A011945 -id:A011945 - cp's OEIS frontend

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

2, 4, 14, 52, 194, 724, 2702, 10084, 37634, 140452, 524174, 1956244, 7300802, 27246964, 101687054, 379501252, 1416317954, 5285770564, 19726764302, 73621286644, 274758382274, 1025412242452, 3826890587534, 14282150107684, 53301709843202, 198924689265124

Offset: 0

N. J. A. Sloane

a(n) gives values of x satisfying x^2 - 3*y^2 = 4; corresponding y values are given by 2*A001353(n).
If M is any given term of the sequence, then the next one is 2*M + sqrt(3*M^2 - 12). - Lekraj Beedassy, Feb 18 2002
For n > 0, the three numbers a(n) - 1, a(n), and a(n) + 1 form a Fleenor-Heronian triangle, i.e., a Heronian triangle with consecutive sides, whose area A(n) may be obtained from the relation [4*A(n)]^2 = 3([a(2n)]^2 - 4); or A(n) = 3*A001353(2*n)/2 and whose semiperimeter is 3*a[n]/2. The sequence is symmetrical about a[0], i.e., a[-n] = a[n].
For n > 0, a(n) + 2 is the number of dimer tilings of a 2*n X 2 Klein bottle (cf. A103999).
Tsumura shows that, for prime p, a(p) is composite (contrary to a conjecture of Juricevic). - Charles R Greathouse IV, Apr 13 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 12 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 192 = 0. - Colin Barker, Feb 16 2014
A268281(n) - 1 is a member of this sequence iff A268281(n) is prime. - Frank M Jackson, Feb 27 2016
a(n) gives values of x satisfying 3*x^2 - 4*y^2 = 12; corresponding y values are given by A005320. - Sture Sjöstedt, Dec 19 2017
Middle side lengths of almost-equilateral Heronian triangles. - Wesley Ivan Hurt, May 20 2020
For all elements k of the sequence, 3*(k-2)*(k+2) is a square. - Davide Rotondo, Oct 25 2020

B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 82.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p.91.
Michael P. Cohen, Generating Heronian Triangles With Consecutive Integer Sides. Journal of Recreational Mathematics, vol. 30 no. 2 1999-2000 p. 123.
L. E. Dickson, History of The Theory of Numbers, Vol. 2 pp. 197;198;200;201. Chelsea NY.
Charles R. Fleenor, Heronian Triangles with Consecutive Integer Sides, Journal of Recreational Mathematics, Volume 28, no. 2 (1996-7) 113-115.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
V. D. To, "Finding All Fleenor-Heronian Triangles", Journal of Recreational Mathematics vol. 32 no.4 2003-4 pp. 298-301 Baywood NY.

T. D. Noe, Table of n, a(n) for n=0..200
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
R. A. Beauregard and E. R. Suryanarayan, The Brahmagupta Triangles, The College Mathematics Journal 29(1) 13-7 1998 MAA.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Daniel Birmajer, Juan B. Gil and Michael D. Weiner, Linear recurrence sequences with indices in arithmetic progression and their sums, arXiv preprint arXiv:1505.06339 [math.NT], 2015.
K. S. Brown's Mathpages, Some Properties of the Lucas Sequence(2, 4, 14, 52, 194, ...)
H. W. Gould, A triangle with integral sides and area, Fib. Quart., 11 (1973), 27-39.
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.
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Hideyuki Ohtskua, proposer, Problem B-1351, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 258.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Yu Tsumura, On compositeness of special types of integers, arXiv:1004.1244 [math.NT], 2010.
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001075, A001353, A001835.
Cf. A011945 (areas), A334277 (perimeters).
Cf. this sequence (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).
Cf. A001570, A002530, A005320, A006051, A048788, A174500, A268281.
Cf. A011943, A102344, A019673, A105531.

a003500 n = a003500_list !! n
a003500_list = 2 : 4 : zipWith (-)
   (map (* 4) $ tail a003500_list) a003500_list
-- Reinhard Zumkeller, Dec 17 2011

I:=[2,4]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2018

A003500 := proc(n) option remember; if n <= 1 then 2*n+2 else 4*procname(n-1)-procname(n-2); fi;
end proc;

a[0]=2; a[1]=4; a[n_]:= a[n]= 4a[n-1] -a[n-2]; Table[a[n], {n, 0, 23}]
LinearRecurrence[{4,-1},{2,4},30] (* Harvey P. Dale, Aug 20 2011 *)
Table[Round@LucasL[2n, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

x='x+O('x^99); Vec(-2*(-1+2*x)/(1-4*x+x^2)) \\ Altug Alkan, Apr 04 2016

[lucas_number2(n,4,1) for n in range(0, 24)] # Zerinvary Lajos, May 14 2009

a(n) = ( 2 + sqrt(3) )^n + ( 2 - sqrt(3) )^n.
a(n) = 2*A001075(n).
G.f.: 2*(1 - 2*x)/(1 - 4*x + x^2). Simon Plouffe in his 1992 dissertation.
a(n) = A001835(n) + A001835(n+1).
a(n) = trace of n-th power of the 2 X 2 matrix [1 2 / 1 3]. - Gary W. Adamson, Jun 30 2003 [corrected by Joerg Arndt, Jun 18 2020]
From the addition formula, a(n+m) = a(n)*a(m) - a(m-n), it is easy to derive multiplication formulas, such as: a(2*n) = (a(n))^2 - 2, a(3*n) = (a(n))^3 - 3*(a(n)), a(4*n) = (a(n))^4 - 4*(a(n))^2 + 2, a(5*n) = (a(n))^5 - 5*(a(n))^3 + 5*(a(n)), a(6*n) = (a(n))^6 - 6*(a(n))^4 + 9*(a(n))^2 - 2, etc. The absolute values of the coefficients in the expansions are given by the triangle A034807. - John Blythe Dobson, Nov 04 2007
a(n) = 2*A001353(n+1) - 4*A001353(n). - R. J. Mathar, Nov 16 2007
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n=0..infinity} (1 + x^(4*n + 1))/(1 + x^(4*n + 3)). Let alpha = 2 - sqrt(3). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.24561 99455 06551 88869 ... = 2 + 1/(4 + 1/(14 + 1/(52 + ...))). Cf. A174500.
Also F(-alpha) = 0.74544 81786 39692 68884 ... has the continued fraction representation 1 - 1/(4 - 1/(14 - 1/(52 - ...))) and the simple continued fraction expansion 1/(1 + 1/((4 - 2) + 1/(1 + 1/((14 - 2) + 1/(1 + 1/((52 - 2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((4^2 - 4) + 1/(1 + 1/((14^2 - 4) + 1/(1 + 1/((52^2 - 4) + 1/(1 + ...))))))).
(End)
a(2^n) = A003010(n). - John Blythe Dobson, Mar 10 2014
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 12*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
E.g.f.: 2*exp(2*x)*cosh(sqrt(3)*x). - Ilya Gutkovskiy, Apr 27 2016
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*n*(n - k - 1)!/(k!*(n - 2*k)!)*4^(n - 2*k) for n >= 1. - Peter Luschny, May 10 2016
From Peter Bala, Oct 15 2019: (Start)
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; -1, 4].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
2*Sum_{n >= 1} 1/( a(n) - 6/a(n) ) = 1.
6*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 2/a(n) ) = 1.
8*Sum_{n >= 1} 1/( a(n) + 24/(a(n) - 12/(a(n))) ) = 1.
8*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 8/(a(n) + 4/(a(n))) ) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/2 - 6*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 6)),
Sum_{n >= 1} 1/a(n) = 1/8 + 24*Sum_{n >= 1} 1/(a(n)*(a(n)^2 + 12)),
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/6 + 2*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 2)) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 12)).
Sum_{n >= 1} 1/a(n) = ( theta_3(2-sqrt(3))^2 - 1 )/4 = 0.34770 07561 66992 06261 .... See Borwein and Borwein, Proposition 3.5 (i), p.91.
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - theta_3(sqrt(3)-2)^2 )/4. Cf. A003499 and A153415. (End)
a(n) = tan(Pi/12)^n + tan(5*Pi/12)^n. - Greg Dresden, Oct 01 2020
From Wolfdieter Lang, Sep 06 2021: (Start)
a(n) = S(n, 4) - S(n-2, 4) = 2*T(n, 2), for n >= 0, with S and T Chebyshev polynomials, with S(-1, x) = 0 and S(-2, x) = -1. S(n, 4) = A001353(n+1), for n >= -1, and T(n, 2) = A001075(n).
a(2*k) = A067902(k), a(2*k+1) = 4*A001570(k+1), for k >= 0. (End)
a(n) = sqrt(2 + 2*A011943(n+1)) = sqrt(2 + 2*A102344(n+1)), n>0. - Ralf Steiner, Sep 23 2021
Sum_{n>=1} arctan(3/a(n)^2) = Pi/6 - arctan(1/3) = A019673 - A105531 (Ohtskua, 2024). - Amiram Eldar, Aug 29 2024

More terms from James Sellers, May 03 2000

Additional comments from Lekraj Beedassy, Feb 14 2002

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

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

A103975 Smaller side in (a,a+1,a+1)-integer triangle with integer area.

Original entry on oeis.org

16, 240, 3360, 46816, 652080, 9082320, 126500416, 1761923520, 24540428880, 341804080816, 4760716702560, 66308229755040, 923554499868016, 12863454768397200, 179164812257692800, 2495443916839302016, 34757050023492535440, 484103256412056194160

Offset: 1

Zak Seidov, Feb 23 2005

Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A102341, A103974, A016064, A011945, A103772.

Corresponding areas are given by A104008.

a[n_] := 1/3 (-4 + (2 - Sqrt[3])^(1 + 2 n) + (2 + Sqrt[3])^(1 + 2 n)); A103975 = Expand[a /@ Range[1, 25]] (* Terentyev Oleg, Nov 12 2009 *)
LinearRecurrence[{15,-15,1},{16,240,3360},30] (* Harvey P. Dale, Apr 25 2012 *)

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3) - Max Alekseyev, May 31 2007
a(n) = 2*A120892(2*n+1) - Max Alekseyev, May 31 2007
a(n) = (1/3)*((2 - sqrt(3))^(1 + 2*n) + (2 + sqrt(3))^(1 + 2*n) - 4). [Terentyev Oleg, Nov 12 2009]
a(n) = (4/3)*(A001570(n+1)-1).
G.f.: -16*x / ((x-1)*(x^2-14*x+1)). - Colin Barker, Apr 09 2013

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

More terms from Colin Barker, Apr 09 2013

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

A334277 Perimeters of almost-equilateral Heronian triangles.

Original entry on oeis.org

12, 42, 156, 582, 2172, 8106, 30252, 112902, 421356, 1572522, 5868732, 21902406, 81740892, 305061162, 1138503756, 4248953862, 15857311692, 59180292906, 220863859932, 824275146822, 3076236727356, 11480671762602, 42846450323052, 159905129529606, 596774067795372, 2227191141651882

Offset: 1

Wesley Ivan Hurt, May 20 2020

			a(1) = 12; there is one Heronian triangle with perimeter 12 whose side lengths are consecutive integers, [3,4,5].
a(2) = 42; there is one Heronian triangle with perimeter 42 whose side lengths are consecutive integers, [13,14,15].

Giovanni Resta, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
Wikipedia, Integer Triangle
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001075.
Cf. A011945 (areas), this sequence (perimeters).
Cf. A003500 (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).

Table[Expand[3 ((2 + Sqrt[3])^n + (2 - Sqrt[3])^n)], {n, 40}]

a(n) = 3*A003500(n).
a(n) = 3 * ((2 + sqrt(3))^n + (2 - sqrt(3))^n).
From Alejandro J. Becerra Jr., Jan 29 2021: (Start)
G.f.: -6*x*(x - 2)/(x^2 - 4*x + 1).
a(n) = 4*a(n-1) - a(n-2). (End)
a(n) = 6 * A001075(n). - Joerg Arndt, Jan 29 2021
E.g.f.: 6*(exp(2*x)*cosh(sqrt(3)*x) - 1). - Stefano Spezia, Jan 29 2021

A335025 Largest side lengths of almost-equilateral Heronian triangles.

Original entry on oeis.org

5, 15, 53, 195, 725, 2703, 10085, 37635, 140453, 524175, 1956245, 7300803, 27246965, 101687055, 379501253, 1416317955, 5285770565, 19726764303, 73621286645, 274758382275, 1025412242453, 3826890587535, 14282150107685, 53301709843203, 198924689265125, 742397047217295, 2770663499604053

Offset: 1

Wesley Ivan Hurt, May 20 2020

			a(1) = 5; there is one Heronian triangle with perimeter 12 whose side lengths are consecutive integers, [3,4,5] and 5 is the largest side length.
a(2) = 15; there is one Heronian triangle with perimeter 42 whose side lengths are consecutive integers, [13,14,15] and 15 is the largest side length.

Giovanni Resta, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
Wikipedia, Integer Triangle
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. a(n) = A003500(n) + 1.
Cf. A011945 (areas), A334277 (perimeters).
Cf. A003500 (middle side lengths), A016064 (smallest side lengths), this sequence (largest side lengths).

Table[Expand[(2 + Sqrt[3])^n + (2 - Sqrt[3])^n + 1], {n, 40}]

a(n) = (2 + sqrt(3))^n + (2 - sqrt(3))^n + 1.
From Alejandro J. Becerra Jr., Feb 12 2021: (Start)
G.f.: x*(3*x^2 - 10*x + 5)/((1 - x)*(x^2 - 4*x + 1)).
a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3). (End)

A144535 Numerators of continued fraction convergents to sqrt(3)/2.

Original entry on oeis.org

0, 1, 6, 13, 84, 181, 1170, 2521, 16296, 35113, 226974, 489061, 3161340, 6811741, 44031786, 94875313, 613283664, 1321442641, 8541939510, 18405321661, 118973869476, 256353060613, 1657092233154, 3570537526921, 23080317394680, 49731172316281, 321467351292366

Offset: 0

N. J. A. Sloane, Dec 29 2008

			0, 1, 6/7, 13/15, 84/97, 181/209, 1170/1351, 2521/2911, 16296/18817, 35113/40545, ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0,14,0,-1).

Cf. A126664, A144536, A002531/A002530.

Bisections give A001570, A011945.

I:=[0, 1, 6, 13]; [n le 4 select I[n] else 14*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Dec 10 2013

with(numtheory); Digits:=200: cf:=convert(evalf(sqrt(3)/2,confrac); [seq(nthconver(cf,i), i=0..100)];

CoefficientList[Series[x (1 + 6 x - x^2)/((1 - 4 x + x^2) (1 + 4 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 10 2013 *)
Numerator[Convergents[Sqrt[3]/2,30]] (* or *) LinearRecurrence[{0,14,0,-1},{0,1,6,13},30] (* Harvey P. Dale, Feb 10 2014 *)

Vec(x*(1+6*x-x^2)/((1-4*x+x^2)*(1+4*x+x^2)) + O(x^30)) \\ Colin Barker, Mar 27 2016

From Colin Barker, Apr 14 2012: (Start)
a(n) = 14*a(n-2) - a(n-4).
G.f.: x*(1 + 6*x - x^2)/((1 - 4*x + x^2)*(1 + 4*x + x^2)). (End)
a(n) = ((-(-2-sqrt(3))^n*(-3+sqrt(3)) + (2-sqrt(3))^n*(-3+sqrt(3)) - (3+sqrt(3))*((-2+sqrt(3))^n - (2+sqrt(3))^n)))/(8*sqrt(3)). - Colin Barker, Mar 27 2016
a(2*n) = 6*a(2*n-1) + a(2*n-2). a(2*n+1) = A003154(A101265(n+1)). - John Elias, Dec 10 2021

A175497 Numbers k with the property that k^2 is a product of two distinct triangular numbers.

Original entry on oeis.org

0, 6, 30, 35, 84, 180, 204, 210, 297, 330, 546, 840, 1170, 1189, 1224, 1710, 2310, 2940, 2970, 3036, 3230, 3900, 4914, 6090, 6930, 7134, 7140, 7245, 7440, 8976, 10710, 12654, 14175, 14820, 16296, 16380, 17220, 19866, 22770, 25172, 25944, 29103

Offset: 1

Zak Seidov, May 30 2010

nonn

From Robert G. Wilson v, Jul 24 2010: (Start)
Terms in the i-th row are products contributed with a factor A000217(i):
(1) 0, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, ...
(2) 30, 297, 2940, 29103, 288090, 2851797, 28229880, ...
(3) 84, 1170, 16296, 226974, 3161340, ...
(4) 180, 3230, 57960, 1040050, 18662940, ...
(5) 330, 7245, 159060, 3492075, 76666590, ...
(6) 546, 14175, 368004, 9553929, ...
(7) 840, 25172, 754320, 22604428, ...
(8) 210, 1224, 7134, 41580, 242346, 1412496, 8232630, 47983284, ...
(9) 1710, 64935, 2465820, 93636225, ...
(10) 2310, 96965, 4070220, ...
(11) 3036, 139590, 6418104, ...
(12) 3900, 194922, 9742200, ...
(13) 4914, 265265, 14319396, ...
(14) 6090, 353115, 20474580, ...
(15) 7440, 461160, 28584480, ...
(End)
Numbers m with property that m^2 is a product of two distinct triangular numbers T(i) and T(j) such that i and j are in the same row of the square array A(n, k) defined in A322699. - Onur Ozkan, Mar 17 2023

R. J. Mathar, Table of n, a(n) for n = 1..1000 replacing a b-file of _Robert G. Wilson v_ of 2010.
R. J. Mathar, OEIS A175497, Mar 16 2023
Project Euler, Problem 833. Square Triangle Products.
M. Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT], 2008. Theorem 5.6.

From Robert G. Wilson v, Jul 24 2010: (Start)
A001109 (with the exception of 1), A011945, A075848 and A055112 are all proper subsets.
Many terms are in common with A147779.
Cf. A001108, A132596, A007654, A001108, A132593. (End)
Cf. A152005 (two distinct tetrahedral numbers).

isA175497 := proc(n)
    local i,Ti,Tj;
    if n = 0 then
        return true;
    end if;
    for i from 1 do
        Ti := i*(i+1)/2 ;
        if Ti > n^2 then
            return false;
        else
            Tj := n^2/Ti ;
            if Tj <> Ti and type(Tj,'integer') then
                if isA000217(Tj) then  # code in A000217
                    return true;
                end if;
            end if;
        end if;
    end do:
end proc:
for n from 0 do
    if isA175497(n) then
        printf("%d,\n",n);
    end if;
end do: # R. J. Mathar, May 26 2016

triangularQ[n_] := IntegerQ[Sqrt[8n + 1]];
okQ[n_] := Module[{i, Ti, Tj}, If[n == 0, Return[True]]; For[i = 1, True, i++, Ti = i(i+1)/2; If[Ti > n^2, Return[False], Tj = n^2/Ti; If[Tj != Ti && IntegerQ[Tj], If[ triangularQ[Tj], Return[True]]]]]];
Reap[For[k = 0, k < 30000, k++, If[okQ[k], Print[k]; Sow[k]]]][[2, 1]] (* Jean-François Alcover, Jun 13 2023, after R. J. Mathar *)

from itertools import count, islice, takewhile
from sympy import divisors
from sympy.ntheory.primetest import is_square
def A175497_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda k:not k or any(map(lambda d: is_square((d<<3)+1) and is_square((k**2//d<<3)+1), takewhile(lambda d:d**2A175497_list = list(islice(A175497_gen(),20)) # Chai Wah Wu, Mar 13 2023

def A175497_list(n):
    def A322699_A(k, n):
        p, q, r, m = 0, k, 4*k*(k+1), 0
        while m < n:
            p, q, r = q, r, (4*k+3)*(r-q) + p
            m += 1
        return p
    def a(k, n, j):
        if n == 0: return 0
        p = A322699_A(k, n)*(A322699_A(k, n)+1)*(2*k+1) - a(k, n-1, 1)
        q = (4*k+2)*p - A322699_A(k, n)*(A322699_A(k, n)+1)//2
        m = 1
        while m < j: p, q = q, (4*k+2)*q - p; m += 1
        return p
    A = set([a(k, 1, 1) for k in range(n+1)])
    k, l, m = 1, 1, 2
    while True:
        x = a(k, l, m)
        if x < max(A):
            A |= {x}
            A  = set(sorted(A)[:n+1])
            m += 1
        else:
            if m == 1 and l == 1:
                if k > n:
                    return sorted(A)
                k += 1
            elif m > 1:
                l += 1; m = 1
            elif l > 1:
                k += 1; l, m = 1, 1
# Onur Ozkan, Mar 15 2023

a(n)^2 = A169836(n). - R. J. Mathar, Mar 12 2023

cp's OEIS Frontend

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

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

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A103975 Smaller side in (a,a+1,a+1)-integer triangle with integer area.

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