cp's OEIS Frontend

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

Showing 1-10 of 13 results. Next

A097826 Partial sums of Chebyshev sequence S(n,11) = U(n,11/2) = A004190(n).

Original entry on oeis.org

1, 12, 132, 1441, 15720, 171480, 1870561, 20404692, 222581052, 2427986881, 26485274640, 288910034160, 3151525101121, 34377866078172, 375005001758772, 4090677153268321, 44622443684192760, 486756203372852040, 5309695793417179681, 57919897524216124452
Offset: 0

Views

Author

Wolfdieter Lang, Aug 31 2004

Keywords

Crossrefs

Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

Programs

  • GAP
    a:=[1,12,132];; for n in [4..30] do a[n]:=12*a[n-1]-12*a[n-2]+ a[n-3]; od; a; # G. C. Greubel, May 24 2019
  • Magma
    I:=[1,12,132]; [n le 3 select I[n] else 12*Self(n-1)-12*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, May 24 2019
    
  • Mathematica
    LinearRecurrence[{12,-12,1}, {1,12,132}, 30] (* G. C. Greubel, May 24 2019 *)
  • PARI
    Vec(1/((1-x)*(1-11*x+x^2)) + O(x^30)) \\ Colin Barker, Jun 15 2015
    
  • Sage
    (1/((1-x)*(1 - 11*x + x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 24 2019
    

Formula

a(n) = Sum_{k=0..n} S(k, 11), with S(k, 11) = U(k, 11/2) = A004190(k) Chebyshev's polynomials of the second kind.
G.f.: 1/((1-x)*(1 - 11*x + x^2)) = 1/(1 - 12*x + 12*x^2 - x^3).
a(n) = 12*a(n-1) - 12*a(n-2) + a(n-3) with n >= 2, a(-1)=0, a(0)=1, a(1)=12.
a(n) = 11*a(n-1) - a(n-2) + 1 with n >= 1, a(-1)=0, a(0)=1.
a(n) = (S(n+1, 11) - S(n, 11) - 1)/9.
a(n) = (2^(-n)*(-13*2^n + (65 - 18*sqrt(13))*(11 - 3*sqrt(13))^n + (11 + 3*sqrt(13))^n*(65 + 18*sqrt(13))))/117. - Colin Barker, Mar 06 2016

A006190 a(n) = 3*a(n-1) + a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, 141481, 467280, 1543321, 5097243, 16835050, 55602393, 183642229, 606529080, 2003229469, 6616217487, 21851881930, 72171863277, 238367471761, 787274278560, 2600190307441, 8587845200883, 28363725910090
Offset: 0

Views

Author

Keywords

Comments

Denominators of continued fraction convergents to (3+sqrt(13))/2. - Benoit Cloitre, Jun 14 2003
a(n) and A006497(n) occur in pairs: (a,b): (1,3), (3,11), (10,36), (33,119), (109,393), ... such that b^2 - 13a^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003
Form the 4-node graph with matrix A = [1,1,1,1; 1,1,1,0; 1,1,0,1; 1,0,1,1]. Then this sequence counts the walks of length n from the vertex with degree 5 to one (any) of the other vertices. - Paul Barry, Oct 02 2004
a(n+1) is the binomial transform of A006138. - Paul Barry, May 21 2006
a(n+1) is the diagonal sum of the exponential Riordan array (exp(3x),x). - Paul Barry, Jun 03 2006
Number of paths in the right half-plane from (0,0) to the line x=n-1, consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0). Example: a(3)=10 because we have hh, H, UD, DU, hU, Uh, UU, hD, Dh and DD. - Emeric Deutsch, Sep 03 2007
Equals INVERT transform of A000129. Example: a(5) = 109 = (29, 12, 5, 2, 1) dot (1, 1, 3, 10, 33) = (29 + 12 + 15 + 20 + 33). - Gary W. Adamson, Aug 06 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 3's along the main diagonal, 1's along the superdiagonal and subdiagonal, and 0's everywhere else. - John M. Campbell, Jul 08 2011
These numbers could also be called "bronze Fibonacci numbers". Indeed, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio; analogously, for Pell numbers (A000129), or "silver Fibonacci numbers", P(n+1) = ceiling(delta*a(n)), if n is even and P(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1 + sqrt(2) is the silver ratio. Here, for n >= 1, we have a(n+1) = ceiling(c*a(n)), if n is even and a(n+1) = floor(c*a(n)), if n is odd, where c = (3 + sqrt(13))/2 is the bronze ratio (cf. comment in A098316). - Vladimir Shevelev, Feb 23 2013
Let p(n,x) denote the Fibonacci polynomial, defined by p(1,x) = 1, p(2,x) = x, p(n,x) = x*p(n-1,x) + p(n-2,x). Let q(n,x) be the numerator polynomial of the rational function p(n, x + 1 + 1/x). Then q(n,1) = a(n). - Clark Kimberling, Nov 04 2013
The (1,1)-entry of the matrix A^n where A = [0,1,0; 1,2,1; 1,1,2]. - David Neil McGrath, Jul 18 2014
a(n+1) counts closed walks on K2, containing three loops on the other vertex. Equivalently the (1,1)-entry of A^(n+1) where the adjacency matrix of digraph is A = (0,1; 1,3). - David Neil McGrath, Oct 29 2014
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,2,3} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
With offset 1 is the INVERTi transform of A001076. - Gary W. Adamson, Jul 24 2015
Apart from the initial 0, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = 1 - 3 S. See A291219. - Clark Kimberling, Sep 02 2017
From Rogério Serôdio, Mar 30 2018: (Start)
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).
gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)
Numbers of straight-chain fatty acids involving oxo and/or hydroxy groups, if cis-/trans isomerism and stereoisomerism are neglected. - Stefan Schuster, Apr 04 2018
Number of 3-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
From Michael A. Allen, Jan 25 2023: (Start)
Also called the 3-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 3 kinds of squares available. (End)
a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n >= 1, P(k) = A000129(k) is the k-th Pell number (see example below). - Enrique Navarrete, Dec 15 2023

Examples

			From _Enrique Navarrete_, Dec 15 2023: (Start)
From the comment on compositions with Pell number of parts, A000129(k), there are A000129(1)=1 type of 1, A000129(2)=2 types of 2, A000129(3)=5 types of 3, A000129(4)=12 types of 4, A000129(5)=29 types of 5 and A000129(6)=70 types of 6.
The following table gives the number of compositions of n=6:
Composition, number of such compositions, number of compositions of this type:
  6,              1,     70;
  5+1,            2,     58;
  4+2,            2,     48;
  3+3,            1,     25;
  4+1+1,          3,     36;
  3+2+1,          6,     60;
  2+2+2,          1,      8;
  3+1+1+1,        4,     20;
  2+2+1+1,        6,     24;
  2+1+1+1+1,      5,     10;
  1+1+1+1+1+1,    1,      1;
for a total of a(6)=360 compositions of n=6. (End).
		

References

  • H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178.
  • A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 128.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • L.-N. Machaut, Query 3436, L'Intermédiaire des Mathématiciens, 16 (1909), 62-63. - N. J. A. Sloane, Mar 08 2022

Crossrefs

Row sums of Pascal's rhombus (A059317). Also row sums of triangle A054456(n, m).
Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), A000129 (k=2), this sequence (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), A154597 (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).
Cf. A006497, A052906, A175182 (Pisano periods), A201001 (prime subsequence), A092936 (squares).
Cf. A243399.

Programs

  • GAP
    a:=[0,1];; for n in [3..30] do a[n]:=3*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018
  • Haskell
    a006190 n = a006190_list !! n
    a006190_list = 0 : 1 : zipWith (+) (map (* 3) $ tail a006190_list) a006190_list
    -- Reinhard Zumkeller, Feb 19 2011
    
  • Magma
    [ n eq 1 select 0 else n eq 2 select 1 else 3*Self(n-1)+Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011
    
  • Maple
    a[0]:=0: a[1]:=1: for n from 2 to 35 do a[n]:= 3*a[n-1]+a[n-2] end do: seq(a[n],n=0..30); # Emeric Deutsch, Sep 03 2007
    A006190:=-1/(-1+3*z+z**2); # Simon Plouffe in his 1992 dissertation, without the leading 0
    seq(combinat[fibonacci](n,3),n=0..30); # R. J. Mathar, Dec 07 2011
  • Mathematica
    a[n_] := (MatrixPower[{{1, 3}, {1, 2}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, -1, 24}] (* Robert G. Wilson v, Jan 13 2005 *)
    LinearRecurrence[{3,1},{0,1},30] (* or *) CoefficientList[Series[x/ (1-3x-x^2), {x,0,30}], x] (* Harvey P. Dale, Apr 20 2011 *)
    Table[If[n==0, a1=1; a0=0, a2=a1; a1=a0; a0=3*a1+a2], {n, 0, 30}] (* Jean-François Alcover, Apr 30 2013 *)
    Table[Fibonacci[n, 3], {n, 0, 30}] (* Vladimir Reshetnikov, May 08 2016 *)
  • PARI
    a(n)=if(n<1,0,contfracpnqn(vector(n,i,2+(i>1)))[2,1])
    
  • PARI
    a(n)=([1,3;1,2]^n)[2,1] \\ Charles R Greathouse IV, Mar 06 2014
    
  • PARI
    concat([0],Vec(x/(1-3*x-x^2)+O(x^30))) \\ Joerg Arndt, Apr 30 2013
    
  • Sage
    [lucas_number1(n,3,-1) for n in range(0, 30)] # Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: x/(1 - 3*x - x^2).
From Benoit Cloitre, Jun 14 2003: (Start)
a(3*n) = 2*A041019(5*n-1), a(3*n+1) = A041019(5*n), a(3*n+2) = A041019(5*n+3).
a(2*n) = 3*A004190(n-1); a(3*n) = 10*A041613(n-1) for n >= 1. (End)
From Gary W. Adamson, Jun 15 2003: (Start)
a(n-1) + a(n+1) = A006497(n).
A006497(n)^2 - 13*a(n)^2 = 4(-1)^n. (End)
a(n) = U(n-1, (3/2)i)(-i)^(n-1), i^2 = -1. - Paul Barry, Nov 19 2003
a(n) = Sum_{k=0..n-1} binomial(n-k-1,k)*3^(n-2*k-1). - Paul Barry, Oct 02 2004
a(n) = F(n, 3), the n-th Fibonacci polynomial evaluated at x=3.
Let M = {{0, 1}, {1, 3}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = Abs[v[n][[1]]]. - Roger L. Bagula, May 29 2005 [Or a(n) = [M^(n+1)]{1,1}. - _L. Edson Jeffery, Aug 27 2013]
From Paul Barry, May 21 2006: (Start)
a(n+1) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(k-j).
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(n-j-k).
a(n+1) = Sum_{k=0..floor(n/2)} C(n-k,k)*3^(n-2*k).
a(n) = Sum_{k=0..n} C(k,n-k)*3^(2*k-n). (End)
E.g.f.: exp(3*x/2)*sinh(sqrt(13)*x/2)/(sqrt(13)/2). - Paul Barry, Jun 03 2006
a(n) = (ap^n - am^n)/(ap - am), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2.
Let C = (3 + sqrt(13))/2 = exp arcsinh(3/2) = 3.3027756377... Then C^n, n > 0 = a(n)*(1/C) + a(n+1). Let X = the 2 X 2 matrix [0, 1; 1, 3]. Then X^n = [a(n-1), a(n); a(n), a(n+1)]. - Gary W. Adamson, Dec 21 2007
1/3 = 3/(1*10) + 3/(3*33) + 3/(10*109) + 3/(33*360) + 3/(109*1189) + ... . - Gary W. Adamson, Mar 16 2008
a(n) = ((3 + sqrt(13))^n - (3 - sqrt(13))^n)/(2^n*sqrt(13)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
a(p) == 13^((p-1)/2) mod p, for odd primes p. - Gary W. Adamson, Feb 22 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
Limit_{k->oo} a(n+k)/a(k) = (A006497(n) + a(n)*sqrt(13))/2.
Limit_{n->oo} A006497(n)/a(n) = sqrt(13). (End)
Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (sqrt(13)-3)/2. - Vladimir Shevelev, Feb 23 2013
From Vladimir Shevelev, Feb 24 2013: (Start)
(1) Expression a(n+1) via a(n): a(n+1) = (3*a(n) + sqrt(13*a(n)^2 + 4*(-1)^n))/2;
(2) a^2(n+1) - a(n)*a(n+2) = (-1)^n;
(3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(4) a(n)/a(n+1) = (sqrt(13)-3)/2 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
a(n) = sqrt(13*(A006497(n))^2 + (-1)^(n-1)*52)/13. - Vladimir Shevelev, Mar 13 2013
Sum_{n >= 1} 1/( a(2*n) + 1/a(2*n) ) = 1/3; Sum_{n >= 1} 1/( a(2*n + 1) - 1/a(2*n + 1) ) = 1/9. - Peter Bala, Mar 26 2015
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)*a(n+1) = 3*Sum_{k=1..n} a(k)^2;
(2) a(n)^2 + a(n+1)^2 = a(2*n+1);
(3) a(n)^2 - a(n-2)^2 = 3*a(n-1)*(a(n) + a(n-2));
(4) a(m*(p+1)) = a(m*p)*a(m+1) + a(m*p-1)*a(m);
(5) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(6) a(2*n) = a(n)*(3*a(n) + 2*a(n-1));
(7) 3*Sum_{k=2..n+1} a(k)*a(k-1) is equal to a(n+1)^2 if n odd, and is equal to a(n+1)^2 - 1 if n is even;
(8) a(n) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = ap^(2*k-1) + am^(2*k-1), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2;
(9) 131|Sum_{k=n..n+9} a(k), for all positive n. (End)
From Kai Wang, Feb 10 2020: (Start)
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2 - Catalan's identity.
arctan(1/a(2n)) - arctan(1/a(2n+2)) = arctan(a(2)/a(2n+1)).
arctan(1/a(2n)) = Sum_{m>=n} arctan(a(2)/a(2m+1)).
The same formula holds for Fibonacci numbers and Pell numbers. (End)
a(n+2) = 3^(n+1) + Sum_{k=0..n} a(k)*3^(n-k). - Greg Dresden and Gavron Campbell, Feb 22 2022
G.f. = 1/(1-Sum_{k>=1} P(k)*x^k), P(k)=A000129(k) (with a(0)=1). - Enrique Navarrete, Dec 17 2023
G.f.: x/(1 - 3*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 3 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024
Sum_{n>=0} a(n)/phi^(3*n) = 1, where phi = A001622 is the golden ratio. - Diego Rattaggi, Apr 07 2025

Extensions

Second formula corrected by Johannes W. Meijer, Jun 02 2010

A097783 Chebyshev polynomials S(n,11) + S(n-1,11) with Diophantine property.

Original entry on oeis.org

1, 12, 131, 1429, 15588, 170039, 1854841, 20233212, 220710491, 2407582189, 26262693588, 286482047279, 3125039826481, 34088956044012, 371853476657651, 4056299287190149, 44247438682433988, 482665526219583719, 5265073349732986921, 57433141320843272412
Offset: 0

Views

Author

Wolfdieter Lang, Aug 31 2004

Keywords

Comments

All positive integer solutions of Pell equation (3*a(n))^2 - 13*b(n)^2 = -4 together with b(n)=A078922(n+1), n>=0.

Examples

			All positive solutions to the Pell equation x^2 - 13*y^2 = -4 are (3=3*1,1), (36=3*12,10), (393=3*131,109), (4287=3*1429,1189 ), ...
		

Crossrefs

Programs

  • Magma
    I:=[1,12]; [n le 2 select I[n] else 11*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015
  • Mathematica
    CoefficientList[Series[(1 + x) / (1 - 11 x + x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 22 2015 *)
  • PARI
    Vec((1+x)/(1-11*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015
    
  • Sage
    [(lucas_number2(n,11,1)-lucas_number2(n-1,11,1))/9 for n in range(1, 19)] # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) = S(n, 11) + S(n-1, 11) = S(2*n, sqrt(13)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x) = 0 = U(-1, x).
a(n) = (-2/3)*i*((-1)^n)*T(2*n+1, 3*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-11*x+x^2).
a(n) = L(n,-11)*(-1)^n, where L is defined as in A108299; see also A078922 for L(n,+11). - Reinhard Zumkeller, Jun 01 2005
a(n) = 11*a(n-1) - a(n-2) with a(0)=1 and a(1)=12. - Philippe Deléham, Nov 17 2008
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 12, 0, 131, 0, 1429, 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 = -9, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = 1/2*( (-1)^n - 1 )*F(n,3) + 1/3*( 1 + (-1)^(n+1) )*F(n+1,3), where F(n,x) is the n-th Fibonacci polynomial. The o.g.f. is x*(1 + x^2)/(1 - 11*x^2 + x^4).
Exp( Sum_{n >= 1} 6*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 6*A006190(n)*x^n.
Exp( Sum_{n >= 1} (-6)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 6*A006190(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = A006497(2n+1)/3. - Adam Mohamed, Aug 22 2024

A057076 A Chebyshev or generalized Fibonacci sequence.

Original entry on oeis.org

2, 11, 119, 1298, 14159, 154451, 1684802, 18378371, 200477279, 2186871698, 23855111399, 260219353691, 2838557779202, 30963916217531, 337764520613639, 3684445810532498, 40191139395243839, 438418087537149731, 4782407823513403202, 52168067971110285491, 569066339858699737199
Offset: 0

Views

Author

Wolfdieter Lang, Oct 31 2002

Keywords

Comments

From Klaus Purath, Sep 25 2024: (Start)
a(n) = (t(i+3n) - t(i))/(t(i+2n) - t(i+n)) - 1 for n >= 1, where (t) is any sequence satisfying t(i) = 12t(i-1) - 12t(i-2) + t(i-3) or t(i) = 11t(i-1) - t(i-2) without regard to initial values and including this sequence itself, as long as t(i+2n) - t(i+n) != 0 for integer i.
a(n) = (t(i+3n) + t(i))/(t(i+2n) + t(i+n)) + 1 for i >= 0, n >= 1, where (t) is any sequence satisfying t(i) = 10t(i-1) + 10t(i-2) - t(i-3) or t(i) = 11t(i-1) - t(i-2) without regard to initial values and including this sequence itself, as long as t(i+2n) + t(i+n) != 0.
a(n) = (t(i-n) + t(i+n))/t(i) for i >= n >= 0, where (t) is any recurrence of the form (11,-1) including this sequence itself, as long as t(i) != 0.
a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/12, where (t) is any third order recurrence with constant coefficients (12,-12,1) and initial values t(0) = x, t(1) = x + 2, t(2) = x + 13 for integer x.
a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/10, where (t) is any third order recurrence with constant coefficients (10,10,-1) and initial values t(0) = x, t(1) = 2 - x, t(2) = x + 9 for integer x. (End)

Examples

			G.f. = 2 + 11*x +119*x^2 + 1298*x^3 + 14159*x^4 + 154451*x^5 + ...
		

Crossrefs

Programs

  • Mathematica
    a[0] = 2; a[1] = 11; a[n_] := 11a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)
    a[ n_] := 2 ChebyshevT[ n, 11/2]; (* Michael Somos, May 28 2014 *)
    LinearRecurrence[{11,-1},{2,11},30] (* Harvey P. Dale, Sep 13 2024 *)
  • PARI
    {a(n) = subst( poltchebi(n), x, 11/2) * 2};
    
  • PARI
    {a(n) = 2 * poltchebyshev(n, 1, 11/2)}; /* Michael Somos, May 28 2014 */
    
  • PARI
    Vec((2-11*x)/(1-11*x+x^2) + O(x^40)) \\ Michel Marcus, Feb 18 2016
    
  • Sage
    [lucas_number2(n,11,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

Formula

a(n) = S(n, 11) - S(n-2, 11) = 2*T(n, 11/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 11)=A004190(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, case. See A049310 and A053120.
G.f.: (2-11*x)/(1-11*x+x^2).
a(n) = ap^n + am^n, with ap := (11+sqrt(117))/2 and am := (11-sqrt(117))/2.
a(n) = sqrt(4+117*A004190(n-1)^2), n>=1.
a(n) = a(-n). - Michael Somos, Apr 25 2003
E.g.f.: 2*exp(11*x/2)*cosh(3*sqrt(13)*x/2). - Stefano Spezia, Aug 07 2024
From Klaus Purath, Sep 25 2024: (Start)
a(n) = (a(n-1)*a(n-2) + 1287)/a(n-3) for integer n.
a(n+1)^2 - a(n)*a(n+2) = -117 for integer n. (End)

A078362 A Chebyshev S-sequence with Diophantine property.

Original entry on oeis.org

1, 13, 168, 2171, 28055, 362544, 4685017, 60542677, 782369784, 10110264515, 130651068911, 1688353631328, 21817946138353, 281944946167261, 3643466354036040, 47083117656301259, 608437063177880327
Offset: 0

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 165*a^2 = +4 with companion sequence b(n)=A078363(n+1), n >= 0.
This is the m=15 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..14 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190 and A004191. The m=1..3 (signed) sequences are A049347, A056594, A010892.
For positive n, a(n) equals the permanent of the tridiagonal matrix of order n with 13's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,12}. - Milan Janjic, Jan 23 2015

Crossrefs

Cf. A078363.

Programs

  • GAP
    a:=[1,13,168];; for n in [4..20] do a[n]:=13*a[n-1]-a[n-2]; od; a; # G. C. Greubel, May 25 2019
  • Magma
    I:=[1, 13, 168]; [n le 3 select I[n] else 13*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012
    
  • Mathematica
    CoefficientList[Series[1/(1 - 13 x + x^2), {x, 0, 20}], x] (* Vincenzo Librandi, Dec 24 2012 *)
    LinearRecurrence[{13,-1},{1,13},20] (* Harvey P. Dale, Feb 07 2019 *)
  • PARI
    my(x='x+O('x^20)); Vec(1/(1-13*x+x^2)) \\ G. C. Greubel, May 25 2019
    
  • Sage
    [lucas_number1(n,13,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008
    

Formula

a(n) = 13*a(n-1) - a(n-2), n >= 1; a(-1)=0, a(0)=1.
a(n) = S(2*n+1, sqrt(15))/sqrt(15) = S(n, 13), where S(n, x) = U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (13+sqrt(165))/2 and am = (13-sqrt(165))/2.
G.f.: 1/(1 - 13*x + x^2).
a(n) = Sum_{k=0..n} A101950(n,k)*12^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = (1/11)*(11 + sqrt(165)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = (1/26)*(11 + sqrt(165)). - Peter Bala, Dec 23 2012
For n >= 1, a(n) = U(n-1,13/2), where U(k,x) represents Chebyshev polynomial of the second order.
a(n) = sqrt((A078363(n+1)^2 - 4)/165), n>=0, (Pell equation d=165, +4).

A078364 A Chebyshev S-sequence with Diophantine property.

Original entry on oeis.org

1, 15, 224, 3345, 49951, 745920, 11138849, 166336815, 2483913376, 37092363825, 553901543999, 8271430796160, 123517560398401, 1844491975179855, 27543862067299424, 411313439034311505
Offset: 0

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 221*a^2 = +4 with companion sequence b(n)=A078365(n+1), n>=0.
This is the m=17 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..16 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362 and A007655. The m=1..3 (signed) sequences are A049347, A056594, A010892.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 15'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,...,14}. - Milan Janjic, Jan 23 2015

Crossrefs

a(n) = sqrt((A078365(n+1)^2 - 4)/221), n>=0, (Pell equation d=221, +4).
Cf. A077428, A078355 (Pell +4 equations).

Programs

  • Mathematica
    LinearRecurrence[{15,-1},{1,15},30] (* Harvey P. Dale, Oct 16 2011 *)
  • Sage
    [lucas_number1(n,15,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

Formula

a(n) = 15*a(n-1) - a(n-2), n>= 1; a(-1)=0, a(0)=1.
a(n) = S(2*n+1, sqrt(17))/sqrt(17) = S(n, 15); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (15+sqrt(221))/2 and am = (15-sqrt(221))/2.
G.f.: 1/(1 - 15*x + x^2). - Philippe Deléham, Nov 17 2008
a(n) = Sum_{k=0..n} A101950(n,k)*14^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/13*(13 + sqrt(221)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/30*(13 + sqrt(221)). - Peter Bala, Dec 23 2012
For n>=1, a(n) = U(n-1,15/2), where U(k,x) is Chebyshev polynomial of the second kind. - Milan Janjic, Jan 23 2015

A078366 A Chebyshev S-sequence with Diophantine property.

Original entry on oeis.org

1, 17, 288, 4879, 82655, 1400256, 23721697, 401868593, 6808044384, 115334885935, 1953885016511, 33100710394752, 560758191694273, 9499788548407889, 160935647131239840, 2726406212682669391, 46187969968474139807, 782469083251377707328, 13255786445304946884769
Offset: 0

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 285*a^2 = +4 with companion sequence b(n)=A078367(n+1), n >= 0.
This is the m=19 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..18 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362, A007655, A078364 and A077412. The m=1..3 (signed) sequences are A049347, A056594, A010892.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 17'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,...,16}. - Milan Janjic, Jan 23 2015

Crossrefs

Cf. A077428, A078355 (Pell +4 equations). Cf. A078367.

Programs

  • GAP
    a:=[1,17,288];; for n in [4..20] do a[n]:=17*a[n-1]-a[n-2]; od; a; # G. C. Greubel, May 25 2019
  • Magma
    I:=[1, 17, 288]; [n le 3 select I[n] else 17*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012
    
  • Mathematica
    CoefficientList[Series[1/(1 - 17 x + x^2), {x, 0, 20}], x] (* Vincenzo Librandi, Dec 24 2012 *)
    LinearRecurrence[{17,-1},{1,17},20] (* Harvey P. Dale, Aug 02 2018 *)
  • PARI
    my(x='x+O('x^20)); Vec(1/(1-17*x+x^2)) \\ G. C. Greubel, May 25 2019
    
  • Sage
    [lucas_number1(n,17,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008
    

Formula

a(n) = 17*a(n-1) - a(n-2), n >= 1; a(-1)=0, a(0)=1.
a(n) = S(2*n+1, sqrt(19))/sqrt(19) = S(n, 17), where S(n, x) = U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (17+sqrt(285))/2 and am = (17-sqrt(285))/2.
G.f.: 1/(1-17*x+x^2).
a(n) = Sum_{k=0..n} A101950(n,k)*16^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = (1/15)*(15 + sqrt(285)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = (1/34)*(15 + sqrt(285)). - Peter Bala, Dec 23 2012
For n >= 1, a(n) = U(n-1,13/2), where U(k,x) represents Chebyshev polynomial of the second order. - Milan Janjic, Jan 23 2015
a(n) = sqrt((A078367(n+1)^2 - 4)/285), n>=0, (Pell equation d=285, +4).
E.g.f.: exp(17*x/2)*(sqrt(285)*cosh(sqrt(285)*x/2) + 17*sinh(sqrt(285)*x/2))/sqrt(285). - Stefano Spezia, Aug 19 2023

A200441 Expansion of g.f. 1/(1 - 33*x + x^2).

Original entry on oeis.org

1, 33, 1088, 35871, 1182655, 38991744, 1285544897, 42383989857, 1397386120384, 46071357982815, 1518957427312511, 50079523743330048, 1651105326102579073, 54436396237641779361, 1794749970516076139840, 59172312630792870835359, 1950891566845648661427007
Offset: 0

Views

Author

Bruno Berselli, Nov 18 2011

Keywords

Comments

A Diophantine property of these numbers: (a(n+1)-a(n-1))^2 - 1085*a(n)^2 = 4.
More generally, for t(m)=(m+sqrt(m^2-4))/2 and u(n)=(t(m)^(n+1)-1/t(m)^(n+1))/(t(m)-1/t(m)), we can verify that (u(n+1)-u(n-1))^2-(m^2-4)*u(n)^2=4.
a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,32}. - Milan Janjic, Jan 26 2015

Crossrefs

Programs

  • Magma
    Z:=PolynomialRing(Integers()); N:=NumberField(x^2-1085); S:=[(((33+r)/2)^n-1/((33+r)/2)^n)/r: n in [1..17]]; [Integers()!S[j]: j in [1..#S]];
    
  • Mathematica
    LinearRecurrence[{33, -1}, {1, 33}, 17]
  • Maxima
    makelist(sum((-1)^k*binomial(n-k,k)*33^(n-2*k),k,0,floor(n/2)),n,0,16);
  • PARI
    Vec(1/(1-33*x+x^2)+O(x^17))
    

Formula

a(n) = 33*a(n-1)-a(n-2) with a(0)=1, a(1)=33.
a(n) = -a(-n-2) = (t^(n+1)-1/t^(n+1))/(t-1/t), where t=(33+sqrt(1085))/2.
a(n) = sum((-1)^k*binomial(n-k, k)*33^(n-2k), k=0..floor(n/2)).
a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*32^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 0} (1 + 1/a(n)) = 1/31*(31 + sqrt(1085)).
Product {n >= 1} (1 - 1/a(n)) = 1/66*(31 + sqrt(1085)). (End)
E.g.f.: exp(33*x/2)*cosh(sqrt(1085)*x/2) + 33*exp(33*x/2)*sinh(sqrt(1085)*x/2)/sqrt(1085). - Stefano Spezia, Apr 16 2023

A078368 A Chebyshev S-sequence with Diophantine property.

Original entry on oeis.org

1, 19, 360, 6821, 129239, 2448720, 46396441, 879083659, 16656193080, 315588584861, 5979526919279, 113295422881440, 2146633507828081, 40672741225852099, 770635449783361800, 14601400804658022101
Offset: 0

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 357*a^2 =+4 with companion sequence b(n)=A078369(n+1), n>=0.
This is the m=21 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..20 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362, A007655, A078364, A077412, A078366 and A049660. The m=1..3 (signed) sequences are A049347, A056594, A010892.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 19'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,...,18}. Milan Janjic, Jan 25 2015

Crossrefs

a(n) = sqrt((A078369(n+1)^2 - 4)/357), n>=0, (Pell equation d=357, +4).
Cf. A077428, A078355 (Pell +4 equations).

Programs

Formula

a(n) = 19*a(n-1)-a(n-2), n >= 1; a(-1)=0, a(0)=1.
a(n) = (ap^(n+1)-am^(n+1))/(ap-am) with ap = (19+sqrt(357))/2 and am = (19-sqrt(357))/2.
a(n) = S(2*n+1, sqrt(21))/sqrt(21) = S(n, 19); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
G.f.: 1/(1-19*x+x^2).
a(n) = Sum_{k=0..n} A101950(n,k)*18^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/17*(17 + sqrt(357)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/38*(17 + sqrt(357)). - Peter Bala, Dec 23 2012

A073134 Table by antidiagonals of T(n,k)=n*T(n,k-1)-T(n,k-2) starting with T(n,1)=1.

Original entry on oeis.org

1, 1, 1, 0, 2, 1, -1, 3, 3, 1, -1, 4, 8, 4, 1, 0, 5, 21, 15, 5, 1, 1, 6, 55, 56, 24, 6, 1, 1, 7, 144, 209, 115, 35, 7, 1, 0, 8, 377, 780, 551, 204, 48, 8, 1, -1, 9, 987, 2911, 2640, 1189, 329, 63, 9, 1, -1, 10, 2584, 10864, 12649, 6930, 2255, 496, 80, 10, 1, 0, 11, 6765, 40545, 60605, 40391, 15456, 3905, 711, 99, 11, 1, 1, 12
Offset: 1

Views

Author

Henry Bottomley, Jul 16 2002

Keywords

Examples

			Rows start:
  1, 1,  0, -1,  -1,   0,    1, ...;
  1, 2,  3,  4,   5,   6,    7, ...;
  1, 3,  8, 21,  55, 144,  377, ...;
  1, 4, 15, 56, 209, 780, 2911, ...;
  ...
		

Crossrefs

Rows include A010892, A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190. Columns include (with some gaps) A000012, A000027, A005563, A057722.
Cf. A094954.

Programs

Formula

T(n, k) = A073133(n, k)-2*A073135(n, k-2).
T(n, k) = Sum_{j=0..k-1} A049310(k-1, j)*n^j.
Showing 1-10 of 13 results. Next