cp's OEIS Frontend

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

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A138288 a(n) = A054320(n) - A001078(n).

Original entry on oeis.org

1, 9, 89, 881, 8721, 86329, 854569, 8459361, 83739041, 828931049, 8205571449, 81226783441, 804062262961, 7959395846169, 78789896198729, 779939566141121, 7720605765212481, 76426118085983689, 756540575094624409, 7488979632860260401, 74133255753507979601, 733843577902219535609
Offset: 0

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Author

Reinhard Zumkeller, Mar 12 2008

Keywords

Comments

Numbers k such that 6*k^2 - 2 is a square. - Bruno Berselli, Feb 10 2014

Examples

			1 + 9*x + 89*x^2 + 881*x^3 + 8721*x^4 + 86329*x^5 + ...
		

References

  • H. Brocard, Note #2049, L'Intermédiaire des Mathématiciens, 8 (1901), pp. 212-213. - N. J. A. Sloane, Mar 02 2022

Crossrefs

Cf. similar sequences listed in A238379.

Programs

  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 10 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)
    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 A041007, A041039 *)
    a[6, 20] (* Gerry Martens, Jun 07 2015 *)
  • PARI
    {a(n) = subst( poltchebi(n+1) + poltchebi(n), x, 5) / 6} /* Michael Somos, Jan 25 2013 */
  • Sage
    [lucas_number1(n,10,1)-lucas_number1(n-1,10,1) for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) = A072256(n+1).
a(n) = A001079(n) + 2*A001078(n).
a(n) = 10*a(n-1) - a(n-2). a(-1) = a(0) = 1.
(sqrt(2)+sqrt(3))^(2*n+1) = A054320(n-1)*sqrt(2) + a(n)*sqrt(3).
From Michael Somos, Jan 25 2013: (Start)
G.f.: (1 - x) / (1 - 10*x + x^2).
a(-1-n) = a(n). (End)
a(n) = sqrt(2+(5-2*sqrt(6))^(1+2*n)+(5+2*sqrt(6))^(1+2*n))/(2*sqrt(3)). - Gerry Martens, Jun 04 2015
E.g.f.: exp(5*x)*(3*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/3. - Stefano Spezia, May 16 2023

A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1
Offset: 0

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Author

Keywords

Comments

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

Examples

			Triangle T(n, k) begins:
  n\k     0     1     2     3     4     5    6   7   8  9
  0:      1
  1:      1     1
  2:      2     3     1
  3:      5     9     5     1
  4:     14    28    20     7     1
  5:     42    90    75    35     9     1
  6:    132   297   275   154    54    11    1
  7:    429  1001  1001   637   273    77   13   1
  8:   1430  3432  3640  2548  1260   440  104  15   1
  9:   4862 11934 13260  9996  5508  2244  663 135  17  1
  ... Reformatted by _Wolfdieter Lang_, Dec 21 2015
From _Paul Barry_, Feb 17 2011: (Start)
Production matrix begins
  1, 1,
  1, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =
  2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)
		

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. 796.
  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

Crossrefs

Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).

Programs

  • Magma
    /* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015
    
  • Maple
    T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006
    T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end:
    seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023
  • Mathematica
    Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *)
    Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *)
    Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)
  • PARI
    a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)
    trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))
    trianglerows(10) \\ Felix Fröhlich, Jun 24 2016
  • Sage
    # Algorithm of L. Seidel (1877)
    # Prints the first n rows of the triangle
    def A039599_triangle(n) :
        D = [0]*(n+2); D[1] = 1
        b = True ; h = 1
        for i in range(2*n-1) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1
            else :
                for k in range(1,h, 1) : D[k] += D[k+1]
            if b : print([D[z] for z in (1..h-1)])
            b = not b
    A039599_triangle(10)  # Peter Luschny, May 01 2012
    

Formula

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Extensions

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009

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

Original entry on oeis.org

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

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Author

Keywords

Comments

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

Examples

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

References

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

Crossrefs

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

Programs

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

Formula

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

A108299 Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].

Original entry on oeis.org

1, 1, -1, 1, -1, -1, 1, -1, -2, 1, 1, -1, -3, 2, 1, 1, -1, -4, 3, 3, -1, 1, -1, -5, 4, 6, -3, -1, 1, -1, -6, 5, 10, -6, -4, 1, 1, -1, -7, 6, 15, -10, -10, 4, 1, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1, 1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1, 1, -1, -11, 10, 45, -36, -84, 56, 70
Offset: 0

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Author

Reinhard Zumkeller, Jun 01 2005

Keywords

Comments

Matrix inverse of A124645.
Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:
L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));
Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);
Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);
abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);
T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;
T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;
T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;
T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;
T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;
T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;
T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;
T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;
T(n,n) = L(n,0) = (-1)^floor((n+1)/2);
L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);
L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;
L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;
L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;
L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;
L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;
L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;
L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;
L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;
L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;
L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;
L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;
L(n,13) = A085260(n);
L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;
L(n,n) = A108366(n); L(n,-n) = A108367(n).
Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005
From L. Edson Jeffery, Mar 12 2011: (Start)
Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form
G_N=A_{N,1}=
(0 1 0 ... 0)
(1 0 1 0 ... 0)
(0 1 0 1 0 ... 0)
...
(0 ... 0 1 0 1)
(0 ... 0 1 1),
with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,
G_7=A_{7,1}=
(0 1 0)
(1 0 1)
(0 1 1).
We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)
The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011
The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

Examples

			Triangle begins:
  1;
  1,  -1;
  1,  -1,  -1;
  1,  -1,  -2,   1;
  1,  -1,  -3,   2,   1;
  1,  -1,  -4,   3,   3,  -1;
  1,  -1,  -5,   4,   6,  -3,  -1;
  1,  -1,  -6,   5,  10,  -6,  -4,   1;
  1,  -1,  -7,   6,  15, -10, -10,   4,   1;
  1,  -1,  -8,   7,  21, -15, -20,  10,   5,  -1;
  1,  -1,  -9,   8,  28, -21, -35,  20,  15,  -5,  -1;
  1,  -1, -10,   9,  36, -28, -56,  35,  35, -15,  -6,   1;
  ...
		

References

  • Friedrich L. Bauer, 'De Moivre und Lagrange: Cosinus eines rationalen Vielfachen von Pi', Informatik Spektrum 28 (Springer, 2005).
  • Jay Kappraff, S. Jablan, G. Adamson, & R. Sazdonovich: "Golden Fields, Generalized Fibonacci Sequences, & Chaotic Matrices"; FORMA, Vol 19, No 4, (2005).

Crossrefs

Cf. A049310, A039961, A124645 (matrix inverse).
Triangle sums (see the comments): A193884 (Kn11), A154955 (Kn21), A087960 (Kn22), A000007 (Kn3), A010892 (Fi1), A134668 (Fi2), A078031 (Ca2), A193669 (Gi1), A001519 (Gi3), A193885 (Ze1), A050935 (Ze3). - Johannes W. Meijer, Aug 08 2011
Cf. A003558.

Programs

  • Haskell
    a108299 n k = a108299_tabl !! n !! k
    a108299_row n = a108299_tabl !! n
    a108299_tabl = [1] : iterate (\row ->
       zipWith (+) (zipWith (*) ([0] ++ row) a033999_list)
                   (zipWith (*) (row ++ [0]) a059841_list)) [1,-1]
    -- Reinhard Zumkeller, May 06 2012
  • Maple
    A108299 := proc(n,k): binomial(n-floor((k+1)/2), floor(k/2))*(-1)^floor((k+1)/2) end: seq(seq(A108299 (n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011
  • Mathematica
    t[n_, k_?EvenQ] := I^k*Binomial[n-k/2, k/2]; t[n_, k_?OddQ] := -I^(k-1)*Binomial[n+(1-k)/2-1, (k-1)/2]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 16 2013 *)
  • PARI
    {T(n,k)=polcoeff(polcoeff((1-x*y)/(1-x+x^2*y^2+x^2*O(x^n)),n,x)+y*O(y^k),k,y)} (Hanna)
    

Formula

T(n,k) = binomial(n-floor((k+1)/2),floor(k/2))*(-1)^floor((k+1)/2).
T(n+1, k) = if sign(T(n, k-1))=sign(T(n, k)) then T(n, k-1)+T(n, k) else -T(n, k-1) for 0 < k < n, T(n, 0) = 1, T(n, n) = (-1)^floor((n+1)/2).
G.f.: A(x, y) = (1 - x*y)/(1 - x + x^2*y^2). - Paul D. Hanna, Jun 12 2005
The generating polynomial (in z) of row n >= 0 is (u^(2*n+1) + v^(2*n+1))/(u + v), where u and v are defined by u^2 + v^2 = 1 and u*v = z. - Emeric Deutsch, Jun 16 2011
From Johannes W. Meijer, Aug 08 2011: (Start)
abs(T(n,k)) = A065941(n,k) = abs(A187660(n,n-k));
T(n,n-k) = A130777(n,k); abs(T(n,n-k)) = A046854(n,k) = abs(A066170(n,k)). (End)

Extensions

Corrected and edited by Philippe Deléham, Oct 20 2008

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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Author

Keywords

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

A072256 a(n) = 10*a(n-1) - a(n-2) for n > 1, a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 9, 89, 881, 8721, 86329, 854569, 8459361, 83739041, 828931049, 8205571449, 81226783441, 804062262961, 7959395846169, 78789896198729, 779939566141121, 7720605765212481, 76426118085983689, 756540575094624409, 7488979632860260401, 74133255753507979601, 733843577902219535609
Offset: 0

Views

Author

Lekraj Beedassy, Jul 08 2002

Keywords

Comments

Any k in the sequence is followed by 5*k + 2*sqrt(2*(3*k^2 - 1)).
Gives solutions for x in 3*x^2 - 2*y^2 = 1. Corresponding y is given by A054320(n-1). [corrected by Jon E. Schoenfield, Jun 08 2018]
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7,8,9} which do not end in 0. - Tanya Khovanova, Jan 10 2007
For n >= 2, a(n) equals the permanent of the (2n-2) X (2n-2) tridiagonal matrix with sqrt(8)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 8 = 0. - Colin Barker, Feb 09 2014
The aerated sequence [b(n)]n>=1 = [1, 0, 9, 0, 89, 0, 881, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -12, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. - Peter Bala, May 12 2025

Crossrefs

Row 10 of array A094954.
First differences of A004189.
Essentially the same as A138288.

Programs

  • GAP
    a:=[1,1];; for n in [3..20] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2020
  • Magma
    [n le 2 select 1 else 10*Self(n-1)-Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 10 2014
    
  • Maple
    seq( simplify(ChebyshevU(n,5) -9*ChebyshevU(n-1,5)), n=0..20); # G. C. Greubel, Jan 14 2020
  • Mathematica
    a[n_]:= a[n]= 10a[n-1] -a[n-2]; a[0]=a[1]=1; Table[ a[n], {n, 0, 20}]
    CoefficientList[Series[(1-9x)/(1-10x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2014 *)
    Table[ChebyshevU[n, 5] -9*ChebyshevU[n-1, 5], {n,0,20}] (* G. C. Greubel, Jan 14 2020 *)
    LinearRecurrence[{10,-1},{1,1},20] (* Harvey P. Dale, Jun 17 2022 *)
  • PARI
    a(n)=([0,1; -1,10]^n*[1;1])[1,1] \\ Charles R Greathouse IV, May 10 2016
    
  • PARI
    vector(21, n, polchebyshev(n-1,2,5) -9*polchebyshev(n-2,2,5) ) \\ G. C. Greubel, Jan 14 2020
    

Formula

a(n) = (3-sqrt(6))/6 * (5+2*sqrt(6))^n + (3+sqrt(6))/6 * (5-2*sqrt(6))^n.
a(n) = (2*A031138(n) + 1)/3 = sqrt((2*A054320(n-1)^2 + 1)/3), n >= 1.
a(n) = U(n-1, 5)-U(n-2, 5) = T(2*n-1, sqrt(3))/sqrt(3) with Chebyshev's U- and T- polynomials and U(-1, x) := 0, U(-2, x) := -1, T(-1, x) := x.
G.f.: (1-9*x)/(1-10*x+x^2).
6*a(n)^2 - 2 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 10 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 8) = a(n+1). - Benoit Cloitre, Nov 10 2002
a(n)*a(n+3) = 80 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = L(n-1,10), where L is defined as in A108299; see also A054320 for L(n,-10). - Reinhard Zumkeller, Jun 01 2005
a(n) = A138288(n-1) for n > 0. - Reinhard Zumkeller, Mar 12 2008
a(n) = sqrt(A046172(n)). - Paul Weisenhorn, May 15 2009
a(n) = ceiling(((3-sqrt(6))*(5+2*sqrt(6))^n)/6). - Paul Weisenhorn, May 23 2020
E.g.f.: exp(5*x)*(3*cosh(2*sqrt(6)*x) - sqrt(6)*sinh(2*sqrt(6)*x))/3. - Stefano Spezia, Oct 25 2023
From Peter Bala, May 08 2025: (Start)
a(n) = (-1)^n * Dir(n-1, -5), where Dir(n, x) denotes the n-th row polynomial of A244419.
For arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = -8 with a(n) := (3-sqrt(6))/6 * (5+2*sqrt(6))^n + (3+sqrt(6))/6 * (5-2*sqrt(6))^n as given above (the particular case x = 0 is noted in the Comments section).
a(n+1/2) = 1/sqrt(3) * A001079(n).
a(n+3/4) + a(n+1/4) = sqrt(2/3) * sqrt(1 + sqrt(3)) * A001079(n).
a(n+3/4) - a(n+1/4) = 4 * sqrt(sqrt(3) - 1) * A004189(n).
a(n) divides a(3*n-1); a(n) divides a(5*n-2); in general, for k >= 0, a(n) divides a((2*k+1)*n - k).
Sum_{n >= 2} 1/(a(n) - 1/a(n)) = 1/8 (telescoping series: for n >= 2, 1/(a(n) - 1/a(n)) = 1/A291181(n-2) - 1/A291181(n-1).)
Product_{n >= 2} ((a(n) + 1)/(a(n) - 1))^(-1)^n = sqrt(3/2) (telescoping product: Product_{n = 2..k} (((a(n) + 1)/(a(n) - 1))^(-1)^n)^2 = 3/2 * (1 - (-1)^(k+1)/(3*A098308(k))).) (End)

Extensions

Edited by Robert G. Wilson v, Jul 17 2002

A146312 a(n) = -cos((2*n-1)*arcsin(sqrt(3)))^2 = -1 + cosh((2*n-1)*arcsinh(sqrt(2)))^2.

Original entry on oeis.org

2, 242, 23762, 2328482, 228167522, 22358088722, 2190864527282, 214682365584962, 21036680962799042, 2061380051988721202, 201994208413931878802, 19793371044513335401442, 1939548368153892937462562, 190055946708036994535929682, 18623543229019471571583646322
Offset: 1

Views

Author

Artur Jasinski, Oct 29 2008

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Round[ -N[Cos[(2 n - 1) ArcSin[Sqrt[3]]], 300]^2], {n, 1, 50}]
    LinearRecurrence[{99, -99, 1}, {2, 242, 23762}, 50] (* G. C. Greubel, Jul 03 2017 *)
  • PARI
    Vec(-2*x*(x^2+22*x+1) / ((x-1)*(x^2-98*x+1)) + O(x^100)) \\ Colin Barker, Oct 26 2014

Formula

General formula: cosh((2*n-1)*arcsinh(sqrt(2)))^2 + cos((2*n-1)*arcsin(sqrt(3)))^2 = 1.
a(n) = A146313(n) - 1.
From Colin Barker, Oct 26 2014: (Start)
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3).
G.f.: -2*x*(x^2+22*x+1) / ((x-1)*(x^2-98*x+1)). (End)
a(n) = 2*A054320(n-1)^2. - Jon E. Schoenfield, Jun 08 2018

Extensions

More terms from Colin Barker, Oct 26 2014

A294099 Rectangular array read by (upward) antidiagonals: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)*binomial(k-floor((j+1)/2), floor(j/2))*n^(k-j), n >= 1, k >= 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 5, -1, 1, 5, 11, 7, -2, 1, 6, 19, 29, 9, -1, 1, 7, 29, 71, 76, 11, 1, 1, 8, 41, 139, 265, 199, 13, 2, 1, 9, 55, 239, 666, 989, 521, 15, 1, 1, 10, 71, 377, 1393, 3191, 3691, 1364, 17, -1, 1, 11, 89, 559, 2584, 8119, 15289, 13775, 3571, 19, -2
Offset: 1

Views

Author

Keywords

Comments

This array is used to compute A269254: A269254(n) = least k such that A(n,k) is a prime, or -1 if no such k exists.
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018
The array can be extended to k<0 with A(n, k) = -A(n, -k-1) for all k in Z. - Michael Somos, Jun 19 2023

Examples

			Array begins:
  1   2    1    -1     -2      -1        1         2          1          -1
  1   3    5     7      9      11       13        15         17          19
  1   4   11    29     76     199      521      1364       3571        9349
  1   5   19    71    265     989     3691     13775      51409      191861
  1   6   29   139    666    3191    15289     73254     350981     1681651
  1   7   41   239   1393    8119    47321    275807    1607521     9369319
  1   8   55   377   2584   17711   121393    832040    5702887    39088169
  1   9   71   559   4401   34649   272791   2147679   16908641   133121449
  1  10   89   791   7030   62479   555281   4935050   43860169   389806471
  1  11  109  1079  10681  105731  1046629  10360559  102558961  1015229051
		

Crossrefs

Programs

  • Mathematica
    (* Array: *)
    Grid[Table[LinearRecurrence[{n, -1}, {1, 1 + n}, 10], {n, 10}]]
    (* Array antidiagonals flattened (gives this sequence): *)
    A294099[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] n^(k - j), {j, 0, k}]; Flatten[Table[A294099[n - k, k], {n, 11}, {k, 0, n - 1}]]
  • PARI
    {A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*n^(k-j))}; /* Michael Somos, Jun 19 2023 */

Formula

A(n,0) = 1, A(n,1) = n + 1, A(n,k) = n*A(n,k-1) - A(n,k-2), n >= 1, k >= 2.
G.f. for row n: (1 + x)/(1 - n*x + x^2), n >= 1.
A(n, k) = B(-n, k) where B = A299045. - Michael Somos, Jun 19 2023

A001078 a(n) = 10*a(n-1) - a(n-2) with a(0) = 0, a(1) = 2.

Original entry on oeis.org

0, 2, 20, 198, 1960, 19402, 192060, 1901198, 18819920, 186298002, 1844160100, 18255302998, 180708869880, 1788833395802, 17707625088140, 175287417485598, 1735166549767840, 17176378080192802, 170028614252160180, 1683109764441408998
Offset: 0

Views

Author

Keywords

Comments

Also 6*x^2+1 is a square. - Cino Hilliard, Mar 08 2003
This sequence has the following property: For each n, if A = a(n), B = 2*a(n+1), C = 3*a(n+1) then A*B+1, A*C+1, B*C+1 are perfect squares. - Deshpande M.N. (dpratap_ngp(AT)sancharnet.in), Sep 22 2004
n such that 6*n^2 = floor(sqrt(6)*n*ceiling(sqrt(6)*n)). - Benoit Cloitre, May 10 2003
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 19 2005
(sqrt(2) + sqrt(3))^(2*n) = A001079(n) + a(n)*sqrt(6); a(n) = A054320(n) + A138288(n). - Reinhard Zumkeller, Mar 12 2008
Numbers m such that A000217(m) plus A000326(m) equals an octagonal number (A000567). For a(3)=198, A000217(198)=19701, A000326(198)=58707, therefore 19701 + 58707 = 78408 = A000567(162). - Bruno Berselli, Apr 15 2013

References

  • O. Bottema: Verscheidenheden XXVI. Het vraagstuk van Malfatti, Euclides 25 (1949-50), pp. 144-149 [in Dutch].
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (pp. 283, 302, P_{16}).
  • 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

  • Haskell
    a001078 n = a001078_list !! n
    a001078_list =
       0 : 2 : zipWith (-) (map (10*) $ tail a001078_list) a001078_list
    -- Reinhard Zumkeller, Mar 18 2011
    
  • Magma
    I:=[0, 2]; [n le 2 select I[n] else 10*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017
  • Maple
    A001078 := proc(n) option remember; if n=0 then 0 elif n=1 then 2 else 10*A001078(n-1)-A001078(n-2); fi; end;
    A001078:=2*z/(1-10*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    a[0]=0; a[1]=2; a[n_] := a[n] = 10*a[n-1] - a[n-2]; Table[a[n],{n,0,19}] (* Jean-François Alcover, Mar 18 2011 *)
    LinearRecurrence[{10,-1},{0,2},20] (* Harvey P. Dale, Jun 23 2011 *)
    CoefficientList[Series[2*x/(1 - 10*x + x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)
  • PARI
    nxsqp1(m,n) = { for(x=1,m, y = n*x*x+1; if(issquare(y),print1(x" ")) ) }
    
  • PARI
    a(n)=imag((5+2*quadgen(24))^n) /* Michael Somos, Jul 05 2005 */
    
  • PARI
    a(n)=subst(poltchebi(n+1)-5*poltchebi(n),x,5)/12 /* Michael Somos, Jul 05 2005 */
    
  • PARI
    x='x+O('x^30); concat([0], Vec(2*x/(1 - 10*x + x^2))) \\ G. C. Greubel, Dec 19 2017
    

Formula

From Emeric Deutsch, Jun 19 2005: (Start)
a(n) = ((5 + 2*sqrt(6))^n - (5 - 2*sqrt(6))^n)/(2*sqrt(6)).
G.f.: 2*z/(1 - 10*z + z^2). (End)
a(-n) = -a(n).
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 9*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 11*(a(n-1) - a(n-2)) + a(n-3). (End)
a(n+1) = A054320(n) + A138288(n). - Reinhard Zumkeller, Mar 12 2008
a(n) = sinh(2n*arcsinh(sqrt(2)))/sqrt(6). - Herbert Kociemba, Apr 24 2008
a(n) = 2*A004189(n). - R. J. Mathar, Oct 26 2009
E.g.f.: 2*exp(5*x)*sinh(2*sqrt(6)*x)/(2*sqrt(6)). - Stefano Spezia, May 16 2023

Extensions

Thanks to Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr) and Floor van Lamoen for the Bottema references.

A077417 Chebyshev T-sequence with Diophantine property.

Original entry on oeis.org

1, 11, 131, 1561, 18601, 221651, 2641211, 31472881, 375033361, 4468927451, 53252096051, 634556225161, 7561422605881, 90102515045411, 1073668757939051, 12793922580223201, 152453402204739361, 1816646903876649131, 21647309444315050211
Offset: 0

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

7*a(n)^2 - 5*b(n)^2 = 2 with companion sequence b(n) = A077416(n), n>=0.
a(n) = L(n,12), where L is defined as in A108299; see also A077416 for L(n,-12). - Reinhard Zumkeller, Jun 01 2005
[a(n), A004191(n)] = the 2 X 2 matrix [1,10; 1,11]^(n+1) * [1,0]. - Gary W. Adamson, Mar 19 2008
Hankel transform of A174227. - Paul Barry, Mar 12 2010
Alternate denominators of the continued fraction convergents to sqrt(35), see A041059. - James R. Buddenhagen, May 20 2010
For positive n, a(n) equals the permanent of the (2n)X(2n) tridiagonal matrix with sqrt(10)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Positive values of x (or y) satisfying x^2 - 12xy + y^2 + 10 = 0. - Colin Barker, Feb 09 2014
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jun 29 2019

Examples

			G.f. = 1 + 11*x + 131*x^2 + 1561*x^3 + 18601*x^4 221651*x^5 + 2641211*x^6 + ...
		

Crossrefs

Cf. A072256(n) with companion A054320(n-1), n>=1.
Row 12 of array A094954.
Cf. A004191.
Cf. A041059. [James R. Buddenhagen, May 20 2010]
Cf. similar sequences listed in A238379.

Programs

  • Magma
    I:=[1,11]; [n le 2 select I[n] else 12*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 10 2014
    
  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 12 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2014 *)
    LinearRecurrence[{12,-1},{1,11},30] (* Harvey P. Dale, Apr 09 2015 *)
    a[ n_] := With[{x = Sqrt[7/2]}, ChebyshevT[2 n + 1, x]/x] // Expand; (* Michael Somos, Jun 29 2019 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-x)/(1-12*x+x^2)) \\ G. C. Greubel, Jan 18 2018
    
  • PARI
    {a(n) = my(x = quadgen(56)/2); simplify(polchebyshev(2*n + 1, 1, x)/x)}; /* Michael Somos, Jun 29 2019 */

Formula

a(n) = 12*a(n-1) - a(n-2), a(-1)=1, a(0)=1.
a(n) = S(n, 12) - S(n-1, 12) = T(2*n+1, sqrt(14)/2)/(sqrt(14)/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, 12)=A004191(n).
G.f.: (1-x)/(1-12*x+x^2).
a(n) = (ap^(2*n+1) + am^(2*n+1))/sqrt(14) with ap := (sqrt(7)+sqrt(5))/sqrt(2) and am := (sqrt(7)-sqrt(5))/sqrt(2).
a(n) = sqrt((5*A077416(n)^2 + 2)/7).
a(n)*a(n+3) = 120 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
E.g.f.: exp(6*x)*(7*cosh(sqrt(35)*x) + sqrt(35)*sinh(sqrt(35)*x))/7. - Stefano Spezia, Aug 29 2025

Extensions

More terms from Vincenzo Librandi, Feb 10 2014
Showing 1-10 of 32 results. Next