448(n+1); a(n+1) = A097924(n) + 2*A001076(n) = A097924(n) + 2(A048876(n) - A048875(n)). - Creighton Dement, Mar 19 2005
a(n) = F(3*n)/2 + F(3*n-1) where F() = Fibonacci numbers A000045. - Gerald McGarvey, Apr 28 2007
a(n) = A000032(3*n)/2.
For n >= 1: a(n) = (1/2)*Fibonacci(6*n)/Fibonacci(3*n) and a(n) = integer part of (2 + sqrt(5))^n. - Artur Jasinski, Nov 28 2011
a(n) = Sum_{k=0..n} A201730(n,k)*4^k. - Philippe Deléham, Dec 06 2011
a(n) = A001076(n) + A015448(n). - R. J. Mathar, Jul 06 2012
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
a(n) is the (1,1)-entry of the matrix W^n with W=[2, sqrt(5); sqrt(5), 2]. - Carmine Suriano, Mar 21 2014
From Rigoberto Florez, Apr 03 2019: (Start)
a(n) = A099919(n) + A049651(n) if n > 0.
a(n) = 1 + Sum_{k=0..n-1} L(3*k + 1) if n >= 0, L(n) = n-th Lucas number (A000032). (End)
From Christopher Hohl, Aug 22 2021: (Start)
For n >= 2, a(2n-1) = A079962(6n-9) + A079962(6n-3).
For n >= 1, a(2n) = sqrt(20*A079962(6n-3)^2 + 1). (End)
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*4^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*4^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = 4^n*Sum_{k=0..n} A374439(n, k)*(-1/4)^k. - Peter Luschny, Jul 26 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) + 5*(-1)^(n+1)/a(n)) = 3/8, since 1/(a(n) + 5*(-1)^(n+1)/a(n)) = b(n) - b(n+1), where b(n) = (1/4) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 5*(-1)^(n+1)/a(n)) = 1/8, since 1/(a(n) + 5*(-1)^(n+1)/a(n)) = c(n) + c(n+1), where c(n) = (1/4) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

Extensions

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A026150 a(0) = a(1) = 1; a(n+2) = 2*a(n+1) + 2*a(n).

Original entry on oeis.org

1, 1, 4, 10, 28, 76, 208, 568, 1552, 4240, 11584, 31648, 86464, 236224, 645376, 1763200, 4817152, 13160704, 35955712, 98232832, 268377088, 733219840, 2003193856, 5472827392, 14952042496, 40849739776
Offset: 0

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Comments

a(n+1)/A002605(n) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003
a(n+1)/a(n) converges to 1 + sqrt(3) = 2.732050807568877293.... - Philippe Deléham, Jul 03 2005
Binomial transform of expansion of cosh(sqrt(3)x) (A000244 with interpolated zeros); inverse binomial transform of A001075. - Philippe Deléham, Jul 04 2005
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 3 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(3). - Cino Hilliard, Sep 25 2005
Inverse binomial transform of A001075: (1, 2, 7, 26, 97, 362, ...). - Gary W. Adamson, Nov 23 2007
Starting (1, 4, 10, 28, 76, ...), the sequence is the binomial transform of [1, 3, 3, 9, 9, 27, 27, 81, 81, ...], and inverse binomial transform of A001834: (1, 5, 19, 71, 265, ...). - Gary W. Adamson, Nov 30 2007
[1, 3; 1, 1]^n * [1,0] = [a(n), A002605(n)]. - Gary W. Adamson, Mar 21 2008
(1 + sqrt(3))^n = a(n) + A002605(n)*(sqrt(3)). - Gary W. Adamson, Mar 21 2008
Equals right border of triangle A143908. Also, starting (1, 4, 10, 28, ...) = row sums of triangle A143908 and INVERT transform of (1, 3, 3, 3, ...). - Gary W. Adamson, Sep 06 2008
a(n) is the number of compositions of n when there are 1 type of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010
An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 85, 277, 337 and 340, lead to this sequence (without the first leading 1). For the corner squares these vectors lead to the companion sequence A002605 (without the leading 0). - Johannes W. Meijer, Aug 15 2010
Pisano period lengths: 1, 1, 1, 1, 24, 1, 48, 1, 3, 24, 10, 1, 12, 48, 24, 1,144, 3,180, 24, ... - R. J. Mathar, Aug 10 2012
(1 + sqrt(3))^n = a(n) + A002605(n)*sqrt(3), for n >= 0; integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 10 2018
a(n) is also the number of solutions for cyclic three-dimensional stable matching instances with master preference lists of size n (Escamocher and O'Sullivan 2018). - Guillaume Escamocher, Jun 15 2018
Starting from a(1), first differences of A005665. - Ivan N. Ianakiev, Nov 22 2019
Number of 3-permutations of n elements avoiding the patterns 231, 312. See Bonichon and Sun. - Michel Marcus, Aug 19 2022

Examples

			G.f. = 1 + x + 4*x^2 + 10*x^3 + 28*x^4 + 76*x^5 + 208*x^6 + 568*x^7 + ...
		

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

First differences of A002605.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Haskell
    a026150 n = a026150_list !! n
    a026150_list = 1 : 1 : map (* 2) (zipWith (+) a026150_list (tail
    a026150_list))
    -- Reinhard Zumkeller, Oct 15 2011
    
  • Magma
    [n le 2 select 1 else 2*Self(n-1) + 2*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 07 2018
  • Maple
    with(combstruct):ZL0:=S=Prod(Sequence(Prod(a, Sequence(b))), a):ZL1:=Prod(begin_blockP, Z, end_blockP):ZL2:=Prod(begin_blockLR, Z, Sequence(Prod(mu_length, Z), card>=1), end_blockLR): ZL3:=Prod(begin_blockRL, Sequence(Prod(mu_length, Z), card>=1), Z, end_blockRL):Q:=subs([a=Union(ZL2,ZL2,ZL2), b=ZL1], ZL0), begin_blockP=Epsilon, end_blockP=Epsilon, begin_blockLR=Epsilon, end_blockLR=Epsilon, begin_blockRL=Epsilon, end_blockRL=Epsilon, mu_length=Epsilon:temp15:=draw([S, {Q}, unlabelled], size=15):seq(count([S, {Q}, unlabelled], size=n)/3, n=2..27); # Zerinvary Lajos, Mar 08 2008
  • Mathematica
    Expand[Table[((1 + Sqrt[3])^n + (1 - Sqrt[3])^n)/(2), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *)
    LinearRecurrence[{2, 2}, {1, 1}, 30] (* T. D. Noe, Mar 25 2011 *)
    Round@Table[LucasL[n, Sqrt[2]] 2^(n/2 - 1), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 15 2016 *)
  • Maxima
    a(n) := if n<=1 then 1 else 2*a(n-1)+2*a(n-2);
    makelist(a(n),n,0,20); /* Emanuele Munarini, Apr 14 2017 */
    
  • PARI
    {a(n) = if( n<0, 0, real((1 + quadgen(12))^n))};
    
  • Sage
    from sage.combinat.sloane_functions import recur_gen2; it = recur_gen2(1,1,2,2); [next(it) for i in range(30)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [lucas_number2(n,2,-2)/2 for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009
    

Formula

a(n) = (1/2)*((1 + sqrt(3))^n + (1 - sqrt(3))^n). - Benoit Cloitre, Oct 28 2002
G.f.: (1 - x)/(1 - 2*x - 2*x^2).
a(n) = a(n-1) + A083337(n-1). A083337(n)/a(n) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003
From Paul Barry, May 15 2003: (Start)
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k)*3^k;
E.g.f.: exp(x)*cosh(sqrt(3)x). (End)
a(n) = Sum_{k=0..n} A098158(n,k)*3^(n - k). - Philippe Deléham, Dec 26 2007
a(n) = upper left and lower right terms of [1, 1; 3, 1]^n. (1 + sqrt(3))^n = a(n) + A083337(n)/(sqrt(3)). - Gary W. Adamson, Mar 12 2008
a(n) = A080040(n)/2. - Philippe Deléham, Nov 19 2008
If p[1] = 1, and p[i] = 3, (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) = det A. - Milan Janjic, Apr 29 2010
a(n) = 2 * A052945(n-1). - Vladimir Joseph Stephan Orlovsky, Mar 24 2011
a(n) = round((1 + sqrt(3))^n/2) for n > 0. - Bruno Berselli, Feb 04 2013
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k - 1)/(x*(3*k + 2) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
a(n) = (-sqrt(2)*i)^n*T(n,sqrt(2)*i/2), with i = sqrt(-1) and the Chebyshev T-polynomials (A053120). - Wolfdieter Lang, Feb 10 2018

A001079 a(n) = 10*a(n-1) - a(n-2); a(0) = 1, a(1) = 5.

Original entry on oeis.org

1, 5, 49, 485, 4801, 47525, 470449, 4656965, 46099201, 456335045, 4517251249, 44716177445, 442644523201, 4381729054565, 43374646022449, 429364731169925, 4250272665676801, 42073361925598085, 416483346590304049
Offset: 0

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Comments

Also gives solutions to the equation x^2-1=floor(x*r*floor(x/r)) where r=sqrt(6). - 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(6). - Benoit Cloitre, Feb 24 2004
a(n) and b(n) (A004189) are the nonnegative proper solutions to the Pell equation a(n)^2 - 6*(2*b(n))^2 = +1, n >= 0. The formula given below by Gregory V. Richardson follows. - Wolfdieter Lang, Jun 26 2013
a(n) are the integer square roots of (A032528 + 1). They are also the values of m where (A032528(m) - 1) has integer square roots. See A122653 for the integer square roots of (A032528 - 1), and see A122652 for the values of m where (A032528(m) + 1) has integer square roots. - Richard R. Forberg, Aug 05 2013
a(n) are also the values of m where floor(2m^2/3) has integer square roots, excluding m = 0. The corresponding integer square roots are given by A122652(n). - Richard R. Forberg, Nov 21 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 24 = 0. - Colin Barker, Feb 09 2014
Dickson on page 384 gives the Diophantine equation "24x^2 + 1 = y^2" and later states "y_{n+1} = 10y_n - y_{n-1}" where y_n is this sequence. - Michael Somos, Jun 19 2023

Examples

			Pell equation: n = 0: 1^2 - 24*0^2 = +1, n = 1: 5^2 - 6*(1*2)^2 = 1, n = 2: 49^2 - 6*(2*10)^2 = +1. - _Wolfdieter Lang_, Jun 26 2013
G.f. = 1 + 5*x + 49*x^2 + 485*x^3 + 4801*x^4 + 47525*x^5 + 470449*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) - From N. J. A. Sloane, May 30 2012
  • L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, p. 384.
  • L. Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 374.
  • 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).
  • V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 281.

Crossrefs

Programs

  • Magma
    I:=[1,5]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 10 2016
    
  • Maple
    A001079 := proc(n)
        option remember;
        if n <= 1 then
            op(n+1,[1,5]) ;
        else
            10*procname(n-1)-procname(n-2) ;
        end if;
    end proc:
    seq(A001079(n),n=0..20) ; # R. J. Mathar, Apr 30 2017
  • Mathematica
    Table[(-1)^n Round[N[Cos[2 n ArcSin[Sqrt[3]]], 50]], {n, 0, 20}] (* Artur Jasinski, Oct 29 2008 *)
    a[ n_] := ChebyshevT[n, 5]; (* Michael Somos, Aug 24 2014 *)
    CoefficientList[Series[(1-5*x)/(1-10*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
    a[n_] := 3^n*Sum[(2/3)^k*Binomial[2*n, 2*k], {k,0,n}]; Flatten[Table[a[n], {n,0,18}]] (* Detlef Meya, May 21 2024 *)
  • PARI
    {a(n) = subst(poltchebi(n), 'x, 5)}; /* Michael Somos, Sep 05 2006 */
    
  • PARI
    {a(n) = real((5 + 2*quadgen(24))^n)}; /* Michael Somos, Sep 05 2006 */
    
  • PARI
    {a(n) = n = abs(n); polsym(1 - 10*x + x^2, n)[n+1] / 2}; /* Michael Somos, Sep 05 2006 */
    
  • PARI
    x='x+O('x^30); Vec((1-5*x)/(1-10*x+x^2)) \\ G. C. Greubel, Dec 20 2017

Formula

For all members x of the sequence, 6*x^2 -6 is a square. Limit_{n->infinity} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 5) = (S(n, 10)-S(n-2, 10))/2 with S(n, x) := U(n, x/2) and T(n), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(n, 10) = A004189(n+1).
a(n) = sqrt(1+24*A004189(n)^2) (cf. Richardson comment).
a(n)*a(n+3) - a(n+1)*a(n+2) = 240. - Ralf Stephan, Jun 06 2005
Chebyshev's polynomials T(n,x) evaluated at x=5.
G.f.: (1-5*x)/(1-10*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n)= ((5+2*sqrt(6))^n + (5-2*sqrt(6))^n)/2.
a(-n) = a(n).
a(n+1) = 5*a(n) + 2*(6*a(n)^2-6)^(1/2) - Richard Choulet, Sep 19 2007
(sqrt(2)+sqrt(3))^(2*n)=a(n)+A001078(n)*sqrt(6). - Reinhard Zumkeller, Mar 12 2008
a(n+1) = 2*A054320(n) + 3*A138288(n). - Reinhard Zumkeller, Mar 12 2008
a(n) = cosh(2*n* arcsinh(sqrt(2))). - Herbert Kociemba, Apr 24 2008
a(n) = (-1)^n * cos(2*n* arcsin(sqrt(3))). - Artur Jasinski, Oct 29 2008
a(n) = cos(2*n* arccos(sqrt(3))). - Artur Jasinski, Sep 10 2016
a(n) = A142238(2n-1) = A041006(2n-1) = A041038(2n-1), for all n > 0. - M. F. Hasler, Feb 14 2009
2*a(n)^2 = 3*A122652(n)^2 + 2. - Charlie Marion, Feb 01 2013
E.g.f.: cosh(2*sqrt(6)*x)*exp(5*x). - Ilya Gutkovskiy, Sep 10 2016
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 10*x + sqrt(1 + 96*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 20*x + 4*x^2) is the g.f. of A098270.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - 3/a(n)) = 1/4.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 2/a(n)) = 1/6.
Sum_{n >= 1} 1/(a(n)^2 - 3) = 1/4 - 1/sqrt(24). (End)
a(n) = 3^n*Sum_{k=0..n} (2/3)^k*binomial(2*n, 2*k). - Detlef Meya, May 21 2024

Extensions

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

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

Original entry on oeis.org

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

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Comments

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

Examples

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

References

  • E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - N. J. A. Sloane, Mar 03 2022
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

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

Programs

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

Formula

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

A092184 Sequence S_6 of the S_r family.

Original entry on oeis.org

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

Views

Author

Rainer Rosenthal, Apr 03 2004

Keywords

Comments

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

Examples

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

Crossrefs

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

Programs

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

Formula

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

Extensions

Extension and Chebyshev comments from Wolfdieter Lang, Sep 10 2004
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