A052530 -id:A052530 - cp's OEIS frontend

A082630 Limit of the sequence obtained from S(0) = (1,1) and, for n > 0, S(n) = I(S(n-1)), where I consists of inserting, for i = 1, 2, 3..., the term a(i) + a(i+1) between any two terms for which 7a(i+1) <= 11a(i).

1, 2, 5, 8, 19, 30, 71, 112, 265, 418, 989, 1560, 3691, 5822, 13775, 21728, 51409, 81090, 191861, 302632, 716035, 1129438, 2672279, 4215120, 9973081, 15731042, 37220045, 58709048, 138907099, 219105150, 518408351, 817711552, 1934726305, 3051741058, 7220496869

Offset: 1

John W. Layman, May 23 2003

nonn

The bisection {1,5,19,265,...} appears to be A001834 and to satisfy the recurrence a(n) = 4*a(n-1) - a(n-2) and the condition that 3*a(n)^2 + 6 is a square. The other bisection {2,8,30,112,...} appears to be A052530 and one-half of this bisection, {1,4,15,56,...}, appears to be A001353 and to satisfy a(n) = 4*a(n-1) - a(n-2) and the condition that 3*a(n)^2 + 1 is a square.
Conjecturally, a(n) = x + y, where these values solve x^2 - floor(y^2/3) = 1, see related sequences and formula below. - Richard R. Forberg, Sep 08 2013
Let alpha be an algebraic integer and define a sequence of integers a(alpha,n) by the condition a(alpha,n) = max { integer d : alpha^n = = 1 (mod d)}. Silverman shows that a(alpha,n) is a strong-divisibility sequence, that is gcd(a(n), a(m)) = a(gcd(n, m)) for all n and m in N; in particular, if n divides m then a(n) divides a(m). This sequence appears to be the strong divisibility sequence a(2 + sqrt(3),n) (Silverman, Example 4). - Peter Bala, Jan 10 2014
This sequence appears as the coefficients of the defining inequalities of a polyhedral realization of the B(infinity) crystal of the Kac-Moody Lie algebra with Cartan matrix [2,-2;-3,2] (see Nakashima-Zelevinsky reference). - Paul E. Gunnells, May 05 2019
From Zhuorui He, Jul 16 2025: (Start)
This sequence is Ratio-determined insertion sequence I(7/11) (see the Layman link below).
If S(0) in the definition is (1,1,a,b,c...) (all numbers >= 0) instead of (1,1), the resulting sequence is still the same.
For a finite sequence S, let k be the least i such that 7*S(i+1) <= 11*S(i). If k didn't exist then I(S)=S. Else, let k' be the least i such that 7*I(S)(i+1) <= 11*I(S)(i). Then k <= k' <= k+1.
This sequence can be generated by this process:
Step 1: Let X=1 and Y=1.
Step 2: If 7*(X+Y)<=11*X, then Y:=X+Y, repeat this step. Else go to step 3.
Step 3: Append X to the sequence. Let X:=X+Y, go back to step 2. (End)

			Let S(0) = (1,1). Since 7*1 <= 11*1 we obtain S(1) = (1,2,1). Then since 7*2 > 11*1 and 7*1 <= 11*2, we obtain S(2) = (1,2,3,1). Continuing, we get S(3) = (1,2,5,3,4,1), S(4) = (1,2,5,8,3,7,4,5,1), S(5) = (1,2,5,8,11,3,...), S(6) =(1,2,5,8,19,11,...), etc.

Zhuorui He, Table of n, a(n) for n = 1..1000
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types [local copy, corrected]
Toshiki Nakashima and Andrei Zelevinsky, Polyhedral realizations of crystal bases for quantized Kac-Moody algebras, Adv. Math. 131 (1997), no. 1, 253-278.
J. H. Silverman, Divisibility sequences and powers of algebraic integers, Documenta Mathematica, Extra Volume: John H. Coates' Sixtieth Birthday (2006) 711-727.

Cf. A001353, A005246, A001834, A001835, A052530.

def A082630_list(n):
  a = []
  x = y = 1
  while len(a) < n:
    a.append(x)
    while 7*(x+y) <= 11*x:
      y += x
    x += y
  return a # Zhuorui He, Jul 16 2025

The sequence appears to satisfy a(n) = 4*a(n-2) - a(n-4).
Apparently a(n)*a(n+3) = -2 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
Conjecturally, a(n) = A143643(n-1) + A005246(n), for n => 2, as derived from comment above. - Richard R. Forberg, Sep 08 2013
If the above conjectures are true, then a(n) = A001353(n)/A005246(n+1). - Andrey Zabolotskiy, Sep 26 2024

Edited by M. F. Hasler, Nov 06 2018

A221255 T(n,k)=Number of nXk arrays of occupancy after each element moves to some horizontal or antidiagonal neighbor, with no occupancy greater than 2.

Original entry on oeis.org

0, 1, 0, 2, 4, 0, 4, 8, 16, 0, 8, 81, 30, 64, 0, 16, 392, 2116, 112, 256, 0, 32, 2500, 22268, 58081, 418, 1024, 0, 64, 14112, 488601, 1319409, 1612900, 1560, 4096, 0, 128, 83521, 7356420, 103489929, 78633756, 44903401, 5822, 16384, 0, 256, 484128, 132273001

Offset: 1

R. H. Hardin Jan 06 2013

Table starts
.0.....1.....2...........4..............8.................16
.0.....4.....8..........81............392...............2500
.0....16....30........2116..........22268.............488601
.0....64...112.......58081........1319409..........103489929
.0...256...418.....1612900.......78633756........22466712321
.0..1024..1560....44903401.....4684821704......4919350997521
.0..4096..5822..1250753956...278802326676...1080371161000000
.0.16384.21728.34842328921.16577412686597.237502965724417600
Even columns are perfect squares

			Some solutions for n=3 k=4
..1..2..1..0....0..2..2..0....0..2..2..1....0..2..2..1....1..1..0..1
..0..2..1..2....0..2..1..0....0..1..1..1....1..0..1..1....1..1..1..1
..2..0..1..0....2..1..1..1....2..1..1..0....1..1..1..1....1..1..2..1

R. H. Hardin, Table of n, a(n) for n = 1..126

Column 2 is A000302(n-1)

Column 3 is A052530

A061167 a(n) = n^5 - n.

Original entry on oeis.org

0, 0, 30, 240, 1020, 3120, 7770, 16800, 32760, 59040, 99990, 161040, 248820, 371280, 537810, 759360, 1048560, 1419840, 1889550, 2476080, 3199980, 4084080, 5153610, 6436320, 7962600, 9765600, 11881350, 14348880, 17210340, 20511120, 24299970, 28629120

Offset: 0

Henry Bottomley, Apr 18 2001

(b^2+c^2)/(bc+1) is an integer if {b,c} are of the form {0,n}, {n,n^3}, {n^3,n^5-n}, {n^5-n,n^7-2n^3}, {n^7-2n^3,n^9-3n^5+n}, etc. for some n, in which case the division results in n^2. Cf. A052530.

Convolution of A033429 by A033581. - R. J. Mathar, Aug 19 2008

			a(2) = 32 - 2 = 30.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. Zagier, Problems posed at the St Andrews Colloquium, 1996
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Subsequence of A249674.

Cf. A000584, A024002, A033429, A033455, A033581, A052530.

[n^5-n: n in [0..40]]; // Vincenzo Librandi, May 02 2011

Table[n^5 - n, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)

a(n) = 30*A033455(n-1). [Corrected by Bernard Schott, Mar 16 2021]
a(n) = -n*A024002(n).
a(n) = A000584(n) - n.
O.g.f.: 30x^2(1+x)^2/(1-x)^6. - R. J. Mathar, Aug 19 2008
a(n) = n * (n-1) * (n+1) * (n^2+1). - Bernard Schott, Mar 16 2021
E.g.f.: exp(x)*x^2*(15 + 25*x + 10*x^2 + x^3). - Stefano Spezia, Dec 27 2021

A065100 a(n+2) = 9*a(n+1) - a(n), a(0) = 3, a(1) = 27.

Original entry on oeis.org

3, 27, 240, 2133, 18957, 168480, 1497363, 13307787, 118272720, 1051146693, 9342047517, 83027280960, 737903481123, 6558104049147, 58285032961200, 518007192601653, 4603779700453677, 40916010111481440, 363640311302879283

Offset: 0

N. J. A. Sloane, Nov 12 2001

Original definition: a(0) = c, a(1) = p*c^3; a(n+2) = p*c^2*a(n+1) - a(n), for p = 1, c = 3.

The sequence could have started with a(0) = 0, then a(1) = 3 etc. Any of the family of sequences x = (0, c, c^3, c^5 - c, ...) with x(n+1) = c^2*x(n) - x(n-1), c >= 1, provides a subset of solutions to A115169 (for n >= 2), their union yields all the solutions. See A052530 for c = 2. - M. F. Hasler, Jun 12 2019

			From _Vincenzo Librandi_, Aug 07 2010: (Start)
a(2) = 9*27 - 3 = 240;
a(3) = 9*240 - 27 = 2133;
a(4) = 9*2133 - 240 = 18957. (End)

Harry J. Smith, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
J.-P. Ehrmann et al., Problem POLYA002, Integer pairs (x,y) for which (x^2+y^2)/(1+pxy) is an integer.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A052530 (analog for c = 2).

a[0] = c; a[1] = p*c^3; a[n_] := a[n] = p*c^2*a[n - 1] - a[n - 2]; p = 1; c = 3; Table[ a[n], {n, 0, 20} ]
LinearRecurrence[{9,-1},{3,27},30] (* Harvey P. Dale, Sep 22 2016 *)

polya002(1,3,20) \\ See A052530 for definition of function polya002().

{ p=1; c=3; k=p*c^2; for (n=0, 100, if (n>1, a=k*a1 - a2; a2=a1; a1=a, if (n, a=a1=k*c, a=a2=c)); write("b065100.txt", n, " ", a) ) } \\ Harry J. Smith, Oct 07 2009

G.f.: 3/(1 - 9*x + x^2). - Floor van Lamoen, Feb 07 2002
a(n) = 3*A018913(n+1). - R. J. Mathar, Oct 26 2009
a(n) = 9*a(n-1) - a(n-2) (with a(0)=3, a(1)=27). - Vincenzo Librandi, Aug 07 2010
E.g.f.: 3*exp(9*x/2)*(77*cosh(sqrt(77)*x/2) + 9*sqrt(77)*sinh(sqrt(77)*x/2))/77. - Stefano Spezia, Feb 23 2025

Definition simplified by M. F. Hasler, Jun 12 2019

A071954 a(n) = 4*a(n-1) - a(n-2) - 4, with a(0) = 2, a(1) = 4.

Original entry on oeis.org

2, 4, 10, 32, 114, 420, 1562, 5824, 21730, 81092, 302634, 1129440, 4215122, 15731044, 58709050, 219105152, 817711554, 3051741060, 11389252682, 42505269664, 158631825970, 592022034212, 2209456310874, 8245803209280, 30773756526242, 114849222895684

Offset: 0

Lekraj Beedassy, Jun 25 2002

a(n) gives the side of a cube having a square number of cubes in its two outermost layers, i.e., solutions p to the equation p^3 - (p - 4)^3 = q^2. The corresponding q is given by 4*A001075(n).

			G.f. = 2 + 4*x + 10*x^2 + 32*x^3 + 114*x^4 + 420*x^5 + 1562*x^6 + ...

M. E. Larsen, "Four Cubes" in Puzzler's Tribute, Ed. D. Wolfe & T. Rodgers, pp. 69-70, A. K. Peters, MA, 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..100 from Reinhard Zumkeller)
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Equals A052530(n) + 2, n > 0.

Cf. A001075, A003699, A072110, A263942.

a:=[2,4,10];; for n in [4..30] do a[n]:=5*a[n-1]-5*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Feb 25 2019

a071954 n = a071954_list !! n
a071954_list = 2 : 4 : zipWith (-)
               (map ((4 *) . pred) (tail a071954_list)) a071954_list
-- Reinhard Zumkeller, Aug 11 2011

I:=[2,4,10]; [n le 3 select I[n] else 5*Self(n-1) -5*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Feb 25 2019

a[n_]:= a[n] = 4*a[n-1] -a[n-2] -4; a[0]=2; a[1]=4; Table[a[n], {n,0,30}]
LinearRecurrence[{5,-5,1},{2,4,10},30] (* Harvey P. Dale, May 05 2011 *)

Vec((2-6*x)/(1-5*x+5*x^2-x^3)+O(x^30)) \\ Charles R Greathouse IV, Feb 09 2012

{a(n) = my(w=quadgen(12)); simplify( 2 + ((2+w)^n - (2-w)^n) / w)}; /* Michael Somos, Nov 03 2016 */

(2*(1-3*x)/((1-x)*(1-4*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) for n > 2, with a(0) = 2, a(1) = 4, a(2) = 10.
G.f.: 2*(1 - 3*x)/((1-x)*(1 -4*x +x^2)). - Harvey P. Dale, May 05 2011
a(n) = (2 + (-(2 - sqrt(3))^n + (2 + sqrt(3))^n)/sqrt(3)). - Colin Barker, Nov 03 2016
A263942(n) = -a(-1-n) for all n in Z. - Michael Somos, Nov 03 2016
E.g.f.: (2/3)*(3*exp(x) + sqrt(3)*exp(2*x)*sinh(sqrt(3)*x)). - Franck Maminirina Ramaharo, Nov 14 2018
From G. C. Greubel, Feb 25 2019: (Start)
a(n) = 2*A072110(n).
a(n) = 2*(1 - (-i)^(n+1)*F(n, 4*i)), where i=sqrt(-1) and F(n,x) is the Fibonacci polynomial. (End)

Edited by Robert G. Wilson v, Jun 27 2002

A060801 Invert transform of odd numbers: a(n) = Sum_{k=1..n} (2k+1)a(n-k), a(0)=1.

Original entry on oeis.org

1, 3, 14, 64, 292, 1332, 6076, 27716, 126428, 576708, 2630684, 12000004, 54738652, 249693252, 1138988956, 5195558276, 23699813468, 108107950788, 493140127004, 2249484733444, 10261143413212, 46806747599172, 213511451169436

Offset: 0

Vladeta Jovovic, Apr 27 2001

a(n) is the number of generalized compositions of n when there are 2*i+1 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010

Colin Barker, Table of n, a(n) for n = 0..1000
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, (an + b)-color compositions, arXiv:1707.07798 [math.CO], 2017.
Stéphane Ouvry and Alexios P. Polychronakos, Signed area enumeration for lattice walks, Séminaire Lotharingien de Combinatoire (2023) Vol. 87B.
N. J. A. Sloane, Transforms
Index entries for linear recurrences with constant coefficients, signature (5,-2).

Cf. A001906, A052530, A033453, A030017, A052913 (partial sums).

Join[{1}, LinearRecurrence[{5, -2}, {3, 14}, 22]] (* Jean-François Alcover, Aug 07 2018 *)

Vec((1 - x)^2 / (1 - 5*x + 2*x^2) + O(x^25)) \\ Colin Barker, Mar 19 2019

G.f.: (x^2-2*x+1)/(2*x^2-5*x+1).
G.f.: 1 / (1 - 3*x - 5*x^2 - 7*x^3 - 9*x^4 - 11*x^5 - ...). - Gary W. Adamson, Jul 27 2009
a(n) = 5*a(n-1) - 2*a(n-2) with a(1) = 3, a(2) = 14, for n >= 3. - Taras Goy, Mar 19 2019
a(n) = (2^(-2-n)*((5-sqrt(17))^n*(-7+sqrt(17)) + (5+sqrt(17))^n*(7+sqrt(17)))) / sqrt(17) for n > 0. - Colin Barker, Mar 19 2019
a(n) = A052913(n)-A052913(n-1). - R. J. Mathar, Sep 20 2020

A067902 a(n) = 14*a(n-1) - a(n-2); a(0) = 2, a(1) = 14.

Original entry on oeis.org

2, 14, 194, 2702, 37634, 524174, 7300802, 101687054, 1416317954, 19726764302, 274758382274, 3826890587534, 53301709843202, 742397047217294, 10340256951198914, 144021200269567502, 2005956546822746114, 27939370455248878094, 389145229826661547202, 5420093847118012782734

Offset: 0

Lekraj Beedassy, May 13 2003

Solves for x in x^2 - 3*y^2 = 4. [Complete nonnegative solutions are in A003500 and A052530. - Wolfdieter Lang, Sep 05 2021]
For n>0, a(n)+2 is the number of dimer tilings of a 4 X 2n Klein bottle (cf. A103999).
This is the Lucas sequence V(14,1). In addition to the comment above: If x = a(n) then y(n) = (a(n+1) - a(n-1))/24, n >= 1. - Klaus Purath, Aug 17 2021

			G.f. = 2 + 14*x + 194*x^2 + 2702*x^3 + 37634*x^4 + 524174*x^5 + ...

G. C. Greubel, Table of n, a(n) for n = 0..850
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (14,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A067900, A302332.
Row 2 * 2 of array A188644.
Cf. A001075, A001353, A003500, A052530.

m:=7;; a:=[2,14];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

[Floor((2+Sqrt(3))^(2*n)+(1+Sqrt(3))^(-n)): n in [0..19]]; // Vincenzo Librandi, Mar 31 2011

a := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(14) fi: 14*a(n-1)-a(n-2): end: for n from 0 to 30 do printf(`%d,`,a(n)) od:
seq( simplify(2*ChebyshevT(n, 7)), n=0..20); # G. C. Greubel, Dec 23 2019

a[0]=2; a[1]=14; a[n_]:= 14a[n-1] -a[n-2]; Table[a[n], {n,0,20}] (* Robert G. Wilson v, Jan 30 2004 *)
2*ChebyshevT[Range[21] -1, 7] (* G. C. Greubel, Dec 23 2019 *)

vector( 21, n, 2*polchebyshev(n-1, 1, 7) ) \\ G. C. Greubel, Dec 23 2019

[lucas_number2(n,14,1) for n in range(0,20)] # Zerinvary Lajos, Jun 26 2008

[2*chebyshev_T(n,7) for n in (0..20)] # G. C. Greubel, Dec 23 2019

G.f.: 2*(1-7*x)/(1-14*x+x^2). - N. J. A. Sloane, Nov 22 2006
a(n) = p^n + q^n, where p = 7 + 4*sqrt(3) and q = 7 - 4*sqrt(3). - Tanya Khovanova, Feb 06 2007
a(n) = 2*A011943(n+1). - R. J. Mathar, Sep 27 2014
From Peter Bala, Oct 16 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 7 - 4*sqrt(3). This sequence gives the partial denominators in the simple continued fraction expansion of 1 + F(alpha) = 2.07140228197873197080... = 2 + 1/(14 + 1/(194 + 1/(2702 + ...))). Cf. A005248.
12*Sum_{n >= 1} 1/(a(n) - 16/a(n)) = 1.
16*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 12/a(n)) = 1.
Series acceleration formula for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/12 - 16*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 16)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(7-4*sqrt(3)))^2 - 1 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.
(End)
From Klaus Purath, Aug 17 2021: (Start)
a(n) = (a(n-1)*a(n-2) + 2688)/a(n-3), n >= 3.
a(n) = (a(n-1)^2 + 192)/a(n-2), n >= 2.
a(2*n) = A302332(n-1) + A302332(n), n >= 1.
a(2*n+1) = 14*A302332(n). (End)
a(n) = A003500(2*n) = S(2*n,4) - S(2*n-2, 4) = 2*T(2*n,2), for n >= 0, with Chebyshev S and T. S(n, 4) = A001353(n+1) and T(n, 2) = A001075(n). - Wolfdieter Lang, Sep 06 2021

A103772 Larger of two sides in a (k,k,k-1)-integer-sided triangle with integer area.

Original entry on oeis.org

1, 17, 241, 3361, 46817, 652081, 9082321, 126500417, 1761923521, 24540428881, 341804080817, 4760716702561, 66308229755041, 923554499868017, 12863454768397201, 179164812257692801, 2495443916839302017, 34757050023492535441, 484103256412056194161

Offset: 1

Zak Seidov, Feb 23 2005

Corresponding areas are 0, 120, 25080, 4890480, 949077360, 184120982760, ...
Values of (x^2 + y^2)/2, where the pair (x, y) satisfies x^2 - 3*y^2 = -2, i.e., a(n) = {(A001834(n))^2 + (A001835(n))^2}/2 = {(A001834(n))^2 + A046184(n)}/2. - Lekraj Beedassy, Jul 13 2006
The heights of these triangles are given in A028230. (A028230(n), A045899(n), A103772(n)) forms a primitive Pythagorean triple.
Shortest side of (k,k+2,k+3) triangle such that median to longest side is integral. Sequence of such medians is A028230. - James R. Buddenhagen, Nov 22 2013
Numbers n such that (n+1)*(3n-1) is a square. - James R. Buddenhagen, Nov 22 2013

Colin Barker, Table of n, a(n) for n = 1..850
J. B. Cosgrave, The Gauss-Factorial Motzkin connection (Maple worksheet, change suffix to .mw)
J. B. Cosgrave and K. Dilcher, An Introduction to Gauss Factorials, The American Mathematical Monthly, 118 (Nov. 2011), 812-829.
Project Euler, Problem 94: Almost Equilateral Triangles.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A102341, A103974, A016064, A011945, A028230.

I:=[1,17]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+4: n in [1..20]]; // Vincenzo Librandi, Mar 05 2016

a[1] = 1; a[2] = 17; a[3] = 241; a[n_] := a[n] = 15a[n - 1] - 15a[n - 2] + a[n - 3]; Table[ a[n] - 1, {n, 17}] (* Robert G. Wilson v, Mar 24 2005 *)
LinearRecurrence[{15,-15,1},{1,17,241},20] (* Harvey P. Dale, Jan 02 2016 *)
RecurrenceTable[{a[1] == 1, a[2] == 17, a[n] == 14 a[n-1] - a[n-2] + 4}, a, {n, 20}] (* Vincenzo Librandi, Mar 05 2016 *)

Vec(x*(1+x)^2/((1-x)*(1-14*x+x^2)) + O(x^25)) \\ Colin Barker, Mar 05 2016

a(n) = (4*A001570(n+1) - 1)/3, n > 0. - Ralf Stephan, May 20 2007
a(n) = A052530(n-1)*A052530(n) + 1. - Johannes Boot, May 21 2011
G.f.: x*(1+x)^2/((1-x)*(1-14*x+x^2)). - Colin Barker, Apr 09 2012
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3); a(1)=1, a(2)=17, a(3)=241. - Harvey P. Dale, Jan 02 2016
a(n) = (-1+(7-4*sqrt(3))^n*(2+sqrt(3))-(-2+sqrt(3))*(7+4*sqrt(3))^n)/3. - Colin Barker, Mar 05 2016
a(n) = 14*a(n-1) - a(n-2) + 4. - Vincenzo Librandi, Mar 05 2016
a(n) = A001353(n)^2 + A001353(n-1)^2. - Antonio Alberto Olivares, Apr 06 2020

More terms from Robert G. Wilson v, Mar 24 2005

A210564 Triangle of coefficients of polynomials v(n,x) jointly generated with A210563; see the Formula section.

Original entry on oeis.org

1, 2, 3, 2, 7, 8, 2, 8, 22, 21, 2, 8, 29, 67, 55, 2, 8, 30, 101, 200, 144, 2, 8, 30, 111, 341, 588, 377, 2, 8, 30, 112, 404, 1122, 1708, 987, 2, 8, 30, 112, 417, 1442, 3613, 4913, 2584, 2, 8, 30, 112, 418, 1543, 5044, 11425, 14018, 6765, 2, 8, 30, 112, 418

Offset: 1

Clark Kimberling, Mar 23 2012

Last terms in rows: even-indexed Fibonacci numbers
Limiting row:  A052530
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
2...3
2...7...8
2...8...22...21
2...8...29...67...55
First three polynomials v(n,x): 1, 2 + 3x , 2 + 8x + 22x^2.

Cf. A210563, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*u[n - 1, x] + 2 x*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A210563 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210564 *)

u(n,x)=x*u(n-1,x)+x*v(n-1,x)+1,
v(n,x)=(x+1)*u(n-1,x)+2x*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A217233 Expansion of (1-2x+x^2)/(1-3x-3*x^2+x^3).

Original entry on oeis.org

1, 1, 7, 23, 89, 329, 1231, 4591, 17137, 63953, 238679, 890759, 3324361, 12406681, 46302367, 172802783, 644908769, 2406832289, 8982420391, 33522849271, 125108976697, 466913057513, 1742543253359, 6503259955919, 24270496570321, 90578726325361

Offset: 0

Bruno Berselli, Sep 28 2012

Numbers with the property a(n)^2+a(n-1)^2 = 2*(a(n)-a(n-1)-(-1)^n)^2.

			a(3)=23, a(2)=7: 23^2+7^2 = 2*(23-7-(-1)^3)^2 = 578;
a(6)=1231, a(5)=329: 1231^2+329^2 = 2*(1231-329-(-1)^6)^2 = 1623602.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Index entries for linear recurrences with constant coefficients, signature (3,3,-1).

Cf. A109437 (1/(1-3*x-3*x^2+x^3)), A006253 ((1-x)/(1-3*x-3*x^2+x^3)).

m:=26; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-2*x+x^2)/(1-3*x-3*x^2+x^3)));

CoefficientList[Series[(1 - 2 x + x^2)/(1 - 3 x - 3 x^2 + x^3), {x, 0, 25}], x]

makelist(coeff(taylor((1-2*x+x^2)/(1-3*x-3*x^2+x^3), x, 0, n), x, n), n, 0, 25);

Vec((1-2*x+x^2)/(1-3*x-3*x^2+x^3)+O(x^26))

G.f.: (1-x)^2/((1+x)*(1-4*x+x^2)).
a(n) = (4*(-2)^n+(1-sqrt(3))^(2*n+1)+(1+sqrt(3))^(2*n+1))/(6*2^n).
a(n) = -a(-n-1) = 3*a(n-1)+3*a(n-2)-a(n-3) = 4*a(n-1)-a(n-2)+4*(-1)^n.
a(n)+a(n-1) = A052530(n) with a(-1)=-1.
a(n)-a(n-2) = A003699(n) with n>1.
Sum(a(i), i=0..n) = A006253(n).

cp's OEIS Frontend

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A082630 Limit of the sequence obtained from S(0) = (1,1) and, for n > 0, S(n) = I(S(n-1)), where I consists of inserting, for i = 1, 2, 3..., the term a(i) + a(i+1) between any two terms for which 7a(i+1) <= 11a(i).

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