A188644 -id:A188644 - cp's OEIS frontend

A056220 a(n) = 2*n^2 - 1.

-1, 1, 7, 17, 31, 49, 71, 97, 127, 161, 199, 241, 287, 337, 391, 449, 511, 577, 647, 721, 799, 881, 967, 1057, 1151, 1249, 1351, 1457, 1567, 1681, 1799, 1921, 2047, 2177, 2311, 2449, 2591, 2737, 2887, 3041, 3199, 3361, 3527, 3697, 3871, 4049, 4231, 4417, 4607, 4801

Offset: 0

N. J. A. Sloane, Aug 06 2000

Image of squares (A000290) under "little Hankel" transform that sends [c_0, c_1, ...] to [d_0, d_1, ...] where d_n = c_n^2 - c_{n+1}*c_{n-1}. - Henry Bottomley, Dec 12 2000
Surround numbers of an n X n square. - Jason Earls, Apr 16 2001
Numbers n such that 2*n + 2 is a perfect square. - Cino Hilliard, Dec 18 2003, Juri-Stepan Gerasimov, Apr 09 2016
The sums of the consecutive integer sequences 2n^2 to 2(n+1)^2-1 are cubes, as 2n^2 + ... + 2(n+1)^2-1 = (1/2)(2(n+1)^2 - 1 - 2n^2 + 1)(2(n+1)^2 - 1 + 2n^2) = (2n+1)^3. E.g., 2+3+4+5+6+7 = 27 = 3^3, then 8+9+10+...+17 = 125 = 5^3. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 29 2005
X values (other than 0) of solutions to the equation 2*X^3 + 2*X^2 = Y^2. To find Y values: b(n) = 2n*(2*n^2 - 1). - Mohamed Bouhamida, Nov 06 2007
Average of the squares of two consecutive terms is also a square. In fact: (2*n^2 - 1)^2 + (2*(n+1)^2 - 1)^2 = 2*(2*n^2 + 2*n + 1)^2. - Matias Saucedo (solomatias(AT)yahoo.com.ar), Aug 18 2008
Equals row sums of triangle A143593 and binomial transform of [1, 6, 4, 0, 0, 0, ...] with n > 1. - Gary W. Adamson, Aug 26 2008
Start a spiral of square tiles. Trivially the first tile fits in a 1 X 1 square. 7 tiles fit in a 3 X 3 square, 17 tiles fit in a 5 X 5 square and so on. - Juhani Heino, Dec 13 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = coeff(charpoly(A,x),x^(n-2)). - Milan Janjic, Jan 26 2010
For each n > 0, the recursive series, formula S(b) = 6*S(b-1) - S(b-2) - 2*a(n) with S(0) = 4n^2-4n+1 and S(1) = 2n^2, has the property that every even term is a perfect square and every odd term is twice a perfect square. - Kenneth J Ramsey, Jul 18 2010
Fourth diagonal of A154685 for n > 2. - Vincenzo Librandi, Aug 07 2010
First integer of (2*n)^2 consecutive integers, where the last integer is 3 times the first + 1. As example, n = 2: term = 7; (2*n)^2 = 16; 7, 8, 9, ..., 20, 21, 22: 7*3 + 1 = 22. - Denis Borris, Nov 18 2012
Chebyshev polynomial of the first kind T(2,n). - Vincenzo Librandi, May 30 2014
For n > 0, number of possible positions of a 1 X 2 rectangle in a (n+1) X (n+2) rectangular integer lattice. - Andres Cicuttin, Apr 07 2016
This sequence also represents the best solution for Ripà's n_1 X n_2 X n_3 dots problem, for any 0 < n_1 = n_2 < n_3 = floor((3/2)*(n_1 - 1)) + 1. - Marco Ripà, Jul 23 2018

			a(0) = 0^2-1*1 = -1, a(1) = 1^2 - 4*0 = 1, a(2) = 2^2 - 9*1 = 7, etc.
a(4) = 31 = (1, 3, 3, 1) dot (1, 6, 4, 0) = (1 + 18 + 12 + 0). - _Gary W. Adamson_, Aug 29 2008

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq., Vol. 13 (2010), Article # 10.7.8.
Mitch Phillipson, Manda Riehl, and Tristan Williams, Enumeration of Wilf classes in Sn ~ Cr for two patterns of length 3, PU. M. A., Vol. 21, No. 2 (2010), pp. 321-338.
Marco Ripà, The rectangular spiral or the n1 X n2 X ... X nk Points Problem , Notes on Number Theory and Discrete Mathematics, Vol. 20, No. 1 (2014), pp. 59-71.
Leo Tavares, Illustration: Twin Squares
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000105, A000217, A000290, A000384, A001082, A001653, A001844, A002378, A002593, A003215, A005563, A028347, A036666, A046092, A047875, A062717, A069074, A077585, A087475, A119258, A143593, A154685, A162610, A188653, A225227.
Cf. A066049 (indices of prime terms)
Column 2 of array A188644 (starting at offset 1).

List([0..50], n-> 2*n^2-1); # Muniru A Asiru, Jul 24 2018

[2*n^2-1 : n in [0..50]]; // Vincenzo Librandi, May 30 2014

A056220:=n->2*n^2-1; seq(A056220(n), n=0..50); # Wesley Ivan Hurt, Jun 16 2014

Array[2 #^2 - 1 &, 50, 0] (* Robert G. Wilson v, Jul 23 2018 *)
CoefficientList[Series[(x^2 +4x -1)/(1-x)^3, {x, 0, 50}], x] (* or *)
LinearRecurrence[{3, -3, 1}, {-1, 1, 7}, 51] (* Robert G. Wilson v, Jul 24 2018 *)

PARI
```
a(n)=2*n^2-1;
```

[2*n^2-1 for n in (0..50)] # G. C. Greubel, Jul 07 2019

G.f.: (-1 + 4*x + x^2)/(1-x)^3. - Henry Bottomley, Dec 12 2000
a(n) = A119258(n+1,2) for n > 0. - Reinhard Zumkeller, May 11 2006
From Doug Bell, Mar 08 2009: (Start)
a(0) = -1,
a(n) = sqrt(A001844(n)^2 - A069074(n-1)),
a(n+1) = sqrt(A001844(n)^2 + A069074(n-1)) = sqrt(a(n)^2 + A069074(n-1)*2). (End)
a(n) + a(n+1) + 1 = (2n+1)^2. - Doug Bell, Mar 09 2009
a(n) = a(n-1) + 4*n - 2 (with a(0)=-1). - Vincenzo Librandi, Dec 25 2010
a(n) = A188653(2*n) for n > 0. - Reinhard Zumkeller, Apr 13 2011
a(n) = A162610(2*n-1,n) for n > 0. - Reinhard Zumkeller, Jan 19 2013
a(n) = ( Sum_{k=0..2} (C(n+k,3)-C(n+k-1,3))*(C(n+k,3)+C(n+k+1,3)) ) - (C(n+2,3)-C(n-1,3))*(C(n,3)+C(n+3,3)), for n > 3. - J. M. Bergot, Jun 16 2014
a(n) = j^2 + k^2 - 2 or 2*j*k if n >= 2 and j = n + sqrt(2)/2 and k = n - sqrt(2)/2. - Avi Friedlich, Mar 30 2015
a(n) = A002593(n)/n^2. - Bruce J. Nicholson, Apr 03 2017
a(n) = A000384(n) + n - 1. - Bruce J. Nicholson, Nov 12 2017
a(n)*a(n+k) + 2k^2 = m^2 (a perfect square), m = a(n) + (2n*k), for n>=1. - Ezhilarasu Velayutham, May 13 2019
From Amiram Eldar, Aug 10 2020: (Start)
Sum_{n>=1} 1/a(n) = 1/2 - sqrt(2)*Pi*cot(Pi/sqrt(2))/4.
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(2)*Pi*csc(Pi/sqrt(2))/4 - 1/2. (End)
From Amiram Eldar, Feb 04 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(2))*csc(Pi/sqrt(2)).
Product_{n>=2} (1 - 1/a(n)) = (Pi/(4*sqrt(2)))*csc(Pi/sqrt(2)). (End)
a(n) = A003215(n) - A000217(n-2)*2. - Leo Tavares, Jun 29 2021
Let T(n) = n*(n+1)/2. Then a(n)^2 = T(2n-2)*T(2n+1) + n^2. - Charlie Marion, Feb 12 2023
E.g.f.: exp(x)*(2*x^2 + 2*x - 1). - Stefano Spezia, Jul 08 2023

A011943 Numbers k such that any group of k consecutive integers has integral standard deviation (viz. A011944(k)).

Original entry on oeis.org

1, 7, 97, 1351, 18817, 262087, 3650401, 50843527, 708158977, 9863382151, 137379191137, 1913445293767, 26650854921601, 371198523608647, 5170128475599457, 72010600134783751, 1002978273411373057, 13969685227624439047, 194572614913330773601, 2710046923559006391367

Offset: 1

E. K. Lloyd

If k is in the sequence, then it has successor 7*k + 4*sqrt(3*(k^2 - 1)). - Lekraj Beedassy, Jun 28 2002
Chebyshev's polynomials T(n,x) evaluated at x=7.
a(n+1) give all (nontrivial) solutions of Pell equation a(n+1)^2 - 48*b(n+1)^2 = +1 with b(n+1)=A007655(n+2), n >= 0.
Also all solutions for x in Pell equation x^2 - 12*y^2 = 1. A011944 gives corresponding values for y. - Herbert Kociemba, Jun 05 2022
Also numbers x of the form 3j+1 such that x^2 = 3m^2+1. Also solutions of x in x^2 - 3*y^2 = 1 in A001075 if x = 3j+1, j=1,2,... - Cino Hilliard, Mar 05 2005
In addition to having integral standard deviation, these k consecutive integers also have integral mean. This question was posed by Jim Delany of Cal Poly in 1989. The solution appeared in the American Mathematical Monthly Vol. 97, No. 5, (May, 1990), pp. 432 as problem E3302. - Ronald S. Tiberio, Jun 23 2008
Lebl and Lichtblau give the formula a(d) = ((7+4*sqrt(3))^d + (7-4*sqrt(3))^d)/2 in Theorem 1.2(iii), p. 4. - Jonathan Vos Post, Aug 05 2008
In a (square pyramidal) pile of cannonballs, paint the ball at the top and the balls on 2 opposite sides of the base. The sequence, after its 1st term, gives the numbers of painted balls, k, in piles where the total number of balls is twice a square multiple of k. - Peter Munn, Feb 06 2025

P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238. - N. J. A. Sloane, Mar 03 2022

Robert Israel, Table of n, a(n) for n = 1..788
Jim Delany, Roger Douglass, Mike Breen and Roger B. Eggleton, Problem E 3302: Averaging to Integers, The American Mathematical Monthly, Vol. 97, No. 5 (May, 1990), p. 432.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Tanya Khovanova, Recursive Sequences
Jiri Lebl and Daniel Lichtblau, Uniqueness of certain polynomials constant on a hyperplane, arXiv:0808.0284 [math.CV], 2008-2010.
E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
Ronald S. Tiberio, Solution to Problem E 3302 [Broken link]
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (14,-1).

a(n) = A001075(2n).
Row 2 of array A188644
Cf. A000330, A007654, A007655, A011944, A102344.

I:=[1,7]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Apr 19 2015

seq(orthopoly[T](n,7), n = 0..50); # Robert Israel, Jun 02 2015
a := n -> (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4):
seq(simplify(a(n)), n=1..20); # Peter Luschny, Jul 26 2020

LinearRecurrence[{14,-1},{1,7},30] (* Harvey P. Dale, Dec 16 2013 *)
a[n_]:=1/2((7-4 Sqrt[3])^n+(7+4 Sqrt[3])^n); Table[a[n] // Simplify,{n,0,20}] (* Gerry Martens, May 30 2015 *)

PARI
```
a(n)=if(n<0,0,subst(poltchebi(n),x,7))
```

g(n) = forstep(x=1,n,3,y=(x^2-1)/3;if(issquare(y),print1(x","))) \\ Cino Hilliard, Mar 05 2005

a(n) = 14*a(n-1) - a(n-2).
a(n) = sqrt(12*A011944(n)^2 + 1).
a(n) ~ (1/2)*(2 + sqrt(3))^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = T(n, 7) = (S(n, 14)-S(n-2, 14))/2 = T(2*n, 2) with S(n, x) := U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 14)=A007655(n+2).
a(n) = ((7+4*sqrt(3))^n + (7-4*sqrt(3))^n)/2.
a(n) = sqrt(48*A007655(n+1)^2 + 1).
G.f.: (1-7*x)/(1-14*x+x^2).
a(n) = cosh(2n*arcsinh(sqrt(3))). - Herbert Kociemba, Apr 24 2008
a(n) = (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4). - Peter Luschny, Jul 26 2020
E.g.f.: exp(7*x)*cosh(4*sqrt(3)*x). - Stefano Spezia, Dec 12 2022

Better description from Lekraj Beedassy, Jun 27 2002
Chebyshev comments from Wolfdieter Lang, Nov 08 2002
More terms from Vincenzo Librandi, Apr 19 2015

A060626 Number of right triangles of a given area required to form successively larger squares.

Original entry on oeis.org

2, 14, 34, 62, 98, 142, 194, 254, 322, 398, 482, 574, 674, 782, 898, 1022, 1154, 1294, 1442, 1598, 1762, 1934, 2114, 2302, 2498, 2702, 2914, 3134, 3362, 3598, 3842, 4094, 4354, 4622, 4898, 5182, 5474, 5774, 6082, 6398, 6722, 7054, 7394, 7742, 8098, 8462, 8834, 9214

Offset: 0

Jason Earls, Apr 13 2001

a(n) = number of row of Pascal's triangle in which three consecutive entries appear in the ratio n : n+1 : n+2 (valid for n = 0 if you consider a position of -1 to have value 0). E.g., entries in the ratio 1:2:3 appear in row 14 (1001, 2002, 3003); entries in the ratio 2:3:4 appear in row 34 (927983760, 1391975640, 1855967520); and so on. (The position within the row is given by A091823). - Howard A. Landman, Mar 08 2004
a(n)*(a(n)+1) is an oblong number (Cf. A002378) with the property that the product with the oblong numbers n*(n+1) or (n+1)*(n+2) both are again oblong numbers. Example: For n=3 we have (62*63)*(3*4) = 216*217 and (62*63)*(4*5) = 279*280. - Herbert Kociemba, Apr 13 2008
For n > 0, Hermite polynomial H_2(n) = 4*n^2 - 2. - Vincenzo Librandi, Aug 07 2010
The identity (4*n^2-2)^2 - (n^2-1)*(4*n)^2 = 4 can be written as a(n+1)^2 - A132411(n+2)*A008586(n+2)^2 = 4. - Vincenzo Librandi, Jun 16 2014
Equivalently: positive integers k congruent to 2 mod 4 (A016825) such that k$ / (k/2+1)! is a square when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692, A349496 and A349766 for further information). Integers k multiple of 4 such that that k$ / (k/2+1)! is a square are in A035008. - Bernard Schott, Dec 05 2021

Harry J. Smith, Table of n, a(n) for n=0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000178, A002378, A007318, A008586, A016777, A035008, A056220, A091823.
Cf. A132411, A348692.
Twice Column 2 of array A188644.
Subsequence of A016825.
Equals disjoint union of A349496 and A349766.

for n from 0 to 80 do printf(`%d,`,4*n^2+8*n+2) od:

Table[4*n*(n + 2) + 2, {n, 0, 100}] (* Paolo Xausa, Aug 08 2024 *)

a(n) = { 4*n^2 + 8*n + 2 } \\ Harry J. Smith, Jul 08 2009

a(n) = 4*n^2 + 8*n + 2.
a(n) = 8*n + a(n-1) + 4 with n > 0, a(0)=2. - Vincenzo Librandi, Aug 07 2010
G.f.: 2*(1 + 4*x - x^2)/(1-x)^3. - Colin Barker, Jun 28 2012
a(n) = 4*(n+1)^2 - 2 = 2*A056220(n+1). - Bruce J. Nicholson, Aug 31 2017
a(n) + a(n-1) + (n-1)^2 = (3*n + 1)^2 = A016777(n)^2. - Ezhilarasu Velayutham, May 23 2019
From Elmo R. Oliveira, Oct 31 2024: (Start)
E.g.f.: 2*exp(x)*(1 + 6*x + 2*x^2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

More terms from James Sellers, Apr 14 2001

A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.

Original entry on oeis.org

1, 17, 577, 19601, 665857, 22619537, 768398401, 26102926097, 886731088897, 30122754096401, 1023286908188737, 34761632124320657, 1180872205318713601, 40114893348711941777, 1362725501650887306817, 46292552162781456490001

Offset: 0

Henry Bottomley, Aug 16 2000

The sequence satisfies the Pell equation a(n)^2 - 18 * A202299(n+1)^2 = 1. - Vincenzo Librandi, Dec 19 2011
Also numbers n such that n - 1 and 2*n + 2 are squares. - Arkadiusz Wesolowski, Mar 15 2015
And they, n - 1 and 2*n + 2, are the squares of A005319 and A003499. - Michel Marcus, Mar 15 2015
This sequence {a(n)} gives all the nonnegative integer solutions of the Pell equation a(n)^2 - 32*(3*A091761(n))^2 = +1. - Wolfdieter Lang, Mar 09 2019

			G.f. = 1 + 17*x + 577*x^2 + 19601*x^3 + 665857*x^4 + 22619537*x^5 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..600 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A001075, A001541, A001091, A001079, A023038, A011943, A001081, A023039, A001085 and note relationship with square triangular number sequences A001110 and A001109. A091761.

Row 3 of array A188644.

I:=[1, 17]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 18 2011

LinearRecurrence[{34,-1},{1,17},30] (* Vincenzo Librandi, Dec 18 2011 *)
a[ n_] := ChebyshevT[ 2 n, 3]; (* Michael Somos, May 28 2014 *)

makelist(expand(((17+sqrt(288))^n+(17-sqrt(288))^n))/2, n, 0, 15); /* Vincenzo Librandi, Dec 18 2011 */

{a(n) = polchebyshev( n, 1, 17)}; /* Michael Somos, Apr 05 2019 */

[lucas_number2(n,34,1)/2 for n in range(0,15)] # Zerinvary Lajos, Jun 27 2008

a(n) = (r^n + 1/r^n)/2 with r = 17 + sqrt(17^2-1).
a(n) = 16*A001110(n) + 1 = A001541(2n) = (4*A001109(n))^2 + 1 = 3*A001109(2n-1) - A001109(2n-2) = A001109(2n) - 3*A001109(2n-1).
a(n) = T(n, 17) = T(2*n, 3) with T(n, x) Chebyshev's polynomials of the first kind. See A053120. T(n, 3)= A001541(n).
G.f.: (1-17*x)/(1-34*x+x^2).
G.f.: (1 - 17*x / (1 - 288*x / (17 - x))). - Michael Somos, Apr 05 2019
a(n) = cosh(2n*arcsinh(sqrt(8))). - Herbert Kociemba, Apr 24 2008
a(n) = (a^n + b^n)/2 where a = 17 + 12*sqrt(2) and b = 17 - 12*sqrt(2); sqrt(a(n)-1)/4 = A001109(n). - James R. Buddenhagen, Dec 09 2011
a(-n) = a(n). - Michael Somos, May 28 2014
a(n) = sqrt(1 + 32*9*A091761(n)^2), n >= 0. See one of the Pell comments above. - Wolfdieter Lang, Mar 09 2019

More terms from James Sellers, Sep 07 2000

Chebyshev comments from Wolfdieter Lang, Nov 29 2002

A188645 Array of ((k^n)+(k^(-n)))/2 where k=(sqrt(x^2+1)+x)^2 for integers x>=1.

Original entry on oeis.org

1, 3, 1, 17, 9, 1, 99, 161, 19, 1, 577, 2889, 721, 33, 1, 3363, 51841, 27379, 2177, 51, 1, 19601, 930249, 1039681, 143649, 5201, 73, 1, 114243, 16692641, 39480499, 9478657, 530451, 10657, 99, 1, 665857, 299537289, 1499219281, 625447713, 54100801, 1555849, 19601, 129, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; and constant k=f(x, y); then for all integers x>=1 and y=[+-]1, k may be irrational, but ((k^n)+(k^(-n)))/2 always produces integer sequences; y=1 results shown here; y=-1 results are A188644.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{k}(x), evaluated at x=2*n^2+1. - Seiichi Manyama, Jan 01 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   3,     17,        99,          577, ...
   2 | 1,   9,    161,      2889,        51841, ...
   3 | 1,  19,    721,     27379,      1039681, ...
   4 | 1,  33,   2177,    143649,      9478657, ...
   5 | 1,  51,   5201,    530451,     54100801, ...
   6 | 1,  73,  10657,   1555849,    227143297, ...
   7 | 1,  99,  19601,   3880899,    768398401, ...
   8 | 1, 129,  33281,   8586369,   2215249921, ...
   9 | 1, 163,  53137,  17322499,   5647081537, ...
  10 | 1, 201,  80801,  32481801,  13057603201, ...
  11 | 1, 243, 118097,  57394899,  27893802817, ...
  12 | 1, 289, 167041,  96549409,  55805391361, ...
  13 | 1, 339, 229841, 155831859, 105653770561, ...
  14 | 1, 393, 308897, 242792649, 190834713217, ...
  15 | 1, 451, 406801, 366934051, 330974107201, ...
  ...

Row 1 is A001541, row 2 is A023039, row 3 is A078986, row 4 is A099370, row 5 is A099397, row 6 is A174747, row 8 is A176368, (row 1)*2 is A003499, (row 2)*2 is A087215.
Column 1 is A058331, (column 1)*2 is A005899.
A188644 (f(x, y) as above with y=-1).
Diagonal gives A173128.
Cf. A188647.

max = 9; y = 1; t = Table[k = ((x^2 + y)^(1/2) + x)^2; ((k^n) + (k^(-n)))/2 // FullSimplify, {n, 0, max - 1}, {x, 1, max}]; Table[ t[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)

A(n,k) = (A188647(n,k-1) + A188647(n,k))/2.

A(n,k) = Sum_{j=0..k} binomial(2*k,2*j)*(n^2+1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Edited and extended by Seiichi Manyama, Jan 01 2019

A188646 Array of a(n)=a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2-1)+x)^2 for integers x>=1.

Original entry on oeis.org

1, 1, 1, 1, 13, 1, 1, 181, 33, 1, 1, 2521, 1121, 61, 1, 1, 35113, 38081, 3781, 97, 1, 1, 489061, 1293633, 234361, 9505, 141, 1, 1, 6811741, 43945441, 14526601, 931393, 20021, 193, 1, 1, 94875313, 1492851361, 900414901, 91267009, 2842841, 37441, 253, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=-1 results shown here; y=1 results are A188647.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is (1/n) * T_{2*k+1}(n), with the Chebyshev polynomials of the first kind (type T). - Seiichi Manyama, Jan 01 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   1,      1,         1,            1, ...
   2 | 1,  13,    181,      2521,        35113, ...
   3 | 1,  33,   1121,     38081,      1293633, ...
   4 | 1,  61,   3781,    234361,     14526601, ...
   5 | 1,  97,   9505,    931393,     91267009, ...
   6 | 1, 141,  20021,   2842841,    403663401, ...
   7 | 1, 193,  37441,   7263361,   1409054593, ...
   8 | 1, 253,  64261,  16322041,   4145734153, ...
   9 | 1, 321, 103361,  33281921,  10716675201, ...
  10 | 1, 397, 158005,  62885593,  25028308009, ...
  11 | 1, 481, 231841, 111746881,  53861764801, ...
  12 | 1, 573, 328901, 188788601, 108364328073, ...
  13 | 1, 673, 453601, 305726401, 206059140673, ...
  14 | 1, 781, 610741, 477598681, 373481557801, ...
  15 | 1, 897, 805505, 723342593, 649560843009, ...
  ...

Row 1-8 give A000012, A001570, A077420, A302329, A302330, A302331, A302332, A253880.
Column 1 is A082109(n-1).
Cf. A188644, A188647 (f(x, y) as above with y=1).
Diagonal gives A322904.

A[n_, k_] := 1/n ChebyshevT[2k+1, n];
Table[A[n-k, k], {n, 1, 9}, {k, n-1, 0, -1}] // Flatten (* Jean-François Alcover, Jan 02 2019, after Seiichi Manyama *)

A(n,k) = 2 * A188644(n,k) - A(n,k-1).

A(n,k) = Sum_{j=0..k} binomial(2*k+1,2*j+1)*(n^2-1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Edited and extended by Seiichi Manyama, Jan 01 2019

A089775 Lucas numbers L(12n).

Original entry on oeis.org

2, 322, 103682, 33385282, 10749957122, 3461452808002, 1114577054219522, 358890350005878082, 115561578124838522882, 37210469265847998489922, 11981655542024930675232002, 3858055874062761829426214722, 1242282009792667284144565908482, 400010949097364802732720796316482

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 09 2004

a(n+1)/a(n) converges to (322 + sqrt(103680))/2 = 321.996894379... a(0)/a(1) = 2/322; a(1)/a(2) = 322/103682; a(2)/a(3) = 103682/33385282; a(3)/a(4) = 33385282/10749957122; etc. Lim_{n -> inf} a(n)/a(n+1) = 0.00310562... = 2/(322 + sqrt(103680)) = (322 - sqrt(103680))/2.

			a(4) = 10749957122 = 322*a(3) - a(2) = 322*33385282 - 103682 = ((322 + sqrt(103680))/2)^4 + ((322 - sqrt(103680))/2)^4.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.

Nathaniel Johnston, Table of n, a(n) for n = 0..100
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 (322, -1).

Cf. A000032, A060964.
a(n) = A000032(12n).
Row 9 * 2 of array A188644
Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11).

[ Lucas(12*n) : n in [0..70]]; // Vincenzo Librandi, Apr 15 2011

Table[LucasL[12n], {n, 0, 13}] (* Indranil Ghosh, Mar 15 2017 *)

Vec((2 - 322*x)/(1 - 322*x + x^2) + O(x^14)) \\ Indranil Ghosh, Mar 15 2017

a(n) = 322*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 322
a(n) = ((322 + sqrt(103680))/2)^n + ((322 - sqrt(103680))/2)^n.
(a(n))^2 = a(2n) + 2.
G.f.: (2-322*x)/(1-322*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^6 - 6*Lucas(2*n)^4 + 9*Lucas(2*n)^2 - 2 = 2*T(6, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
320*Sum_{n >= 1} 1/(a(n) - 324/a(n)) = 1: (324 = Lucas(12) + 2 and 320 = Lucas(12) - 2)
324*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 320/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (322 - sqrt(103680))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(103680) - 322)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. (End)

a(11) - a(13) from Vincenzo Librandi, Apr 15 2011

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

A175633 Numbers x such that x^2 - 28*y^2 = 1 for some integer y.

Original entry on oeis.org

1, 127, 32257, 8193151, 2081028097, 528572943487, 134255446617601, 34100354867927167, 8661355881006882817, 2199950293420880308351, 558778713173022591438337, 141927593195654317345029247

Offset: 1

Vincenzo Librandi, Dec 04 2010

nonn

This sequence gives the values of x in solutions of the Pell equation x^2 - 28*y^2 = 1; the corresponding y values are in A175672. [Edited by Jon E. Schoenfield, May 04 2014]

Index entries for linear recurrences with constant coefficients, signature (254,-1).

Cf. A175672.

Row 8 of array A188644.

a(n) = 254*a(n-1) - a(n-2) (with a(1)=1, a(2)=127).

G.f.: x*(1-127*x)/(1-254*x+x^2). - Bruno Berselli, Apr 18 2011

A322899 a(n) = T_{2*n}(n+1) where T_{n}(x) is a Chebyshev polynomial of the first kind.

Original entry on oeis.org

1, 7, 577, 119071, 46099201, 28860511751, 26650854921601, 34100354867927167, 57780789062419261441, 125283240358674708816199, 338393251269110482793304001, 1114259437504123772777608493087, 4394174409561746573589926449440001

Offset: 0

Seiichi Manyama, Dec 30 2018

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..193
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Diagonal of A188644.

a[n_] := ChebyshevT[2n, n+1];
Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Jan 02 2019 *)

PARI
```
{a(n) = polchebyshev(2*n, 1, n+1)}
```

{a(n) = sum(k=0, n, binomial(2*n, 2*k)*(n^2+2*n)^(n-k)*(n+1)^(2*k))}

a(n) = T_{n}(2*n^2+4*n+1).
a(n) = Sum_{k=0..n} binomial(2*n,2*k)*(n^2+2*n)^(n-k)*(n+1)^(2*k).
a(n) ~ exp(2) * 2^(2*n-1) * n^(2*n). - Vaclav Kotesovec, Apr 15 2020

cp's OEIS Frontend

A056220 a(n) = 2*n^2 - 1.

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A011943 Numbers k such that any group of k consecutive integers has integral standard deviation (viz. A011944(k)).

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A060626 Number of right triangles of a given area required to form successively larger squares.

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A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.

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A188645 Array of ((k^n)+(k^(-n)))/2 where k=(sqrt(x^2+1)+x)^2 for integers x>=1.

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A188646 Array of a(n)=a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2-1)+x)^2 for integers x>=1.

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