Displaying 1-10 of 12 results found.
Partial sums of Chebyshev sequence S(n,11) = U(n,11/2) = A004190(n).
+20
7
1, 12, 132, 1441, 15720, 171480, 1870561, 20404692, 222581052, 2427986881, 26485274640, 288910034160, 3151525101121, 34377866078172, 375005001758772, 4090677153268321, 44622443684192760, 486756203372852040, 5309695793417179681, 57919897524216124452
REFERENCES
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., 58:2 (2020), 140-142.
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
MATHEMATICA
LinearRecurrence[{12, -12, 1}, {1, 12, 132}, 30] (* G. C. Greubel, May 24 2019 *)
PROG
(PARI) Vec(1/((1-x)*(1-11*x+x^2)) + O(x^30)) \\ Colin Barker, Jun 15 2015 (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 (Sage) (1/((1-x)*(1 - 11*x + x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 24 2019 (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
CROSSREFS
Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).
a(n) = 3*a(n-1) + a(n-2), with a(0)=0, a(1)=1.
(Formerly M2844)
+10
147
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
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 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 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 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 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
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
LINKS
W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1976), 318-328.
FORMULA
G.f.: x/(1 - 3*x - x^2).
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 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 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 (1) Expression a(n+1) via a(n): a(n+1) = (3*a(n) + sqrt(13*a^2(n) + 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)
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 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) - a(n-2*k+1) = 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)
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.: 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
EXAMPLE
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).
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 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 *)
PROG
(PARI) a(n)=if(n<1, 0, contfracpnqn(vector(n, i, 2+(i>1)))[2, 1])
(Sage) [lucas_number1(n, 3, -1) for n in range(0, 30)] # Zerinvary Lajos, Apr 22 2009 (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 (Haskell)
a006190 n = a006190_list !! n
a006190_list = 0 : 1 : zipWith (+) (map (* 3) $ tail a006190_list) a006190_list
(PARI) concat([0], Vec(x/(1-3*x-x^2)+O(x^30))) \\ Joerg Arndt, Apr 30 2013 (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
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).
Chebyshev polynomials S(n,11) + S(n-1,11) with Diophantine property.
+10
12
1, 12, 131, 1429, 15588, 170039, 1854841, 20233212, 220710491, 2407582189, 26262693588, 286482047279, 3125039826481, 34088956044012, 371853476657651, 4056299287190149, 44247438682433988, 482665526219583719, 5265073349732986921, 57433141320843272412
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.
LINKS
Andersen, K., Carbone, L. and Penta, D., Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
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) = 11*a(n-1) - a(n-2) with a(0)=1 and a(1)=12. - Philippe Deléham, Nov 17 2008 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.
EXAMPLE
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 ), ...
MATHEMATICA
CoefficientList[Series[(1 + x) / (1 - 11 x + x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 22 2015 *)
PROG
(Sage) [(lucas_number2(n, 11, 1)-lucas_number2(n-1, 11, 1))/9 for n in range(1, 19)] # Zerinvary Lajos, Nov 10 2009 (PARI) Vec((1+x)/(1-11*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015 (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
A Chebyshev or generalized Fibonacci sequence.
+10
9
2, 11, 119, 1298, 14159, 154451, 1684802, 18378371, 200477279, 2186871698, 23855111399, 260219353691, 2838557779202, 30963916217531, 337764520613639, 3684445810532498, 40191139395243839, 438418087537149731, 4782407823513403202, 52168067971110285491, 569066339858699737199
COMMENTS
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)
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. E.g.f.: 2*exp(11*x/2)*cosh(3*sqrt(13)*x/2). - Stefano Spezia, Aug 07 2024 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)
EXAMPLE
G.f. = 2 + 11*x +119*x^2 + 1298*x^3 + 14159*x^4 + 154451*x^5 + ...
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 *)
PROG
(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
A Chebyshev S-sequence with Diophantine property.
+10
9
1, 13, 168, 2171, 28055, 362544, 4685017, 60542677, 782369784, 10110264515, 130651068911, 1688353631328, 21817946138353, 281944946167261, 3643466354036040, 47083117656301259, 608437063177880327
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
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).
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).
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 *)
PROG
(Sage) [lucas_number1(n, 13, 1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008 (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 (PARI) my(x='x+O('x^20)); Vec(1/(1-13*x+x^2)) \\ G. C. Greubel, May 25 2019 (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
A Chebyshev S-sequence with Diophantine property.
+10
7
1, 15, 224, 3345, 49951, 745920, 11138849, 166336815, 2483913376, 37092363825, 553901543999, 8271430796160, 123517560398401, 1844491975179855, 27543862067299424, 411313439034311505
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
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.
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
MATHEMATICA
LinearRecurrence[{15, -1}, {1, 15}, 30] (* Harvey P. Dale, Oct 16 2011 *)
PROG
(Sage) [lucas_number1(n, 15, 1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008
CROSSREFS
a(n) = sqrt(( A078365(n+1)^2 - 4)/221), n>=0, (Pell equation d=221, +4).
A Chebyshev S-sequence with Diophantine property.
+10
7
1, 17, 288, 4879, 82655, 1400256, 23721697, 401868593, 6808044384, 115334885935, 1953885016511, 33100710394752, 560758191694273, 9499788548407889, 160935647131239840, 2726406212682669391, 46187969968474139807, 782469083251377707328, 13255786445304946884769
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
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).
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
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 *)
PROG
(Sage) [lucas_number1(n, 17, 1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008 (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 (PARI) my(x='x+O('x^20)); Vec(1/(1-17*x+x^2)) \\ G. C. Greubel, May 25 2019 (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
Expansion of g.f. 1/(1 - 33*x + x^2).
+10
6
1, 33, 1088, 35871, 1182655, 38991744, 1285544897, 42383989857, 1397386120384, 46071357982815, 1518957427312511, 50079523743330048, 1651105326102579073, 54436396237641779361, 1794749970516076139840, 59172312630792870835359, 1950891566845648661427007
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
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)).
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
MATHEMATICA
LinearRecurrence[{33, -1}, {1, 33}, 17]
PROG
(PARI) Vec(1/(1-33*x+x^2)+O(x^17))
(Magma) Z<x>:=PolynomialRing(Integers()); N<r>:=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]];
(Maxima) makelist(sum((-1)^k*binomial(n-k, k)*33^(n-2*k), k, 0, floor(n/2)), n, 0, 16);
A Chebyshev S-sequence with Diophantine property.
+10
4
1, 19, 360, 6821, 129239, 2448720, 46396441, 879083659, 16656193080, 315588584861, 5979526919279, 113295422881440, 2146633507828081, 40672741225852099, 770635449783361800, 14601400804658022101
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
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).
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
MATHEMATICA
LinearRecurrence[{19, -1}, {1, 19}, 20] (* Harvey P. Dale, Feb 10 2019 *)
PROG
(Sage) [lucas_number1(n, 19, 1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008
CROSSREFS
a(n) = sqrt(( A078369(n+1)^2 - 4)/357), n>=0, (Pell equation d=357, +4).
Table by antidiagonals of T(n,k)=n*T(n,k-1)-T(n,k-2) starting with T(n,1)=1.
+10
3
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
FORMULA
T(n, k) = Sum_{j=0..k-1} A049310(k-1, j)*n^j.
EXAMPLE
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.
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