A103974 -id:A103974 - cp's OEIS frontend

A104009 Area of (a,a,a+1)-integer triangle. Corresponding a's are in A103974.

0, 12, 1848, 351780, 68149872, 13219419708, 2564481115560, 497495864091732, 96511629630137568, 18722758603319903340, 3632118656731075949592, 704612296637707504759428, 136691153428925957468727120, 26517379152913151616036727452, 5144234864511696770165553009288

Offset: 1

Zak Seidov, Feb 24 2005

Index entries for linear recurrences with constant coefficients, signature (208,-2718,208,-1).

Cf. A103974.

Block[{a, nmax = 14}, a[n_] := ((-(2 - Sqrt[3])^(2 n) + (2 + Sqrt[3])^( 2 n)) (4 + (2 - Sqrt[3])^(2 n) + (2 + Sqrt[3])^(2 n)))/( 12 Sqrt[3]); Expand[a /@ Range[0, nmax]]] (* Terentyev Oleg, Nov 15 2009 *)
LinearRecurrence[{208,-2718,208,-1},{0,12,1848,351780},20] (* Harvey P. Dale, Mar 02 2018 *)

s(n) = ((a+1)/4)*sqrt((3*a+1)*(a-1)) where a=A103974(n).
a(n+1) = (((2 + sqrt(3))^(2*n) - (2 - sqrt(3))^(2*n))*(4 + (2 - sqrt(3))^(2*n) + (2 + sqrt(3))^(2*n)))/(12*sqrt(3)). - Terentyev Oleg, Nov 15 2009
G.f.: 12*x^2*(x^2-54*x+1) / ((x^2-194*x+1)*(x^2-14*x+1)). - Colin Barker, Apr 10 2013

More terms from Rick L. Shepherd, Sep 05 2005

More terms from Colin Barker, Apr 10 2013

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

A016064 Smallest side lengths of almost-equilateral Heronian triangles (sides are consecutive positive integers, area is a nonnegative integer).

Original entry on oeis.org

1, 3, 13, 51, 193, 723, 2701, 10083, 37633, 140451, 524173, 1956243, 7300801, 27246963, 101687053, 379501251, 1416317953, 5285770563, 19726764301, 73621286643, 274758382273, 1025412242451, 3826890587533, 14282150107683, 53301709843201, 198924689265123, 742397047217293

Offset: 0

Robert G. Wilson v

Least side in a triangle with integer sides (m, m+1, m+2) (m >= 1) and integer area. The degenerate triangle with sides (1,2,3) is included.
Also describes triangles whose sides are consecutive integers and in which the inscribed circle has an integer radius. - Harvey P. Dale, Dec 28 2000 [Then, the length of this inradius is A001353(n). - Bernard Schott, Mar 21 2023]
Equivalently, positive integers m such that (3/16)*m^4 + (3/4)*m^3 + (3/8)*m^2 - (3/4)*m - 9/16 is a square (A000290), a direct result of Heron's formula. - Rick L. Shepherd, Sep 04 2005
"The problem is to find the sides of a triangle that shall have the values n, n + 1, and n + 2 and such that the perpendicular upon the longest side from the opposite vertex shall be rational. Nakane solves it as follows..." (Smith and Mikami, 2004). - Jonathan Sondow, May 09 2013
For n >= 1 all terms are congruent to {1,3} mod 10. Among first 100 terms there are 6 prime numbers: 3, 13, 193, 37633, 7300801, 1416317953. - Zak Seidov, Jun 14 2018
n > 1 is in this sequence if and only if the triangle with sides 4, n, n+2 has integer area (compare with A072221 for sides 3, n, n+1). - Michael Somos, May 11 2019
a(0) = 1 corresponds to the degenerate triangle [1,2,3], with area = 0. - Wesley Ivan Hurt, May 20 2020
Since this is a list it should really have offset 1, but that would require a large number of changes. - N. J. A. Sloane, Feb 04 2021
Least distance from centroid of a triangle to vertices, distances being m, m+1, m+2 and triangle area being a nonnegative integer. - Alexandru Petrescu, Feb 28 2023
Then, in this case, with a(n) = m, the corresponding area of this triangle is 3 * A011945(n+1). - Bernard Schott, Mar 21 2023

			G.f. = 1 + 3*x + 13*x^2 + 51*x^3 + 193*x^4 + 723*x^5 + 2701*x^6 + ... - _Michael Somos_, May 11 2019

Nakane Genkei (Nakane the Elder), Shichijo Beki Yenshiki, 1691.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000 (first 100 terms from Zak Seidov)
David Eugene Smith and Yoshio Mikami, A history of Japanese mathematics, Dover, 2004, p. 168.
Eric Weisstein's World of Mathematics, Heronian Triangle.
Wikipedia, Heronian triangle.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A011945 (areas), A334277 (perimeters) A001353 (inradius).
Cf. A003500 (middle side lengths), this sequence (smallest side lengths), A335025 (largest side lengths).
Cf. A001353, A019973 (2 + sqrt(3)), A102341, A103974, A103975.
Cf. A072221.

I:=[1,3,13]; [n le 3 select I[n] else 4*Self(n-1)-Self(n-2)+2: n in [1..30]]; // Vincenzo Librandi, Nov 13 2018

LinearRecurrence[{5,-5,1},{1,3,13},26] (* Ray Chandler, Jan 27 2014 *)
CoefficientList[Series[(1 - 2 x + 3 x^2) / (1 - 5 x + 5 x^2 - x^3), {x, 0, 33}], x] (* Vincenzo Librandi, Nov 13 2018 *)
a[ n_] := 2 ChebyshevT[n, 2] - 1; (* Michael Somos, May 11 2019 *)

for(a=1,10^9, b=a+1; c=a+2; s=(a+b+c)/2; if(issquare(s*(s-a)*(s-b)*(s-c)), print1(a,","))) \\ Rick L. Shepherd, Feb 18 2007

a(n)=if(n<1,1,-1+ceil((2+sqrt(3))^(n))) \\ Ralf Stephan

is(n)=issquare(3*n^2+6*n-9) \\ Charles R Greathouse IV, May 16 2014

{a(n) = 2 * polchebyshev(n, 1, 2) - 1}; /* Michael Somos, May 11 2019 */

a(n) = 3 + floor((2 + sqrt(3))*a(n-1)), n >= 3. - Rick L. Shepherd, Sep 04 2005
From Paul Barry, Feb 17 2004: (Start)
a(n) = 4*a(n-1) - a(n-2) + 2.
a(n) = (2 + sqrt(3))^n + (2 - sqrt(3))^n - 1.
a(n) = 2*A001075(n) - 1.
G.f.: (1 - 2*x + 3*x^3)/((1 - x)*(1 - 4*x + x^2)) = (1 - 2*x + 3*x^2)/(1 - 5*x + 5*x^2 - x^3). (End)
For n >= 1, a(n) = ceiling((2 + sqrt(3))^n) - 1.
a(n) = A003500(n) - 1. - T. D. Noe, Jun 17 2004
a(n) = [x^n] ( 1 + 2*x + sqrt(1 + 2*x + 3*x^2) )^n. - Peter Bala, Jun 23 2015
E.g.f.: exp((2 + sqrt(3))*x) + exp((2 - sqrt(3))*x) - exp(x). - Franck Maminirina Ramaharo, Nov 12 2018
a(n) = a(-n) for all integer n. - Michael Somos, May 11 2019

More terms from Rick L. Shepherd, Feb 18 2007

Definition revised by N. J. A. Sloane, Feb 04 2021

A011945 Areas of almost-equilateral Heronian triangles (integral side lengths m-1, m, m+1 and integral area).

Original entry on oeis.org

0, 6, 84, 1170, 16296, 226974, 3161340, 44031786, 613283664, 8541939510, 118973869476, 1657092233154, 23080317394680, 321467351292366, 4477462600698444, 62363009058485850, 868604664218103456, 12098102289994962534, 168504827395711372020, 2346969481249964245746

Offset: 1

E. K. Lloyd

Corresponding m's are in A016064. Corresponding values of lesser side give A016064.

Vincenzo Librandi, Table of n, a(n) for n = 1..890
Tanya Khovanova, Recursive Sequences
E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
P. Yiu, Heron triangles with consecutive sides, Recreational Mathematics, Chap. 9.3, pp. 80/360. (This is a download of 360 pages.)
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Equals 6 * A007655(n+1).
Cf. this sequence (areas), A334277 (perimeters).
Cf. A003500 (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).
Cf. A102341, A103974, A103975.

CoefficientList[Series[6 x/(1 - 14 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 15 2013 *)
LinearRecurrence[{14,-1},{0,6},20] (* Harvey P. Dale, Jan 24 2015 *)

s(n) = floor((a+1)/4)*sqrt(3*(a+3)*(a-1)), where a = A016064(n). - Zak Seidov, Feb 23 2005
a(n) = 14*a(n-1) - a(n-2); a(1) = 0, a(2) = 6.
G.f.: 6*x^2/(1 - 14*x + x^2). - Philippe Deléham, Nov 17 2008
a(n) = (s/4)*((7 + 4*s)^n - (7 - 4*s)^n), where s = sqrt(3). - Zak Seidov, Apr 02 2014
E.g.f.: 6 - exp(7*x)*(12*cosh(4*sqrt(3)*x) - 7*sqrt(3)*sinh(4*sqrt(3)*x))/2. - Stefano Spezia, Dec 12 2022

Entry revised by N. J. A. Sloane, Feb 03 2007

A350916 Positive integers k such that (k+1)^4 has a divisor congruent to -1 modulo k.

Original entry on oeis.org

1, 2, 3, 5, 9, 11, 14, 17, 29, 35, 41, 43, 59, 65, 69, 125, 134, 139, 174, 194, 339, 386, 449, 461, 681, 901, 937, 1169, 1322, 1325, 1715, 1971, 2211, 3054, 6395, 7989, 8857, 9077, 10849, 11483, 12545, 13082, 20909, 21506, 23861, 35233, 54734, 62210, 66923, 89045, 129494, 143289, 172899, 174725, 203321, 332315, 375129, 390051, 426389, 493697, 561513, 982094

Offset: 1

Max Alekseyev, Jan 21 2022

nonn

For (k+1)^3 similar sequence is finite {1, 2, 3, 5, 9, 11, 14}, while for (k+1)^2 it is just {1, 2, 3, 5}. Starting with power 4 (this sequence), the number of values of k is infinite. One series of values for power 6 is given by A001570.
Formed by the union of 10 linear recurrent sequences satisfying b(n) = q*b(n-1) - b(n-2) - 4: A350919 (q=3), A350920 (q=4), A350921 (q=6), A350922 (q=7), A350923 (q=10), A103974 (q=14), A350924 (q=16), A350925 (q=16), A350926 (q=23), A350917 (q=23). Each of them give identities (b(n)+1)^4 = (b(n)*b(n-1)-1) * (b(n)*b(n+1)-1).
Only terms 1, 2, 5, 9, 11, 14, 29 are shared between two or more sequences, all others come from exactly one sequence.

Max Alekseyev, Table of n, a(n) for n = 1..5000

Union of A350917, A350919, A350920, A350921, A350922, A350923, A103974, A350924, A350925, A350926.

Cf. A001570.

{ for(k=1,10^6, fordiv((k+1)^4,d, if(Mod(d,k)==-1, print1(k,", "); break)) ); }

A350923 a(0) = 2, a(1) = 2, and a(n) = 10*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

2, 2, 14, 134, 1322, 13082, 129494, 1281854, 12689042, 125608562, 1243396574, 12308357174, 121840175162, 1206093394442, 11939093769254, 118184844298094, 1169909349211682, 11580908647818722, 114639177128975534, 1134810862641936614, 11233469449290390602, 111199883630261969402

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Essentially the same as A157085. - R. J. Mathar, Feb 07 2022

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A031138, A054318, A097784, A253175, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350921, A350922, A350924, A350925, A350926.

LinearRecurrence[{11, -11, 1}, {2, 2, 14}, 25] (* Paolo Xausa, May 30 2025 *)

G.f.: 2*(1 - 10*x + 7*x^2)/((1 - x)*(1 - 10*x + x^2)). - Stefano Spezia, Jan 22 2022
From Hugo Pfoertner, Jan 22 2022: (Start)
a(n) = A031138(n) + 1.
a(n) = 3*A054318(n) - 1.
a(n) = 12*A097784(n-2) + 2 for n >= 2. (End)
a(n) = 2 * A253175(n) for n>=1. - Alois P. Heinz, Jan 22 2022

A350922 a(0) = 2, a(1) = 5, and a(n) = 7*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

2, 5, 29, 194, 1325, 9077, 62210, 426389, 2922509, 20031170, 137295677, 941038565, 6449974274, 44208781349, 303011495165, 2076871684802, 14235090298445, 97568760404309, 668746232531714, 4583654867317685, 31416837838692077, 215334210003526850, 1475922632185995869

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.
From William P. Orrick, Dec 20 2023: (Start)
Every term is a Markov number (see A002559) and, for n > 1, corresponds to a node of the Markov tree A368546 whose sibling and ancestors are all odd-indexed Fibonacci numbers. For n > 1, a(n) is the label of the node obtained from the root by going left n - 2 times and then right. Its Farey index, described in the comments to A368546, is 2 / (2*n - 1).
For instance, a(3) = 194 comes from going left once from the root node of the Markov tree and then right, which corresponds to the sequence of Markov numbers 5, 13, 194.  The corresponding sequence of Farey indices is 1/2, 1/3, 2/5. The sibling of the final node corresponds to Markov number 34 and Farey index 1/4. (End)

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A049684, A064170, A350916, A002559, A000045.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350921, A350923, A350924, A350925, A350926.

CoefficientList[Series[(2 - x)*(1 - 5*x)/((1 - x)*(1 - 7*x + x^2)), {x, 0, 22}],x] (* James C. McMahon, Dec 22 2023 *)

G.f.: (2 - x)*(1 - 5*x)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jan 22 2022
a(n) = 3*A049684(n) + 2 = 3*A064170(n+2) - 1. - Hugo Pfoertner, Jan 22 2022
a(n) = 3*A000045(2*n - 1) * A000045(2*n + 1) - 1 = A000045(2*n - 1)^2 + A000045(2*n + 1)^2. - William P. Orrick, Jan 08 2023

A011922 a(n) = 15a(n-1) - 15a(n-2) + a(n-3) with a(0)=1, a(1)=3, and a(2)=33.

Original entry on oeis.org

1, 3, 33, 451, 6273, 87363, 1216801, 16947843, 236052993, 3287794051, 45793063713, 637815097923, 8883618307201, 123732841202883, 1723376158533153, 24003533378261251, 334326091137124353, 4656561742541479683, 64857538304443591201, 903348974519668797123, 12582028104970919568513

Offset: 0

Mario Velucchi (mathchess(AT)velucchi.it)

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

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 31, 56.
Z. Franusic, On the Extension of the Diophantine Pair {1,3} in Z[surd d], J. Int. Seq. 13 (2010) # 10.9.6.
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001353, A001075, A007655, A011916, A011918, A011920, A011943, A103974.

I:=[1,3,33]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..17]]; // Bruno Berselli, Jul 09 2011

a:= gfun:-rectoproc({a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3), a(0)=1,a(1)=3,a(2)=33},a(n),remember):
map(a,[$0..100]); # Robert Israel, Jul 02 2015

RecurrenceTable[{a[n] == 15 a[n - 1] - 15 a[n - 2] + a[n - 3], a[0] == 1, a[1] == 3, a[2] == 33}, a, {n, 0, 15}] (* Michael De Vlieger, Jul 02 2015 *)
LinearRecurrence[{15,-15,1},{1,3,33},30] (* Harvey P. Dale, Dec 04 2018 *)

a[0]:1$ a[1]:3$ a[2]:33$ a[n]:=15*a[n-1]-15*a[n-2]+a[n-3]$ makelist(a[n], n, 0, 16); /* Bruno Berselli, Jul 09 2011 */

a(n)=([0,1,0; 0,0,1; 1,-15,15]^n*[1;3;33])[1,1] \\ Charles R Greathouse IV, Jul 02 2015

a(n) = (2+sqrt(1+((((2+sqrt(3))^(2*n)-(2-sqrt(3))^(2*n))^2)/4)))/3. [corrected by Francisco Salinas (franciscodesalinas(AT)hotmail.com), Dec 30 2001]
a(n) = ((7+4*sqrt(3))^n+(7-4*sqrt(3))^n+4)/6. - Bruno Berselli, Jul 09 2011
G.f.: (1-12*x+3*x^2)/ ((1-x) * (x^2-14*x+1)). - R. J. Mathar, Apr 15 2010
Sqrt(3) = 1 + Sum_{n>=1} 2/a(n) = 1 + 2/3 + 2/33 + ... - Gary W. Adamson, Jun 12 2003
a(n)^2 = A103974(n+1)^2 - (4*A007655(n+1))^2. - Paul D. Hanna, Mar 06 2005
a(n) = (A011943(n+1) + 2)/3. - Ralf Stephan, Aug 13 2013
a(n) = A001075(n)^2 - A001353(n)^2. - Richard R. Forberg, Aug 24 2013
E.g.f.: exp(x)*(2 + exp(6*x)*cosh(4*sqrt(3)*x))/3. - Stefano Spezia, Dec 11 2022

Recurrence in definition by R. J. Mathar, Apr 15 2010

A350917 a(0) = 1, a(1) = 2, and a(n) = 23*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

1, 2, 41, 937, 21506, 493697, 11333521, 260177282, 5972743961, 137112933817, 3147624733826, 72258255944177, 1658792261982241, 38079963769647362, 874180374439907081, 20068068648348215497, 460691398537569049346, 10575834097715739919457, 242783492848924449098161, 5573444501427546589338242, 127946440039984647105681401, 2937194676418219336841333977

Offset: 0

Max Alekseyev, Jan 21 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.
Other properties for all n:
(a(n)+1)*(a(n+2)+1) = (a(n+1)+1)*(a(n+1)+26);
((105*a(n) - 20)^2 - 50^2) / 21 is an integer square.

Sebastian et al., Are there infinitely many positive integer solutions to (3+3k+l)^2=m(kl-k^3-1)?, MathOverflow, 2022.
Index entries for linear recurrences with constant coefficients, signature (24,-24,1).

Cf. A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350919, A350920, A350921, A350922, A350923, A350924, A350925, A350926.

a(n) = 17/42*A090731(n) - 15/2*A097778(n-1) + 4/21.

G.f.: ( -1+22*x-17*x^2 ) / ( (x-1)*(x^2-23*x+1) ). - R. J. Mathar, Feb 07 2022

A350919 a(0) = 9, a(1) = 9, and a(n) = 3*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

9, 9, 14, 29, 69, 174, 449, 1169, 3054, 7989, 20909, 54734, 143289, 375129, 982094, 2571149, 6731349, 17622894, 46137329, 120789089, 316229934, 827900709, 2167472189, 5674515854, 14856075369, 38893710249, 101825055374, 266581455869, 697919312229, 1827176480814, 4783610130209, 12523653909809, 32787351599214, 85838400887829, 224727851064269

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A032908, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350920, A350921, A350922, A350923, A350924, A350925, A350926.

nxt[{a_,b_}]:={b,3b-a-4}; NestList[nxt,{9,9},40][[;;,1]] (* or *) LinearRecurrence[{4,-4,1},{9,9,14},40] (* Harvey P. Dale, Jul 19 2024 *)

a(n) = 5*A032908(n) - 1. - Hugo Pfoertner, Jan 22 2022
G.f.: (3 - 2*x)*(3 - 7*x)/((1 - x)*(1 - 3*x + x^2)). - Stefano Spezia, Jan 22 2022
a(n) = 5*A001519(n) +4. - R. J. Mathar, Feb 07 2022

cp's OEIS Frontend

A104009 Area of (a,a,a+1)-integer triangle. Corresponding a's are in A103974.

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A007655 Standard deviation of A007654.

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A016064 Smallest side lengths of almost-equilateral Heronian triangles (sides are consecutive positive integers, area is a nonnegative integer).

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A011945 Areas of almost-equilateral Heronian triangles (integral side lengths m-1, m, m+1 and integral area).

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A350916 Positive integers k such that (k+1)^4 has a divisor congruent to -1 modulo k.

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A350923 a(0) = 2, a(1) = 2, and a(n) = 10*a(n-1) - a(n-2) - 4 for n >= 2.

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A350922 a(0) = 2, a(1) = 5, and a(n) = 7*a(n-1) - a(n-2) - 4 for n >= 2.

A011922 a(n) = 15a(n-1) - 15a(n-2) + a(n-3) with a(0)=1, a(1)=3, and a(2)=33.