A001078 -id:A001078 - cp's OEIS frontend

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

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

Offset: 0

Reinhard Zumkeller, Mar 12 2008

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

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

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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jean-Paul Allouche, Jeffrey Shallit, and Manon Stipulanti, Combinatorics on words and generating Dirichlet series of automatic sequences, arXiv:2401.13524 [math.CO], 2024.
Bruno Deschamps, Sur les bonnes valeurs initiales de la suite de Lucas-Lehmer, Journal of Number Theory, Volume 130, Issue 12, December 2010, Pages 2658-2670.
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).
Index entries for linear recurrences with constant coefficients, signature (10, -1).

Cf. A001078, A001079, A072256, A138281.
Cf. similar sequences listed in A238379.
Cf. A041007, A041039, A054320.

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 *)

{a(n) = subst( poltchebi(n+1) + poltchebi(n), x, 5) / 6} /* Michael Somos, Jan 25 2013 */

[lucas_number1(n,10,1)-lucas_number1(n-1,10,1) for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009

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

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

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

A041041 Denominators of continued fraction convergents to sqrt(26).

Original entry on oeis.org

1, 10, 101, 1020, 10301, 104030, 1050601, 10610040, 107151001, 1082120050, 10928351501, 110365635060, 1114584702101, 11256212656070, 113676711262801, 1148023325284080, 11593909964103601, 117087122966320090, 1182465139627304501, 11941738519239365100

Offset: 0

N. J. A. Sloane

Generalized Fibonacci sequence.
Sqrt(26) = 10/2 + 10/101 + 10/(101*10301) + 10/(10301*1050601) + ... - Gary W. Adamson, Jun 13 2008
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 10's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0, 1, ..., 10} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Bruno Berselli, May 03 2018: (Start)
Numbers k for which m*k^2 + (-1)^k is a perfect square:
m =  2: 0, 1, 2, 5, 12, 29, 70, 169, ...   (A000129);
m =  3: 0, 4, 56, 780, 10864, 151316, ...  (4*A007655);
m =  5: 0, 1, 4, 17, 72, 305, 1292, ...    (A001076);
m =  6: 0, 2, 20, 198, 1960, 19402, ...    (A001078);
m =  7: 0, 48, 12192, 3096720, ...         (2*A175672);
m =  8: 0, 6, 204, 6930, 235416, ...       (A082405);
m = 10: 0, 1, 6, 37, 228, 1405, 8658, ...  (A005668);
m = 11: 0, 60, 23880, 9504180, ...         [°];
m = 12: 0, 2, 28, 390, 5432, 75658, ...    (A011944);
m = 13: 0, 5, 180, 6485, 233640, ...       (5*A041613);
m = 14: 0, 4, 120, 3596, 107760, ...       (A068204);
m = 15: 0, 8, 496, 30744, 1905632, ...     [°];
m = 17: 0, 1, 8, 65, 528, 4289, 34840, ... (A041025);
m = 18: 0, 4, 136, 4620, 156944, ...       (A202299);
m = 19: 0, 13260, 1532829480, ...          [°];
m = 20: 0, 2, 36, 646, 11592, 208010, ...  (A207832);
m = 21: 0, 12, 1320, 145188, ...           (A174745);
m = 22: 0, 42, 16548, 6519870, ...         (A174766);
m = 23: 0, 240, 552480, 1271808720, ...    [°];
m = 24: 0, 10, 980, 96030, 9409960, ...    (A168520);
m = 26: 0, 1, 10, 101, 1020, 10301, ...    (this sequence);
m = 27: 0, 260, 702520, 1898208780, ...    [°];
m = 28: 0, 24, 6096, 1548360, ...          (A175672);
m = 29: 0, 13, 1820, 254813, 35675640, ... [°];
m = 30: 0, 2, 44, 966, 21208, 465610, ...  (2*A077421), etc.
[°] apparently without related sequences in the OEIS.
(End)
From Michael A. Allen, Mar 12 2023: (Start)
Also called the 10-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 10 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcón and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
Sergio Falcón and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle Chaos, Solitons & Fractals 2007; 33(1): 38-49.
Sergio Falcón and Ángel Plaza, On k-Fibonacci sequences and polynomials and their derivatives, Chaos, Solitons & Fractals (2007).
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (10,1).

Cf. A099374 (squares).
Cf. A041040.
Cf. A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371, A243399.
Row n=10 of A073133, A172236 and A352361.

I:=[1,10]; [n le 2 select I[n] else 10*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

seq(combinat:-fibonacci(n+1, 10), n=0..19); # Peter Luschny, May 04 2018

Denominator[Convergents[Sqrt[26], 30]] (* Vincenzo Librandi, Dec 10 2013 *)
LinearRecurrence[{10,1}, {1,10}, 30] (* G. C. Greubel, Jan 24 2018 *)

x='x+O('x^30); Vec(1/(1-10*x-x^2)) \\ G. C. Greubel, Jan 24 2018

[lucas_number1(n,10,-1) for n in range(1, 19)] # Zerinvary Lajos, Apr 26 2009

G.f.: 1/(1 - 10*x - x^2).
a(n) = 10*a(n-1) + a(n-2), n>=1; a(-1):=0, a(0)=1.
a(n) = S(n, 10*i)*(-i)^n where i^2:=-1 and S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind. See A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = 5+sqrt(26), am = -1/ap = 5-sqrt(26).
a(n) = F(n+1, 10), the (n+1)-th Fibonacci polynomial evaluated at x=10. - T. D. Noe, Jan 19 2006
a(n) = Sum_{i=0..floor(n/2)} binomial(n-i,i)*10^(n-2*i). - Sergio Falcon, Sep 24 2007

Extended by T. D. Noe, May 23 2011

A036353 Square pentagonal numbers.

Original entry on oeis.org

0, 1, 9801, 94109401, 903638458801, 8676736387298001, 83314021887196947001, 799981229484128697805801, 7681419682192581869134354401, 73756990988431941623299373152801, 708214619789503821274338711878841001, 6800276705461824703444258688161258139001

Offset: 0

Jean-Francois Chariot (jeanfrancois.chariot(AT)afoc.alcatel.fr)

Lim_{n -> oo} a(n)/a(n-1) = (sqrt(2) + sqrt(3))^8 = 4801 + 1960*sqrt(6). - Ant King, Nov 06 2011

Pentagonal numbers (A000326) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 11 2015

Colin Barker, Table of n, a(n) for n = 0..252
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7.
Byungchan Kim, Eunmi Kim, and Jeremy Lovejoy, On weighted overpartitions related to some q-series in Ramanujan's lost notebook, Int'l J. Number Theory (2021). Also at Université de Paris (France, 2020).
Eric Weisstein's World of Mathematics, Pentagonal Square Number
Index entries for linear recurrences with constant coefficients, signature (9603,-9603,1).

Cf. A000326, A001078, A001079, A001110, A046172, A046173, A248205.

Table[Floor[1/96 ( Sqrt[2] + Sqrt[3] ) ^ ( 8*n - 4 ) ] , {n, 0, 9}] (* Ant King, Nov 06 2011 *)
LinearRecurrence[{9603,-9603,1},{0,1,9801,94109401},20] (* Harvey P. Dale, Apr 14 2019 *)

for(n=0,10^9,g=(n*(3*n-1)/2); if(issquare(g),print(g)))

concat(0, Vec(x*(1+198*x+x^2)/((1-x)*(1-9602*x+x^2)) + O(x^20))) \\ Colin Barker, Jun 24 2015

a(n) = 9602*a(n-1) - a(n-2) + 200; g.f.: x*(1+198*x+x^2)/((1-x)*(1-9602*x+x^2)). - Warut Roonguthai, Jan 05 2001
a(n+1) = 4801*a(n)+100+980*(24*a(n)^2+a(n))^(1/2). - Richard Choulet, Sep 21 2007
From Ant King, Nov 06 2011: (Start)
a(n) = floor(1/96*(sqrt(2) + sqrt(3))^(8*n-4)).
a(n) = 9603*a(n-1) - 9603*a(n-2) + a(n-3).
(End)

More terms from Eric W. Weisstein

A084070 a(n) = 38*a(n-1) - a(n-2), with a(0)=0, a(1)=6.

Original entry on oeis.org

0, 6, 228, 8658, 328776, 12484830, 474094764, 18003116202, 683644320912, 25960481078454, 985814636660340, 37434995712014466, 1421544022419889368, 53981237856243781518, 2049865494514843808316, 77840907553707820934490, 2955904621546382351702304

Offset: 0

Benoit Cloitre, May 10 2003

nonn

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 10*y^2 = 1. The corresponding x values are in A078986. - Vincenzo Librandi, Aug 08 2010 [edited by Jon E. Schoenfield, May 04 2014]

			G.f. = 6*x + 228*x^2 + 8658*x^3 + 328776*x^4 + ... - _Michael Somos_, Feb 24 2023

Indranil Ghosh, Table of n, a(n) for n = 0..632
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A001078, A001109, A001353, A001653, A005668, A060645, A084068, A084069, A221874.

Cf. A078986. - Vincenzo Librandi, Apr 14 2010

a:=[0,6];; for n in [3..20] do a[n]:=38*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2020

I:=[0,6]; [n le 2 select I[n] else 38*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Jan 12 2020

seq( simplify(6*ChebyshevU(n-1, 19)), n=0..20); # G. C. Greubel, Jan 12 2020

LinearRecurrence[{38,-1},{0,6},30] (* Harvey P. Dale, Nov 01 2011 *)
6*ChebyshevU[Range[20]-2, 19] (* G. C. Greubel, Jan 12 2020 *)

u=0; v=6; for(n=2,20, w=38*v-u; u=v; v=w; print1(w,","))

vector(21, n, 6*polchebyshev(n-2, 2, 19) ) \\ G. C. Greubel, Jan 12 2020

[6*chebyshev_U(n-1, 19) for n in (0..20)] # G. C. Greubel, Jan 12 2020

Numbers k such that 10*k^2 = floor(k*sqrt(10)*ceiling(k*sqrt(10))).
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 37*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 39*(a(n-1) - a(n-2)) + a(n-3). (End)
From R. J. Mathar, Feb 19 2008: (Start)
O.g.f.: 6*x/(1 - 38*x + x^2).
a(n) = 6*A078987(n-1). (End)
a(n) = 6*ChebyshevU(n-1, 19). - G. C. Greubel, Jan 12 2020
a(n) = A005668(2*n). - Michael Somos, Feb 24 2023

A045502 Numbers k such that 2k+1 and 3k+1 are squares.

Original entry on oeis.org

0, 40, 3960, 388080, 38027920, 3726348120, 365144087880, 35780394264160, 3506113493799840, 343563341998120200, 33665701402321979800, 3298895174085555900240, 323258061358982156243760, 31675991118006165755988280, 3103923871503245261930607720

Offset: 0

Fred Schwab (fschwab(AT)nrao.edu)

Problem 1 for the 3rd grade of the 38th Mathematics Competition of the Republic of Slovenia (1998) was to prove that if k is a natural number such that 2*k+1 and 3*k+1 are perfect squares, then k is divisible by 40 (see link with solution Crux Mathematicorum and formula Mar 25 2021). - Bernard Schott, Mar 25 2021

Colin Barker, Table of n, a(n) for n = 0..500.
John Albert, Pell Equations, Putnam practice, November 17, 2004 (1-2).
R. S. Luthar, Problem E2606, Amer. Math. Monthly, 84 (1977), 823-824.
R. E. Woodrow, Problem 1 for the third grade of the 38th Mathematics competition of the Republic of Slovenia (1998), Crux Mathematicorum, The Olympiad Corner, p. 208, April 1999, Vol. 25, No. 4.
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).
Index to sequences related to Olympiads and other Mathematical Competitions.

Cf. A001078, A245031, A278620.

a:=[0,40,3960];; for n in [4..15] do a[n]:=99*a[n-1]-99*a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Jul 17 2018

I:=[0,40,3960]; [n le 3 select I[n] else 99*Self(n-1) -99*Self(n-2) + Self(n-3): n in [1..15]]; // G. C. Greubel, Jan 13 2020

seq(coeff(series(40*x/((1-x)*(x^2-98*x+1)), x,n+1),x,n),n=0..15); # Muniru A Asiru, Jul 17 2018

f[0]=0; f[1]=2; f[n_]:= f[n]= 10*f[n-1] -f[n-2]; a[n_]:= f[n]*f[n+1];
CoefficientList[Series[40x/((1-x)(1-98x+x^2)), {x,0,15}], x] (* Michael De Vlieger, Jul 20 2018 *)
Table[5*(ChebyshevT[n, 49] +48*ChebyshevU[n-1, 49] -1)/12, {n,0,15}] (* G. C. Greubel, Jan 13 2020 *)
LinearRecurrence[{99,-99,1},{0,40,3960},20] (* Harvey P. Dale, Dec 02 2023 *)

concat(0, Vec(40*x/((1-x)*(1-98*x+x^2))+O(x^20))) \\ Colin Barker, Mar 23 2017

[4*chebyshev_U(n-1,5)*chebyshev_U(n,5) for n in (0..15)] # G. C. Greubel, Jan 13 2020

From Colin Barker, Mar 23 2017: (Start)
O.g.f.: 40*x / ((1 - x)*(1 - 98*x + x^2)).
a(n) = 99*a(n-1)- 99*a(n-2) + a(n-3) for n>2.
a(n) = (-10 + (5 - 2*sqrt(6))*(49 + 20*sqrt(6))^(-n) + (5 + 2*sqrt(6))*(49 + 20*sqrt(6))^n)/24. (End)
From G. C. Greubel, Jan 13 2020: (Start)
a(n) = 5*(ChebyshevT(n, 49) + 48*ChebyshevU(n-1, 48) - 1)/12.
a(n) = 4*ChebyshevU(n-1, 5)*ChebyshevU(n, 5). (End)
a(n) = 40*A278620(n). - Bernard Schott, Mar 25 2021

A122652 a(0) = 0, a(1) = 4; for n > 1, a(n) = 10*a(n-1) - a(n-2).

Original entry on oeis.org

0, 4, 40, 396, 3920, 38804, 384120, 3802396, 37639840, 372596004, 3688320200, 36510605996, 361417739760, 3577666791604, 35415250176280, 350574834971196, 3470333099535680, 34352756160385604, 340057228504320360, 3366219528882817996, 33322138060323859600

Offset: 0

N. J. A. Sloane, Sep 21 2006

Kekulé numbers for the benzenoids P_2(n).
a(n) are the values of m where A032528(m) - 1 has integer square roots. The roots are given by A001079. - Richard R. Forberg, Aug 05 2013
Numbers n such that 6*n^2 + 4 is a square. - Colin Barker, Mar 17 2014

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 283, K{P_2(n)}).

Michael De Vlieger, Table of n, a(n) for n = 0..1004
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.
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.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Cf. A001078, A001079, A032528.

CoefficientList[Series[(4 z)/(z^2 - 10 z + 1), {z, 0, 200}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 11 2011 *)
LinearRecurrence[{10, -1}, {0, 4}, 21] (* Jean-François Alcover, Jan 07 2019 *)

a(n)=if(n<2,(n%2)*4,10*a(n-1)-a(n-2)) \\ Benoit Cloitre, Sep 23 2006

G.f.: 4*x/(1 - 10*x + x^2). - Philippe Deléham, Nov 17 2008
3*a(n)^2 + 2 = 2*A001079(n)^2. - Charlie Marion, Feb 01 2013
a(n) = (2*arcsinh(sqrt(2))*sinh(2*n*arcsinh(sqrt(2)))/log(sqrt(2) + sqrt(3)))/sqrt(6). - Artur Jasinski, Aug 09 2016
a(n) = 2*A001078(n). - Bruno Berselli, Nov 25 2016
E.g.f.: sqrt(6)*exp(5*x)*sinh(2*sqrt(6)*x)/3. - Franck Maminirina Ramaharo, Jan 07 2019

More terms and better definition from Benoit Cloitre, Sep 23 2006

A138281 a(n) = floor((sqrt(2) + sqrt(3))^n).

Original entry on oeis.org

1, 3, 9, 31, 97, 308, 969, 3051, 9601, 30210, 95049, 299052, 940897, 2960313, 9313929, 29304086, 92198401, 290080547, 912670089, 2871501385, 9034502497, 28424933309, 89432354889, 281377831710, 885289046401, 2785353383794, 8763458109129, 27572156006234

Offset: 0

Reinhard Zumkeller, Mar 12 2008

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A089078, A110117, A121503, A135611, A135798.

[Floor((Sqrt(2) + Sqrt(3))^n): n in [0..50]]; // G. C. Greubel, Jan 27 2018

Table[Floor[(Sqrt[2] + Sqrt[3])^n], {n, 0, 50}] (* G. C. Greubel, Jan 27 2018 *)

for(n=0,50, print1(floor((sqrt(2) + sqrt(3))^n), ", ")) \\ G. C. Greubel, Jan 27 2018

a(2*n) = floor(A001079(n) + A001078(n)*sqrt(6));
(sqrt(2) + sqrt(3))^(2*n) = A001079(n) + A001078(n)*sqrt(6);
a(2*n+1) = floor(A054320(n)*sqrt(2) + A138288(n)*sqrt(3));
(sqrt(2)+sqrt(3))^(2*n+1) = A054320(n)*sqrt(2) + A138288(n)*sqrt(3).

Terms a(16) and a(18) corrected, terms a(19) onward added by G. C. Greubel, Jan 27 2018

A153111 Solutions of the Pell-like equation 1 + 6AA = 7BB, with A, B integers.

Original entry on oeis.org

1, 25, 649, 16849, 437425, 11356201, 294823801, 7654062625, 198710804449, 5158826853049, 133930787374825, 3477041644892401, 90269151979827601, 2343520909830625225, 60841274503616428249, 1579529616184196509249, 41006928746285492812225

Offset: 1

Ctibor O. Zizka, Dec 18 2008

B is of the form B(i) = 26*B(i-1) - B(i-2) for B(0) = 1, B(1) = 25 (this sequence).
A is of the form A(i) = 26*A(i-1) - A(i-2) for A(0) = 1, A(1) = 27.
In general a Pell-like equation of the form 1 + X*A*A = (X + 1)*B*B has the solution A(i) = (4*X + 2)*A(i-1) - A(i-2), for A(0) = 1 and A(1) = (4*X + 3), and B(i) = (4*X + 2)*B(i-1) - B(i-2) for B(0) = 1 and B(1) = (4*X + 1).
Examples in the OEIS:
  X = 1 gives A002315 for A(i) and A001653 for B(i);
  X = 2 gives A054320 for A(i) and A072256 for B(i);
  X = 3 gives A028230 for A(i) and A001570 for B(i);
  X = 4 gives A049629 for A(i) and A007805 for B(i);
  X = 5 gives A133283 for A(i) and A157014 for B(i);
  X = 6 gives A157461 for A(i) and this sequence for B(i).
Positive values of x (or y) satisfying x^2 - 26*x*y + y^2 + 24 = 0. - Colin Barker, Feb 20 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Luigi Cimmino, Algebraic relations for recursive sequences, arXiv:math/0510417 [math.NT], 2005-2008.
Jeroen Demeyer, Diophantine sets of polynomials over number fields, arXiv:0807.1970 [math.NT], 2008.
Franz Lemmermeyer, Conics - a Poor Man's Elliptic Curves, arXiv:math/0311306 [math.NT], 2003.
Index entries for linear recurrences with constant coefficients, signature (26,-1).

Cf. A002315, A001653, A054320, A072256, A001078, A028230, A001570, A049629, A007805, A133283, A140480.

Cf. similar sequences listed in A238379.

I:=[1,25]; [n le 2 select I[n] else 26*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 22 2014

CoefficientList[Series[(1 - x)/(x^2 - 26 x + 1), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 22 2014 *)
LinearRecurrence[{26, -1}, {1, 25}, 20] (* Jean-François Alcover, Jan 07 2019 *)

Vec(-x*(x-1)/(x^2-26*x+1) + O(x^100)) \\ Colin Barker, Feb 20 2014

a(n) = 26*a(n-1) - a(n-2). - Colin Barker, Feb 20 2014
G.f.: -x*(x - 1) / (x^2 - 26*x + 1). - Colin Barker, Feb 20 2014
a(n) = (1/14)*(7 - sqrt(42))*(1 + (13 + 2*sqrt(42))^(2*n - 1))/(13 + 2*sqrt(42))^(n - 1). - Bruno Berselli, Feb 25 2014
E.g.f.: (1/7)*(7*cosh(2*sqrt(42)*x) - sqrt(42)*sinh(2*sqrt(42)*x))*exp(13*x) - 1. - Franck Maminirina Ramaharo, Jan 07 2019

More terms from Philippe Deléham, Sep 19 2009; corrected by N. J. A. Sloane, Sep 20 2009

Additional term from Colin Barker, Feb 20 2014

cp's OEIS Frontend

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

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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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A041041 Denominators of continued fraction convergents to sqrt(26).

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A036353 Square pentagonal numbers.

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A084070 a(n) = 38*a(n-1) - a(n-2), with a(0)=0, a(1)=6.

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A045502 Numbers k such that 2k+1 and 3k+1 are squares.

A153111 Solutions of the Pell-like equation 1 + 6AA = 7BB, with A, B integers.