A011922 -id:A011922 - cp's OEIS frontend

A011916 a(n) = ((b(n)-1)+sqrt(3b(n)^2-4b(n)+1))/2, where b(n) is A011922(n).

0, 3, 44, 615, 8568, 119339, 1662180, 23151183, 322454384, 4491210195, 62554488348, 871271626679, 12135248285160, 169022204365563, 2354175612832724, 32789436375292575, 456697933641263328, 6360981634602394019

Offset: 0

Mario Velucchi (mathchess(AT)velucchi.it)

Integers k such that k^2 = Sum_{i=1..x} (k+i) for some value of x. 3 is a term because 3^2=9 and 4+5=9; 44 is a term because 44^2=1936 and the sum of (45,46,47,...,76) = 1936. - Gil Broussard, Dec 23 2008
Also the index of the first of two consecutive octagonal numbers whose sum is equal to the sum of two consecutive squares. - Colin Barker, Dec 20 2014
Also the index of a triangular number included in A239071. - Ivan Neretin, May 31 2015

Mario Velucchi, "Seeing couples" in Recreational and Educational Computing, to appear 1997. [apparently never materialized, Colin Barker, Dec 23 2014]

Colin Barker, Table of n, a(n) for n = 0..874
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A011918, A109437.

RecurrenceTable[{a[n] == 15 a[n - 1] - 15 a[n - 2] + a[n - 3], a[0] == 0, a[1] == 3, a[2] == 44}, a, {n, 0, 17}] (* Michael De Vlieger, Jul 02 2015 *)
LinearRecurrence[{15,-15,1},{0,3,44},30] (* Harvey P. Dale, Jul 26 2018 *)

{a(n) = if( n<0, n = -n; polcoeff( x*(1 - 3*x) / ((x-1) * (x^2 - 14*x + 1)) + x * O(x^n), n), polcoeff( x*(x - 3) / ((x-1) * (x^2 - 14*x + 1)) + x * O(x^n), n))} /* Michael Somos, Jul 27 2012 */

concat(0, Vec(x*(-3+x)/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, Dec 20 2014

From R. J. Mathar, Apr 15 2010: (Start)
a(n) = +15*a(n-1) -15*a(n-2) +a(n-3).
G.f.: x*(-3 + x) / ((x - 1)*(x^2 - 14*x + 1)). (End)
From Michael Somos, Jul 27 2012: (Start)
a(n) = A109437(2*n).
a(-1 - n) = -A109437(2*n + 1). (End)
a(n) = (A001353(n+1)^2 - A001075(n)^2)/4. - Richard R. Forberg, Aug 26 2013
a(n) = (-2-(7-4*sqrt(3))^n*(-1+sqrt(3))+(1+sqrt(3))*(7+4*sqrt(3))^n)/12. - Colin Barker, Mar 05 2016

More terms from R. J. Mathar, Apr 15 2010

Added a(0)=0, Michael Somos, Jul 27 2012

A011918 a(n) = A011916(n) + A011922(n) - 1.

Original entry on oeis.org

5, 76, 1065, 14840, 206701, 2878980, 40099025, 558507376, 7779004245, 108347552060, 1509086724601, 21018866592360, 292755045568445, 4077551771365876, 56792969753553825, 791024024778387680, 11017543377143873701, 153454583255235844140, 2137346622196157944265

Offset: 1

Mario Velucchi (mathchess(AT)velucchi.it)

Also integers n such that n^2+(n+1)^2 is equal to the sum of two consecutive octagonal numbers. - Colin Barker, Dec 20 2014

Mario Velucchi, "Seeing couples", Recreational and Educational Computing, to appear 1997. [apparently never materialized, Joerg Arndt, Aug 14 2013]

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A011916.

CoefficientList[Series[-(5 + x) / ((x - 1) (x^2 - 14 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 13 2013 *)

Vec(-x*(5+x)/((x-1)*(x^2-14*x+1)) + O(x^100)) \\ Colin Barker, Dec 20 2014

a(n) = +15*a(n-1) -15*a(n-2) +a(n-3). G.f.: -x*(5+x)/ ((x-1) * (x^2-14*x+1)). - R. J. Mathar, Apr 15 2010
a(n) = (A007655(n+2) - 3*A007655(n+1) - 1)/2. - Ralf Stephan, Aug 13 2013
a(n) = (-6-(7-4*sqrt(3))^n*(-3+sqrt(3))+(3+sqrt(3))*(7+4*sqrt(3))^n)/12. - Colin Barker, Mar 05 2016

More terms from R. J. Mathar, Apr 15 2010

A007655 Standard deviation of A007654.

Original entry on oeis.org

0, 1, 14, 195, 2716, 37829, 526890, 7338631, 102213944, 1423656585, 19828978246, 276182038859, 3846719565780, 53577891882061, 746243766783074, 10393834843080975, 144767444036350576, 2016350381665827089, 28084137899285228670, 391161580208327374291, 5448177985017298011404

Offset: 1

N. J. A. Sloane

a(n) corresponds also to one-sixth the area of Fleenor-Heronian triangle with middle side A003500(n). - Lekraj Beedassy, Jul 15 2002
a(n) give all (nontrivial, integer) solutions of Pell equation b(n+1)^2 - 48*a(n+1)^2 = +1 with b(n+1)=A011943(n), n>=0.
For n>=3, a(n) equals the permanent of the (n-2) X (n-2) tridiagonal matrix with 14's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,13}. - Milan Janjic, Jan 25 2015
6*a(n)^2 = 6*S(n-1, 14)^2 is the triangular number Tri((T(n, 7) - 1)/2) with Tri = A000217 and T = A053120. This is instance k = 3 of the general k-identity given in a comment to A001109. - Wolfdieter Lang, Feb 01 2016

			G.f. = x^2 + 14*x^3 + 195*x^4 + 2716*x^5 + 37829*x^6 + 526890*x^7 + ...

D. A. Benaron, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Indranil Ghosh, Table of n, a(n) for n = 1..874 (terms 1..100 from T. D. Noe)
R. Flórez, R. A. Higuita and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
D. S. Hale, 3165. Perfect Squares of the Form 48n^2+1, Math. Gaz., Oct. 1966, page 307.
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
Murray S. Klamkin, Perfect Squares of the Form (m^2 - 1)a_n^2 + t, Math. Mag., 1969, page 111.
E. K. Lloyd, The standard deviation of 1, 2, ..., n, Pell's equation and rational triangles, The Mathematical Gazette, Vol. 81, No. 491 (Jul., 1997), pp. 231-243.
Dino Lorenzini and Z. Xiang, Integral points on variable separated curves, Preprint 2016.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Cf. A000217, A001570, A003500, A011922, A011943, A011945, A028230, A046184, A049310, A053120, A055793, A067900, A098301, A101950, A103974.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), this sequence (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

m:=7;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

[n le 2 select n-1 else 14*Self(n-1)-Self(n-2): n in [1..70]]; // Vincenzo Librandi, Feb 02 2016

0,seq(orthopoly[U](n,7),n=0..30); # Robert Israel, Feb 04 2016

Table[GegenbauerC[n, 1, 7], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
LinearRecurrence[{14,-1}, {0,1}, 20] (* Vincenzo Librandi, Feb 02 2016 *)
ChebyshevU[Range[21] -2, 7] (* G. C. Greubel, Dec 23 2019 *)
Table[Sum[Binomial[n, 2 k - 1]*7^(n - 2 k + 1)*48^(k - 1), {k, 1, n}], {n, 0, 15}] (* Horst H. Manninger, Jan 16 2022 *)

concat(0, Vec((x^2/(1-14*x+x^2) + O(x^30)))) \\ Michel Marcus, Feb 02 2016

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

[lucas_number1(n,14,1) for n in range(0,20)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n,7) for n in (-1..20)] # G. C. Greubel, Dec 23 2019

a(n) = 14*a(n-1) - a(n-2).
G.f.: x^2/(1-14*x+x^2).
a(n+1) ~ 1/24*sqrt(3)*(2 + sqrt(3))^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n+1) = S(n-1, 14), n>=0, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
a(n+1) = ( (7+4*sqrt(3))^n - (7-4*sqrt(3))^n )/(8*sqrt(3)).
a(n+1) = sqrt((A011943(n)^2 - 1)/48), n>=0.
Chebyshev's polynomials U(n-2, x) evaluated at x=7.
a(n) = A001353(2n)/4. - Lekraj Beedassy, Jul 15 2002
4*a(n+1) + A046184(n) = A055793(n+2) + A098301(n+1) 4*a(n+1) + A098301(n+1) + A055793(n+2) = A046184(n+1) (4*a(n+1))^2 = A098301(2n+1) (conjectures). - Creighton Dement, Nov 02 2004
(4*a(n))^2 = A103974(n)^2 - A011922(n-1)^2. - Paul D. Hanna, Mar 06 2005
From Mohamed Bouhamida, May 26 2007: (Start)
a(n) = 13*( a(n-1) + a(n-2) ) - a(n-3).
a(n) = 15*( a(n-1) - a(n-2) ) + a(n-3). (End)
a(n) = b such that (-1)^n/4*Integral_{x=-Pi/2..Pi/2} (sin((2*n-2)*x))/(2-sin(x)) dx = c+b*log(3). - Francesco Daddi, Aug 02 2011
a(n+2) = Sum_{k=0..n} A101950(n,k)*13^k. - Philippe Deléham, Feb 10 2012
Product {n >= 1} (1 + 1/a(n)) = 1/3*(3 + 2*sqrt(3)). - Peter Bala, Dec 23 2012
Product {n >= 2} (1 - 1/a(n)) = 1/7*(3 + 2*sqrt(3)). - Peter Bala, Dec 23 2012
a(n) = (A028230(n) - A001570(n))/2. - Richard R. Forberg, Nov 14 2013
E.g.f.: 1 - exp(7*x)*(12*cosh(4*sqrt(3)*x) - 7*sqrt(3)*sinh(4*sqrt(3)*x))/12. - Stefano Spezia, Dec 11 2022

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

A103974 Smaller sides (a) in (a,a,a+1)-integer triangle with integer area.

Original entry on oeis.org

1, 5, 65, 901, 12545, 174725, 2433601, 33895685, 472105985, 6575588101, 91586127425, 1275630195845, 17767236614401, 247465682405765, 3446752317066305, 48007066756522501, 668652182274248705

Offset: 1

Zak Seidov, Feb 23 2005

Corresponding areas are: 0, 12, 1848, 351780, 68149872, 13219419708, 2564481115560 (see A104009).
What is the next term? Is the sequence finite? The possible last two digits of "a" are (it may help in searching for more terms): {01, 05, 09, 15, 19, 25, 29, 33, 35, 39, 45, 49, 51, 55, 59, 65, 69, 75, 79, 83, 85, 89, 95, 99}.
Equivalently, positive integers a such that 3/16*a^4 + 1/4*a^3 - 1/8*a^2 - 1/4*a - 1/16 is a square (A000290), a direct result of Heron's formula. Conjecture: lim_{n->oo} a(n+1)/a(n) = 7 + 4*sqrt(3) (= 7 + A010502). - Rick L. Shepherd, Sep 04 2005
Values x^2 + y^2, where the pair (x, y) solves for x^2 - 3y^2=1, i.e., a(n)= (A001075(n))^2 + (A001353(n))^2 = A055793(n) + A098301(n). - Lekraj Beedassy, Jul 13 2006
Floretion Algebra Multiplication Program, FAMP Code: 1lestes[ 3'i - 2'j + 'k + 3i' - 2j' + k' - 4'ii' - 3'jj' + 4'kk' - 'ij' - 'ji' + 3'jk' + 3'kj' + 4e ]

Christian Aebi, and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Project Euler, Problem 94: Almost Equilateral Triangles.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A011922, A007655, A001353, A102341, A103975, A016064, A011945, A010502 (4*sqrt(3)), A000290 (square numbers), A350916.

A:=rsolve({-A(n+3)+15*A(n+2)-15*A(n+1)+A(n), A(0) = 1, A(1) = 5, A(2)=65}, A(n), makeproc); # Mihailovs

f[n_] := Simplify[((2 + Sqrt[3])^(2n) + (2 - Sqrt[3])^(2n) + 1)/3]; Table[ f[n], {n, 0, 16}] (* Or *)
a[1] = 1; a[2] = 5; a[3] = 65; a[n_] := a[n] = 15a[n - 1] - 15a[n - 2] + a[n - 3]; Table[ a[n], {n, 17}] (* Or *)
CoefficientList[ Series[(1 - 10x + 5x^2)/(1 - 15x + 15x^2 - x^3), {x, 0, 16}], x] (* Or *)
Range[0, 16]! CoefficientList[ Simplify[ Series[(E^x + E^((7 + 4Sqrt[3])x) + E^((7 - 4Sqrt[3])x))/3, {x, 0, 16}]], x] (* Robert G. Wilson v, Mar 24 2005 *)

for(a=1,10^6, b=a; c=a+1; s=(a+b+c)/2; if(issquare(s*(s-a)*(s-b)*(s-c)), print1(a,","))) /* Uses Heron's formula */ \\ Rick L. Shepherd, Sep 04 2005

Composite of comments from Alec Mihailovs (alec(AT)mihailovs.com) and David Terr, Mar 07 2005: (Start)
"a(n)^2 = A011922(n)^2 + (4*A007655(n))^2, so that A011922(n) = 1/2 base of triangles, A007655(n) = 1/4 height of triangles (conjectured by Paul Hanna).
Area is (a+1)/4*sqrt((3*a+1)*(a-1)). If a is even, the numerator is odd and the area is not an integer. That means a=2*k-1. In this case, Area=k*sqrt((3*k-1)*(k-1)).
Solving equation (3*k-1)*(k-1)=y^2, we get k=(2+sqrt(1+3*y^2))/3. That means that 1+3*y^2=x^2 with integer x and y. This is a Pell equation, all solutions of which have the form x=((2+sqrt(3))^n+(2-sqrt(3))^n)/2, y=((2+sqrt(3))^n-(2-sqrt(3))^n)/(2*sqrt(3)). Therefore k=(x+2)/3 is an integer only for even n. Then a=2*k-1=(2*x+1)/3 with even n. Q.E.D.
a(n)=(1/3)*((2+sqrt(3))^(2*n-2)+(2-sqrt(3))^(2*n-2)+1).
Recurrence: a(n+3)=15*a(n+2)-15*a(n+1)+a(n), a(0)=1, a(1)=5, a(2)=65.
G.f.: x*(1-10*x+5*x^2)/(1-15*x+15*x^2-x^3).
E.g.f.: 1/3*(exp(x)+exp((7+4*sqrt(3))*x)+exp((7-4*sqrt(3))*x)).
a(n) = 4U(n)^2 + 1, where U(1) = 0, U(2)=1 and U(n+1) = 4U(n) - U(n-1) for n>1. (U(n), V(n)) is the n-th solution to Pell's equation 3U(n)^2 + 1 = V(n)^2. (U(n) is the sequence A001353.)" (End)
a(n+1) = A098301(n+1) + A055793(n+2) - Creighton Dement, Apr 18 2005
a(n) = floor((7+4*sqrt(3))*a(n-1))-4, n>=3. - Rick L. Shepherd, Sep 04 2005
a(n)= [1+14*A007655(n+2)-194*A007655(n+1)]/3. - R. J. Mathar, Nov 16 2007
For n>=3, a(n) = 14*a(n-1) - a(n-2) - 4. It is one of 10 second-order linear recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916. - Max Alekseyev, Jan 22 2022

More terms from Creighton Dement, Apr 18 2005

Edited by Max Alekseyev, Jan 22 2022

A045899 Numbers k such that k+1 and 3*k+1 are perfect squares.

Original entry on oeis.org

0, 8, 120, 1680, 23408, 326040, 4541160, 63250208, 880961760, 12270214440, 170902040408, 2380358351280, 33154114877520, 461777249934008, 6431727384198600, 89582406128846400, 1247721958419651008, 17378525011746267720, 242051628206028097080, 3371344269872647091408

Offset: 1

Andrej Dujella (duje(AT)math.hr)

Essentially the same as A051047.
It appears that a(n) = A046175(n)-A046174(n), that is, the triangular index of the n-th pentagonal triangular number minus its pentagonal index. - Jonathan Vos Post, Feb 28 2011
Sequence lists the nonnegative x solutions when (x + 1)*(3*x + 1) is a square. Positive x solutions when (x - 1)*(3*x - 1) is a square are in A011922. - Bruno Berselli, Feb 20 2018

G. C. Greubel, Table of n, a(n) for n = 1..870
A. Baker and H. Davenport, The Equations 3x^2-2=y^2 and 8x^2-7=z^2, Quart. J. Math. Oxford 20 (1969).
A. Dujella and A. Pethoe, A generalization of a theorem of Baker and Davenport, Quart. J. Math. Oxford Ser. (2) 49 (1998), 291-306.
A. Dujella, The Problem of Diophantus and Davenport, References
A. Dujella, Publications of Andrej Dujella
Z. Franusic, On the Extension of the Diophantine Pair {1,3} in Z[surd d], J. Int. Seq. 13 (2010) # 10.9.6
P. Gibbs, 1,3,8,120 ... A Diophantine Problem
P. Gibbs, Diophantine quadruples and Cayley's hyperdeterminant, arXiv:math/0107203 [math.NT], 2001.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A051047, A067900. A076139, A076140, A123480, A217855.

Cf. A046184, A245031.

f[n_] := FullSimplify[((Sqrt[3] + 2)*(7 + 4*Sqrt[3])^n - (Sqrt[3] - 2) (7 - 4 Sqrt[3])^n - 4)/6]; Array[f, 18, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006 *)
Rest[CoefficientList[Series[-8*x^2/((x - 1)*(x^2 - 14*x + 1)), {x,0,50}], x]] (* G. C. Greubel, Jun 07 2017 *)
LinearRecurrence[{15,-15,1},{0,8,120},20] (* Harvey P. Dale, Jul 14 2024 *)

x='x+O('x^50); concat([0], Vec(-8*x^2/((x - 1)*(x^2 - 14*x + 1)))) \\ G. C. Greubel, Jun 07 2017

a(n) = A046184(n+1) - 1.
a(n) = 14*a(n-1) - a(n-2) + 8.
a(n) = ((2 + sqrt(3))*(7 + 4*sqrt(3))^n + (2 - sqrt(3))*(7 - 4*sqrt(3))^n - 4)/6. - Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006
a(n) = 8*A076139(n-1) = 4*A217855(n-1) = 2*A123480(n-1) = 8/3*A076140(n-1). - Peter Bala, Dec 31 2012
From Colin Barker, Jul 30 2013: (Start)
G.f.: -8*x^2 / ((x - 1)*(x^2 - 14*x + 1)).
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). (End)
E.g.f.: (-4*exp(x) + (2 + sqrt(3))*exp((7-4*sqrt(3))*x) + (2 - sqrt(3))*exp((7+4*sqrt(3))*x))/6. - Ilya Gutkovskiy, Apr 28 2016

A120892 a(n)=3a(n-1)+3a(n-2)-a(n-3);a(0)=1,a(1)=0,a(2)=3. a(n)=4*{a(n-1)+(-1)^n}-a(n-2);a(0)=1,a(1)=0.

Original entry on oeis.org

1, 0, 3, 8, 33, 120, 451, 1680, 6273, 23408, 87363, 326040, 1216801, 4541160, 16947843, 63250208, 236052993, 880961760, 3287794051, 12270214440, 45793063713, 170902040408, 637815097923, 2380358351280, 8883618307201

Offset: 0

Lekraj Beedassy, Jul 13 2006

For n>1, short leg of primitive Pythagorean triangles having an angle nearing pi/3 with larger values of sides.[Complete triple (X,Y,Z),XA001353(n),Z=A120893(n), with recurrence relations Y(i+1)=2*{Y(i)-(-1)^i} + 3*a(i) ; Z(i+1)=2*{2*Z(i)-a(i-1)} - 3*(-1)^i] A120893(n)=2*a(n)-(-1)^n.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
J. P. Chabert, Right Triangle Applet (Hypotenuse & angles computation, given legs<350)
Index entries for linear recurrences with constant coefficients, signature (3, 3, -1).

LinearRecurrence[{3,3,-1},{1,0,3},30] (* Harvey P. Dale, Mar 05 2014 *)

a(n)=([0,1,0; 0,0,1; -1,3,3]^n*[1;0;3])[1,1] \\ Charles R Greathouse IV, Oct 19 2022

Union of A045899 and A011922.

O.g.f.: -(-1+3*x)/((x+1)*(x^2-4*x+1)). - R. J. Mathar, Nov 23 2007

Corrected and extended by T. D. Noe, Nov 07 2006

A227418 Array A(n,k) with all numbers m such that 3*m^2 +- 3^k is a square and their corresponding square roots, read by downward antidiagonals.

Original entry on oeis.org

0, 1, 1, 0, 2, 4, 3, 3, 7, 15, 0, 6, 12, 26, 56, 9, 9, 21, 45, 97, 209, 0, 18, 36, 78, 168, 362, 780, 27, 27, 63, 135, 291, 627, 1351, 2911, 0, 54, 108, 234, 504, 1086, 2340, 5042, 10864, 81, 81, 189, 405, 873, 1881, 4053, 8733, 18817, 40545

Offset: 0

Richard R. Forberg, Sep 02 2013

Array is analogous to A228405 in goal and structure, with key differences.
Left column is A001353.  Top row (not in OEIS) interleaves 0 with the powers of 3, as: 0, 1, 0, 3, 0, 9, 0, 27, 0, 81.
Either or both may be used as initializing values. See Formula section.
The left column is the second binomial transform of the top row. The intermediate transform sequence is A002605, not present in this array.
The columns of the array hold all values, in sequential order, of numbers m such that 3*m^2 + 3^k or 3*m^2 - 3^k are squares, and their corresponding square roots in the next column, which then form the "next round" of m values for column k+1.
For example: A(n,0) are numbers such that 3*m^2 + 1 are squares, the integer square roots of each are in A(n,1), which are then numbers m such that 3*m^2 - 3 are squares, with those square roots in A(n,2), etc.  The sign alternates for each increment of k, etc. No integer square roots exist for the opposite sign in a given column, regardless of n.
Also, A(n,1) are values of m such that floor(m^2/3) is square, with the corresponding square roots given by A(n,0).
A(n,k)/A(n,k-2) = 3; A(n,k)/A(n,k-1) converges to sqrt(3) for large n.
A(n,k)/A(n-1,k) converges to 2 + sqrt(3) for large n.
Several ways of combining the first few columns give OEIS sequences:
  A(n,0) + A(n,1) = A001835; A(n,1) + A(n,2)= A001834; A(n,2) + A(n,3) = A082841;
  A(n,0)*A(n,1)/2 = A007655(n); A(n+2,0)*A(n+1,1) = A001922(n);
  A(n,0)*A(n+1,1) = A001921(n); A(n,0)^2 + A(n,1)^2 = A103974(n);
  A(n,1)^2 - A(n,0)^2 = A011922(n); (A(n+2,0)^2 + A(n+1,1)^2)/2 = A122770(n) = 2*A011916(n).
The main diagonal (without initial 0) = 2*A090018.  The first subdiagonal =  abs(A099842). First superdiagonal = A141041.
A001353 (in left column) are the only initializing set of numbers where the recursive square root equation (see below) produces exclusively integer values, for all iterations of k.  For any other initial values only even iterations (at k = 2, 4, ...) produce integers.

			The array, A(n, k), begins as:
    0,    1,    0,    3,    0,     9,     0,    27, ... see A000244;
    1,    2,    3,    6,    9,    18,    27,    54, ... A038754;
    4,    7,   12,   21,   36,    63,   108,   189, ... A228879;
   15,   26,   45,   78,  135,   234,   405,   702, ...
   56,   97,  168,  291,  504,   873,  1512,  2619, ...
  209,  362,  627, 1086, 1881,  3258,  5643,  9774, ...
  780, 1351, 2340, 4053, 7020, 12159, 21060, 36477, ...
Antidiagonal triangle, T(n, k), begins as:
   0;
   1,  1;
   0,  2,   4;
   3,  3,   7,  15;
   0,  6,  12,  26,  56;
   9,  9,  21,  45,  97,  209;
   0, 18,  36,  78, 168,  362,  780;
  27, 27,  63, 135, 291,  627, 1351, 2911;
   0, 54, 108, 234, 504, 1086, 2340, 5042, 10864;
  81, 81, 189, 405, 873, 1881, 4053, 8733, 18817, 40545;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A001353, A001075, A005320, A038754, A090018, A099842, A141041, A151961.

Cf. A000244, A084156, A228879.

function A(n,k)
  if k lt 0 then return 0;
  elif n eq 0 then return Round((1/2)*(1-(-1)^k)*3^((k-1)/2));
  elif k eq 0 then return Evaluate(ChebyshevSecond(n), 2);
  else return 2*A(n, k-1) - A(n-1, k-1);
  end if; return A;
end function;
A227418:= func< n,k | A(k, n-k) >;
[A227418(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Oct 09 2022

A[n_, k_]:= If[k<0, 0, If[k==0, ChebyshevU[n-1, 2], 2*A[n, k-1] - A[n-1, k-1]]];
T[n_, k_]:= A[k, n-k];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 09 2022 *)

def A(n,k):
    if (k<0): return 0
    elif (k==0): return chebyshev_U(n-1,2)
    else: return 2*A(n, k-1) - A(n-1, k-1)
def A227418(n, k): return A(k, n-k)
flatten([[A227418(n,k) for k in range(n+1)] for n in range(15)]) # G. C. Greubel, Oct 09 2022

If using the left column and top row to initialize, then: A(n,k) = 2*A(n, k-1) - A(n-1, k-1).
If using only the top row to initialize, then: A(n,k) = 4*A(n-1,k) - A(n-2,k).
If using the left column to initialize, then: A(n,k) = sqrt(3*A(n,k-1) + (-3)^(k-1)), for all n, k > 0.
Other internal relationships that apply are: A(2*n-1, 2*k) = A(n,k)^2 - A(n-1,k)^2;
A(n+1,k) * A(n,k+1) - A(n+1, k+1) * A(n,k) = (-3)^k, for all n, k > 0.
A(n, 0) = A001353(n).
A(n, 1) = A001075(n).
A(n, 2) = A005320(n).
A(n, 3) = A151961(n).
A(1, k) = A038754(k).
A(n, n) = 2*A090018(n), for n > 0 (main diagonal).
A(n, n+1) = A141041(n-1) (superdiagonal).
A(n+1, n) = abs(A099842(n)) (subdiagonal).
From G. C. Greubel, Oct 09 2022: (Start)
T(n, 0) = (1/2)*(1-(-1)^n)*3^((n-1)/2).
T(n, 1) = A038754(n-1).
T(n, 2) = A228879(n-2).
T(2*n-1, n-1) = A141041(n-1).
T(2*n, n) = 2*A090018(n-1), n > 0.
T(n, n-4) = 3*A005320(n-4).
T(n, n-3) = 3*A001075(n-3).
T(n, n-2) = 3*A001353(n-2).
T(n, n-1) = A001075(n-1).
T(n, n) = A001353(n).
Sum_{k=0..n-1} T(n, k) = A084156(n).
Sum_{k=0..n} T(n, k) = A084156(n) + A001353(n). (End)

Offset corrected by G. C. Greubel, Oct 09 2022

A011920 a(n) = b(n)*(b(n)+1) = b(n) + ... + c(n), where b(n) = A011916(n), c(n) = A011918(n).

Original entry on oeis.org

12, 1980, 378840, 73419192, 14241916260, 2762844014580, 535977297450672, 103976830083273840, 20170969020163148220, 3913064012542622257452, 759114247456742016195720, 147264250942490855924510760

Offset: 1

Mario Velucchi (mathchess(AT)velucchi.it)

Mario Velucchi "Seeing couples" in Recreational and Educational Computing, to appear 1997.

Harvey P. Dale, Table of n, a(n) for n = 1..437
Index entries for linear recurrences with constant coefficients, signature (209,-2926,2926,-209,1).

A011922 := proc(n) (2+sqrt(1+((((2+sqrt(3))^(2*n)-(2-sqrt(3))^(2*n))^2)/4)))/3 ; expand(%) ; simplify(%) ; end proc:
A011916 := proc(n) ((A011922(n)-1)+sqrt(3*A011922(n)^2-4*A011922(n)+1))/2 ; end proc:
A011920 := proc(n) A011916(n)*(A011916(n)+1) ; end proc:
seq(A011920(n),n=1..20) ; # R. J. Mathar, Apr 15 2010

LinearRecurrence[{209,-2926,2926,-209,1},{12,1980,378840,73419192,14241916260},20] (* Harvey P. Dale, Jan 01 2021 *)

From R. J. Mathar, Apr 15 2010: (Start)
a(n) = +209*a(n-1) -2926*a(n-2) +2926*a(n-3) -209*a(n-4) +a(n-5).
G.f.: -12*x*(1-44*x+11*x^2)/ ((x-1) * (x^2-14*x+1) * (x^2-194*x+1)). (End)

More terms from R. J. Mathar, Apr 15 2010

cp's OEIS Frontend

A011916 a(n) = ((b(n)-1)+sqrt(3*b(n)^2-4*b(n)+1))/2, where b(n) is A011922(n).

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A011918 a(n) = A011916(n) + A011922(n) - 1.

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A007655 Standard deviation of A007654.

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A103974 Smaller sides (a) in (a,a,a+1)-integer triangle with integer area.

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A045899 Numbers k such that k+1 and 3*k+1 are perfect squares.

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A120892 a(n)=3*a(n-1)+3*a(n-2)-a(n-3);a(0)=1,a(1)=0,a(2)=3. a(n)=4*{a(n-1)+(-1)^n}-a(n-2);a(0)=1,a(1)=0.

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A011916 a(n) = ((b(n)-1)+sqrt(3b(n)^2-4b(n)+1))/2, where b(n) is A011922(n).

A120892 a(n)=3a(n-1)+3a(n-2)-a(n-3);a(0)=1,a(1)=0,a(2)=3. a(n)=4*{a(n-1)+(-1)^n}-a(n-2);a(0)=1,a(1)=0.