A041038 -id:A041038 - cp's OEIS frontend

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

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

Offset: 0

N. J. A. Sloane

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

			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 + ...

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.

T. D. Noe, Table of n, a(n) for n=0..200
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Leonhard Euler, Vollstaendige Anleitung zur Algebra, Zweiter Teil.
Leonhard Euler, De solutione problematum diophanteorum per numeros integros, par. 18.
Tanya Khovanova, Recursive Sequences
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (10,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A004189, A001078, A046173, A046172, A036353, A138281, A004189.

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

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

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 *)

{a(n) = subst(poltchebi(n), 'x, 5)}; /* Michael Somos, Sep 05 2006 */

{a(n) = real((5 + 2*quadgen(24))^n)}; /* Michael Somos, Sep 05 2006 */

{a(n) = n = abs(n); polsym(1 - 10*x + x^2, n)[n+1] / 2}; /* Michael Somos, Sep 05 2006 */

x='x+O('x^30); Vec((1-5*x)/(1-10*x+x^2)) \\ G. C. Greubel, Dec 20 2017

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

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

Original entry on oeis.org

1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, 102558961, 1015229051, 10049731549, 99482086439, 984771132841, 9748229241971, 96497521286869, 955226983626719, 9455772314980321, 93602496166176491, 926569189346784589, 9172089397301669399, 90794324783669909401

Offset: 0

Ignacio Larrosa Cañestro, Feb 27 2000

Chebyshev's even-indexed U-polynomials evaluated at sqrt(3).
a(n)^2 is a star number (A003154).
Any k in the sequence has the successor 5*k + 2*sqrt(3(2*k^2 + 1)). - Lekraj Beedassy, Jul 08 2002
{a(n)} give the values of x solving: 3*y^2 - 2*x^2 = 1. Corresponding values of y are given by A072256(n+1). x + y = A001078(n+1). - Richard R. Forberg, Nov 21 2013
The aerated sequence (b(n))n>=1 = [1, 0, 11, 0, 109, 0, 1079, 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 = -8, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

			a(1)^2 = 121 is the 5th star number (A003154).

G. C. Greubel, Table of n, a(n) for n = 0..1000
K. Andersen, L. Carbone, and D. Penta, 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.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Star Number
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (10,-1).
Index entries for sequences related to Chebyshev polynomials.

A member of the family A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, which are the expansions of (1+x) / (1-kx+x^2) with k = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, May 04 2004
Cf. A003154, A006061, A031138, A007667, A004189.
Cf. A138281. Cf. A100047.
Cf. A142238.

a:=[1,11];; for n in [3..30] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jul 22 2019

I:=[1,11]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

CoefficientList[Series[(1+x)/(1-10x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Mar 22 2015 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Numerator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A142238 *)
a[3/2, 20] (* Gerry Martens, Jun 07 2015 *)

a(n)=subst(poltchebi(n+1)-poltchebi(n),x,5)/4;

(a(n)-1)^2 + a(n)^2 + (a(n)+1)^2 = b(n)^2 + (b(n)+1)^2 = c(n), where b(n) is A031138 and c(n) is A007667.
a(n) = 10*a(n-1) - a(n-2).
a(n) = (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1).
a(n) = U(2*(n-1), sqrt(3)) = S(n-1, 10) + S(n-2, 10) with Chebyshev's U(n, x) and S(n, x) := U(n, x/2) polynomials and S(-1, x) := 0. S(n, 10) = A004189(n+1), n >= 0.
6*a(n)^2 + 3 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then (-1)^n*q(n, -12) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-10)*(-1)^n, where L is defined as in A108299; see also A072256 for L(n,+10). - Reinhard Zumkeller, Jun 01 2005
From Reinhard Zumkeller, Mar 12 2008: (Start)
(sqrt(2) + sqrt(3))^(2*n+1) = a(n)*sqrt(2) + A138288(n)*sqrt(3);
a(n) = A138288(n) + A001078(n).
a(n) = A001079(n) + 3*A001078(n). (End)
a(n) = A142238(2n) = A041006(2n)/2 = A041038(2n)/4. - M. F. Hasler, Feb 14 2009
a(n) = sqrt(A006061(n)). - Zak Seidov, Oct 22 2012
a(n) = sqrt((3*A072256(n)^2 - 1)/2). - T. D. Noe, Oct 23 2012
(sqrt(3) + sqrt(2))^(2*n+1) - (sqrt(3) - sqrt(2))^(2*n+1) = a(n)*sqrt(8). - Bruno Berselli, Oct 29 2019
a(n) = A004189(n)+A004189(n+1). - R. J. Mathar, Oct 01 2021
E.g.f.: exp(5*x)*(2*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/2. - Stefano Spezia, May 16 2023
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 5), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 10*a(n)*a(n+1) + a(n+1)^2 = 12.
More generally, for arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 12 with a(n) := (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1) as given above.
a(n+1/2) = sqrt(3) * A001078(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6)*sqrt(sqrt(3) + 1) * A001078(n+1).
a(n+3/4) - a(n+1/4) = sqrt(sqrt(3) - 1) * A001079(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/12 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A004291(n) + 1/A004291(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3/2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3/2 * (1 - 1/A171640(k+2))). (End)

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A041006 Numerators of continued fraction convergents to sqrt(6).

Original entry on oeis.org

2, 5, 22, 49, 218, 485, 2158, 4801, 21362, 47525, 211462, 470449, 2093258, 4656965, 20721118, 46099201, 205117922, 456335045, 2030458102, 4517251249, 20099463098, 44716177445, 198964172878, 442644523201, 1969542265682, 4381729054565, 19496458483942

Offset: 0

N. J. A. Sloane

Interspersion of 2 sequences, 2*A054320 and A001079. - Gerry Martens, Jun 10 2015

Hugo Pfoertner, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Cf. A041007 (denominators).
Cf. A054320, A001079.
Analog for other sqrt(m): A001333 (m=2), A002531 (m=3), A001077 (m=5), A041008 (m=7), A041010 (m=8), A005667 (m=10), A041014 (m=11), ..., A042936 (m=1000).

I:=[2, 5, 22, 49]; [n le 4 select I[n] else 10*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 10 2015

Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[6],n]]],{n,1,50}] (* Vladimir Joseph Stephan Orlovsky, Mar 16 2011 *)
LinearRecurrence[{0, 10, 0, -1}, {2, 5, 22, 49}, 50] (* Vincenzo Librandi, Jun 10 2015 *)

A41006=contfracpnqn(c=contfrac(sqrt(6)), #c)[1, ][^-1] \\ Discard possibly incorrect last element. NB: a(n)=A41006[n+1]! M. F. Hasler, Nov 01 2019

\\ For correct index & more terms:
A041006(n)={n<#A041006|| A041006=extend(A041006, [2, 10; 4, -1], n\.8); A041006[n+1]}
extend(A, c, N)={for(n=#A+1, #A=Vec(A, N), A[n]=[A[n-i]|i<-c[, 1]]*c[, 2]); A} \\ M. F. Hasler, Nov 01 2019

From M. F. Hasler, Feb 13 2009: (Start)
a(2n) = 2*A142238(2n) = A041038(2n)/2;
a(2n-1) = A142238(2n-1) = A041038(2n-1) = A001079(n). (End)
G.f.: (2 + 5*x + 2*x^2 - x^3)/(1 - 10*x^2 + x^4).
a(n) = ((2 + sqrt(6))^(n+1) + (2 - sqrt(6))^(n+1))/2^(ceiling(n/2) + 1). - Robert FERREOL, Oct 13 2024
E.g.f.: sqrt(2)*sinh(sqrt(2)*x)*(cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) + cosh(sqrt(2)*x)*(2*cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)). - Stefano Spezia, Oct 14 2024

More terms from Vincenzo Librandi, Jun 10 2015

A142238 Numerators of continued fraction convergents to sqrt(3/2).

Original entry on oeis.org

1, 5, 11, 49, 109, 485, 1079, 4801, 10681, 47525, 105731, 470449, 1046629, 4656965, 10360559, 46099201, 102558961, 456335045, 1015229051, 4517251249, 10049731549, 44716177445, 99482086439, 442644523201, 984771132841, 4381729054565, 9748229241971

Offset: 0

N. J. A. Sloane, Oct 05 2008, following a suggestion from Rob Miller (rmiller(AT)AmtechSoftware.net)

From Charlie Marion, Jan 07 2009: (Start)
In general, denominators, a(k,n) and numerators, b(k,n), of continued
fraction convergents to sqrt((k+1)/k) may be found as follows:
a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1)+a(k,2n-2)
and a(k,2n+1)=(2k)*a(k,2n)+a(k,2n-1);
b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1)+b(k,2n-2)
and b(k,2n+1)=(2k)*b(k,2n)+b(k,2n-1).
For example, the convergents to sqrt(3/2) start 1/1, 5/4, 11/9,
49/40, 109/89.
In general, if a(k,n) and b(k,n) are the denominators and numerators,
respectively, of continued fraction convergents to sqrt((k+1)/k)
as defined above, then
k*a(k,2n)^2-a(k,2n-1)*a(k,2n+1)=k=k*a(k,2n-2)*a(k,2n)-a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1)-k*b(k,2n)^2=k+1=b(k,2n-1)^2-k*b(k,2n-2)*b(k,2n);
for example, if k=2 and n=3, then b(2,n)=a(n) and
2*a(2,6)^2-a(2,5)*a(2,7)=2*881^2-396*3920=2;
2*a(2,4)*a(2,6)-a(2,5)^2=2*89*881-396^2=2;
b(2,5)*b(2,7)-2*b(2,6)^2=485*4801-2*1079^2=3;
b(2,5)^2-2*b(2,4)*b(2,6)=485^2-2*109*1079=3.
Cf. A000129, A001333, A142239, A153315-A153318. (End)

			The initial convergents are 1, 5/4, 11/9, 49/40, 109/89, 485/396, 1079/881, 4801/3920, 10681/8721, 47525/38804, 105731/86329, ...

Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Cf. A115754, A142239, A382713.

with(numtheory): cf := cfrac (sqrt(3)/sqrt(2),100): [seq(nthnumer(cf,i), i=0..50)]; [seq(nthdenom(cf,i), i=0..50)]; [seq(nthconver(cf,i), i=0..50)];

Numerator[Convergents[Sqrt[3/2], 30]] (* Bruno Berselli, Nov 11 2013 *)
LinearRecurrence[{0,10,0,-1},{1,5,11,49},30] (* Harvey P. Dale, Dec 30 2017 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,10,0]^n*[1;5;11;49])[1,1] \\ Charles R Greathouse IV, Jun 21 2015

G.f.'s for numerators and denominators are -(1+5*x+x^2-x^3)/(-1-x^4+10*x^2) and -(1+4*x-x^2)/(-1-x^4+10*x^2).

a(2n) = A041006(2n)/2 = A054320(n), a(2n-1) = A041006(2n-1) = A041038(2n-1) = A001079(n). - M. F. Hasler, Feb 14 2009

A041039 Denominators of continued fraction convergents to sqrt(24).

Original entry on oeis.org

1, 1, 9, 10, 89, 99, 881, 980, 8721, 9701, 86329, 96030, 854569, 950599, 8459361, 9409960, 83739041, 93149001, 828931049, 922080050, 8205571449, 9127651499, 81226783441, 90354434940, 804062262961

Offset: 0

N. J. A. Sloane

The following remarks assume an offset of 1. This is the sequence of Lehmer numbers U_n(sqrt(R),Q) for the parameters R = 8 and Q = -1; it is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. Consequently, this is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, May 28 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Eric W. Weisstein, MathWorld: Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Cf. A010480, A041038, A002530.

Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[24],n]]],{n,1,50}] (* Vladimir Joseph Stephan Orlovsky, Mar 18 2011 *)
Denominator[Convergents[Sqrt[24],30]] (* or *) LinearRecurrence[{0,10,0,-1},{1,1,9,10},30] (* Harvey P. Dale, Apr 12 2022 *)

G.f.: (1+x-x^2)/(1-10*x^2+x^4). - Colin Barker, Jan 01 2012
From Peter Bala, May 28 2014: (Start)
The following remarks assume an offset of 1.
Let alpha = sqrt(2) + sqrt(3) and beta = sqrt(2) - sqrt(3) be the roots of the equation x^2 - sqrt(8)*x - 1 = 0. Then a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, while a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even.
a(n) = Product_{k = 1..floor((n-1)/2)} ( 8 + 4*cos^2(k*Pi/n) ).
Recurrence equations: a(0) = 0, a(1) = 1 and for n >= 1, a(2*n) = a(2*n - 1) + a(2*n - 2) and a(2*n + 1) = 8*a(2*n) + a(2*n - 1). (End)

cp's OEIS Frontend

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

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A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

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A041006 Numerators of continued fraction convergents to sqrt(6).

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A142238 Numerators of continued fraction convergents to sqrt(3/2).

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A041039 Denominators of continued fraction convergents to sqrt(24).

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