A003500 -id:A003500 - cp's OEIS frontend

A072330 Common difference n such that primitive triangles exist which are n-arithmetic (i.e., primitive Heronian triangles whose sides in arithmetic progression have common difference n).

1, 11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, 109, 121, 131, 143, 157, 167, 169, 179, 181, 191, 193, 227, 229, 239, 241, 251, 253, 263, 277, 299, 311, 313, 337, 347, 349, 359, 373, 383, 397, 407, 409, 419, 421, 431, 433, 443, 457, 467, 479, 481, 491, 503

Offset: 1

Lekraj Beedassy, Jul 15 2002

nonn

The first entry in particular is associated with sequences A003500 and A007655.

Such a triangle has a middle side 2*x partitioned into x +- 2*n by the corresponding altitude (i.e., median and altitude points are always a distance 2*n apart).

Frank M Jackson, Table of n, a(n) for n = 1..10000
R. A. Beauregard and E. R. Suryanarayan, Arithmetic Triangles, Mathematics Magazine, pp. 105-115 70(2) 1997 MAA.

Cf. A072360, A086909, A089019, A089020, A096672, A096673, A096674.

isA072330 := proc(n)
    if n = 1 then
        true;
    else
        for p in ifactors(n)[2] do
            if not modp(op(1,p),12) in {1,11} then
                return false ;
            end if;
        end do:
        true;
    end if;
end proc:
for n from 1 to 1000 do
    if isA072330(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Feb 26 2017

fac12Q[n_] := And @@ (MemberQ[{1, 11}, #] & /@ Mod[First /@ FactorInteger@ n, 12]); Select[Range[600], fac12Q] (* Frank M Jackson, Apr 09 2016 with simplification by Giovanni Resta *)
okQ[n_] := AllTrue[FactorInteger[n][[All, 1]], MatchQ[Mod[#, 12], 1|11]&];
Select[Range[1000], okQ] (* Jean-François Alcover, Mar 06 2020 *)

n = 1 or a product of primes p congruent to +- 1 (mod 12).

Corrected and extended by Ray Chandler, Jul 02 2004

Incorrect b-file by Carmine Suriano replaced by Frank M Jackson, May 09 2016

A103999 Square array T(M,N) read by antidiagonals: number of dimer tilings of a 2M x 2N Klein bottle.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 16, 34, 1, 1, 54, 196, 198, 1, 1, 196, 1666, 2704, 1154, 1, 1, 726, 16384, 64152, 37636, 6726, 1, 1, 2704, 171394, 1844164, 2549186, 524176, 39202, 1, 1, 10086, 1844164, 57523158, 220581904, 101757654, 7300804, 228486, 1

Offset: 0

Ralf Stephan, Feb 26 2005

			Array begins:
  1,   1,     1,        1,           1,             1,                1, ...
  1,   6,    34,      198,        1154,          6726,            39202, ...
  1,  16,   196,     2704,       37636,        524176,          7300804, ...
  1,  54,  1666,    64152,     2549186,     101757654,       4064620168, ...
  1, 196, 16384,  1844164,   220581904,   26743369156,    3252222705664, ...
  1, 726,171394, 57523158, 21050622914, 7902001927776, 2988827208115522, ...

Cliff, Danny and Zoe Stoll, About Klein bottles
W. T. Lu and F. Y. Wu, Dimer statistics on the Moebius strip and the Klein bottle, arXiv:cond-mat/9906154 [cond-mat.stat-mech], 1999.

Rows include A003499, A067902+2. Columns include A003500+2.
Main diagonal gives A340557.
Cf. A099390, A103997.

T[m_, n_] := Product[4 Sin[(4k-1) Pi/(4n)]^2 + 4 Cos[j Pi/(2m+1)]^2, {j, 1, m}, {k, 1, n}] // Round;
Table[T[m-n, n], {m, 0, 9}, {n, 0, m}] // Flatten (* Jean-François Alcover, Aug 20 2018 *)

default(realprecision, 120);
{T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*sin((4*a-1)*Pi/(4*n))^2+4*sin((2*b-1)*Pi/(2*k))^2)))} \\ Seiichi Manyama, Jan 11 2021

T(M, N) = Product_{m=1..M} Product_{n=1..N} ( 4sin(Pi*(4n-1)/(4N))^2 + 4sin(Pi*(2m-1)/(2M))^2 ).

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

A006051 Square hex numbers.

Original entry on oeis.org

1, 169, 32761, 6355441, 1232922769, 239180661721, 46399815451081, 9001325016847969, 1746210653453054881, 338755865444875798921, 65716891685652451935769, 12748738231151130799740241, 2473189499951633722697670961, 479786014252385791072548426169

Offset: 1

N. J. A. Sloane

Numbers n of the form n = y^2 = 3*x^2 - 3*x + 1.

			G.f. = x + 169*x^2 + 32761*x^3 + 6355441*x^4 + 1232922769*x^5 + ...

M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 19.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 1..435
M. Gardner & N. J. A. Sloane, Correspondence, 1973-74
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
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
Sociedad Magic Penny Patagonia, Leonardo en Patagonia
Eric Weisstein's World of Mathematics, Hex Number.
Index entries for linear recurrences with constant coefficients, signature (195,-195,1).

Cf. A003500.
Intersection of A000290 and A003215.
Values of x are given by A001922, values of y by A001570.

[(7*Evaluate(ChebyshevSecond(n),97) - 7*Evaluate(ChebyshevU(n-1), 97) + 1)/8: n in [1..30]]; // G. C. Greubel, Nov 04 2017; Oct 07 2022

Rest@ CoefficientList[Series[x(1-26x+x^2)/((1-x)(1-194x+x^2)), {x,0,20}], x] (* Michael De Vlieger, Jan 02 2017 *)
LinearRecurrence[{195,-195,1},{1,169,32761},20] (* Harvey P. Dale, Nov 03 2017 *)

{a(n) = sqr( real( (2 + quadgen( 12)) ^ (2*n - 1)) / 2)} /* Michael Somos, Feb 15 2011 */

def A006051(n): return (7*chebyshev_U(n-1,97) - 7*chebyshev_U(n-2,97) + 1)/8
[A006051(n) for n in range(1,31)] # G. C. Greubel, Oct 07 2022

a(n) = A001570(n)^2.
a(1 - n) = a(n).
G.f.: x * (1 - 26*x + x^2) / ((1 - x) * (1 - 194*x + x^2)). - Simon Plouffe in his 1992 dissertation
a(n) = 194*a(n-1) - a(n-2) - 24, a(1)=1, a(2)=169. - James Sellers, Jul 04 2000
a(n+1) = A003215(A001921(n)). - Joerg Arndt, Jan 02 2017
a(n) = (1/8)*(1 + 7*(ChebyshevU(n-1, 97) - ChebyshevU(n-2, 97))). - G. C. Greubel, Oct 07 2022

A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

Original entry on oeis.org

2, 10, 98, 970, 9602, 95050, 940898, 9313930, 92198402, 912670090, 9034502498, 89432354890, 885289046402, 8763458109130, 86749292044898, 858729462339850, 8500545331353602, 84146723851196170, 832966693180608098, 8245520207954884810

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 11 2003

a(n+1)/a(n) converges to (5+sqrt(24)) = 9.8989794... a(0)/a(1)=2/10; a(1)/a(2)=10/98; a(2)/a(3)=98/970; a(3)/a(4)=970/9602; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.10102051... = 1/(5+sqrt(24)) = (5-sqrt(24)).
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 96 = 0. - Colin Barker, Feb 25 2014
A triangle whose sides are a(n) - 1, a(n) and a(n) + 1 is nearly Fleenor-Heronian since its area is the product of an integer and the square root of 2. See A003500. - Charlie Marion, Dec 18 2020

			a(4) = 9602 = 10*a(3) - a(2) = 10*970 - 98 = (5+sqrt(24))^4 + (5-sqrt(24))^4.

T. D. Noe, Table of n, a(n) for n = 0..200
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
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 (10,-1).

Cf. A005248, A006242, A036336, A086927, A135927, A174503.

I:=[2,10]; [n le 2 select I[n] else 10*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Nov 07 2018

a[0] = 2; a[1] = 10; a[n_] := 10a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{10,-1}, {2,10}, 30] (* G. C. Greubel, Nov 07 2018 *)

polsym(x^2 - 10*x + 1,20) \\ Charles R Greathouse IV, Jun 11 2011

{a(n) = 2 * real( (5 + 2 * quadgen(24))^n )}; /* Michael Somos, Feb 25 2014 */

[lucas_number2(n,10,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.
G.f.: (2-10*x)/(1-10*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 5 - sqrt(24). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.09989 80642 72052 68138 ... = 2 + 1/(10 + 1/(98 + 1/(970 + ...))).
Also F(-alpha) = 0.89989 78538 78393 34715 ... has the continued fraction representation 1 - 1/(10 - 1/(98 - 1/(970 - ...))) and the simple continued fraction expansion 1/(1 + 1/((10-2) + 1/(1 + 1/((98-2) + 1/(1 + 1/((970-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((10^2-4) + 1/(1 + 1/((98^2-4) + 1/(1 + 1/((970^2-4) + 1/(1 + ...))))))). Cf. A174503 and A005248. (End)
a(-n) = a(n). - Michael Somos, Feb 25 2014
From Peter Bala, Oct 16 2019: (Start)
8*Sum_{n >= 1} 1/(a(n) - 12/a(n)) = 1.
12*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 8/a(n)) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/8 - 12*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 12))  and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/12 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 8)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(5-sqrt(24)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(sqrt(24)-5))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499. (End)
E.g.f.: 2*exp(5*x)*cosh(2*sqrt(6)*x). - Stefano Spezia, Oct 18 2019
From Peter Bala, Mar 29 2022: (Start)
a(n) = 2*T(n,5), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(2^n) = A135927(n+1) and a(3^n) = A006242(n+1), both for n >= 0. (End)

More terms from Colin Barker, Feb 25 2014

A082840 a(n) = 4*a(n-1) - a(n-2) + 3, with a(0) = -1, a(1) = 1.

Original entry on oeis.org

-1, 1, 8, 34, 131, 493, 1844, 6886, 25703, 95929, 358016, 1336138, 4986539, 18610021, 69453548, 259204174, 967363151, 3610248433, 13473630584, 50284273906, 187663465043, 700369586269, 2613814880036, 9754889933878, 36405744855479, 135868089488041

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Apr 14 2003

Apart from the initial -1, these are the numbers k such that the triangular number k*(k + 1)/2 is the sum of three consecutive triangular numbers - see A129803. - Brian Nowell, Nov 03 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001571, A071954.

a:=[-1,1,8];; for n in [4..30] do a[n]:=5*a[n-1]-5*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Feb 25 2019

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( -(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2)) )); // G. C. Greubel, Feb 25 2019

CoefficientList[Series[(-1+6x-2x^2)/((1-x)(1-4x+x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Apr 15 2014 *)
LinearRecurrence[{5,-5,1}, {-1,1,8}, 30] (* G. C. Greubel, Feb 25 2019 *)

is(n)=ispolygonal(3/2*n*(n+1)+4,3) || n==-1 \\ Charles R Greathouse IV, Apr 14 2014

my(x='x+O('x^30)); Vec(-(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2))) \\ G. C. Greubel, Feb 25 2019

(-(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

a(n) = A001571(n) - 1. - N. J. A. Sloane, Nov 03 2009
G.f.: -(1 -6*x +2*x^2)/((1 - x)*(1 - 4*x + x^2)).
a(n) = -3/2 + (1/12)*( (a -2*b +5)*a^n + (b -2*a +5)*b^n ), with a = 2 + sqrt(3), b = 2 - sqrt(3):.
a(n) = -3/2 + (3/4)*A003500(n) - (1/4)*A003500(n-1).
a(n) = (1/2)*(A001834(n) - 3).
E.g.f.: ((1 + sqrt(3))*exp((2 + sqrt(3))*x) + (1 - sqrt(3))*exp((2 - sqrt(3))*x) - 6*exp(x))/4. - Franck Maminirina Ramaharo, Nov 12 2018

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)

A236330 Positive integers n such that x^2 - 14xy + y^2 + n = 0 has integer solutions.

Original entry on oeis.org

32, 48, 128, 176, 192, 288, 368, 416, 432, 512, 624, 704, 752, 768, 800, 944, 1056, 1136, 1152, 1184, 1200, 1328, 1472, 1568, 1584, 1664, 1712, 1728, 1776, 1952, 2048, 2096, 2208, 2288, 2336, 2352, 2496, 2592, 2672, 2816, 2864, 2928, 3008, 3056, 3072, 3104

Offset: 1

Colin Barker, Feb 16 2014

nonn

			48 is in the sequence because x^2 - 14xy + y^2 + 48 = 0 has integer solutions, for example (x, y) = (2, 26).

Cf. A001835 (n = 32), A001075 (n = 48), A237250 (n = 176), A003500 (n = 192), A082841 (n = 288), A151961 (n = 432), A077238 (n = 624).

Cf. A031363, A084917, A237351, A237599, A237606, A237609, A237610, A236331.

A296795 Numbers k such that m = 2*k is the middle side in a Heronian triangle with sides m - 11, m, m + 11.

Original entry on oeis.org

13, 14, 22, 38, 43, 77, 139, 158, 286, 518, 589, 1067, 1933, 2198, 3982, 7214, 8203, 14861, 26923, 30614, 55462, 100478, 114253, 206987, 374989, 426398, 772486, 1399478, 1591339, 2882957, 5222923, 5938958, 10759342, 19492214, 22164493, 40154411, 72745933

Offset: 0

Sture Sjöstedt, Dec 20 2017

a(n) gives values of x satisfying 3*x^2 - y^2 = 363; the corresponding y values are given by A296796.

			The smallest triangle of this type has following sides: 15, 26, 37 has the altitude 12 and is a part of the Pythagorean triangle with sides : 12, 35, 37.

Colin Barker, Table of n, a(n) for n = 0..1000
Wikipedia, Heronian triangle
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,-1).

Cf. A003500, A005320, A296796.

CoefficientList[Series[(13 + 14 x + 22 x^2 - 14 x^3 - 13 x^4 - 11 x^5)/(1 - 4 x^3 + x^6), {x, 0, 36}], x] (* Michael De Vlieger, Dec 22 2017 *)

Vec((13 + 14*x + 22*x^2 - 14*x^3 - 13*x^4 - 11*x^5) / (1 - 4*x^3 + x^6) + O(x^40)) \\ Colin Barker, Dec 22 2017

From Colin Barker, Dec 22 2017: (Start)
G.f.: (13 + 14*x + 22*x^2 - 14*x^3 - 13*x^4 - 11*x^5) / (1 - 4*x^3 + x^6).
a(n) = 4*a(n-3) - a(n-6) for n>5.
(End)

More terms from Colin Barker, Dec 22 2017

cp's OEIS Frontend

A072330 Common difference n such that primitive triangles exist which are n-arithmetic (i.e., primitive Heronian triangles whose sides in arithmetic progression have common difference n).

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A103999 Square array T(M,N) read by antidiagonals: number of dimer tilings of a 2M x 2N Klein bottle.

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

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A006051 Square hex numbers.

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A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

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A082840 a(n) = 4*a(n-1) - a(n-2) + 3, with a(0) = -1, a(1) = 1.

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