A016064 -id:A016064 - cp's OEIS frontend

A335025 Largest side lengths of almost-equilateral Heronian triangles.

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)

A100233 a(n) = Lucas(3*n) - 1.

Original entry on oeis.org

1, 3, 17, 75, 321, 1363, 5777, 24475, 103681, 439203, 1860497, 7881195, 33385281, 141422323, 599074577, 2537720635, 10749957121, 45537549123, 192900153617, 817138163595, 3461452808001, 14662949395603, 62113250390417, 263115950957275, 1114577054219521

Offset: 0

Paul D. Hanna, Nov 29 2004

Main diagonal of triangle A100232.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Cf. A000032, A100230, A100232, A016064.

I:=[1, 3, 17]; [n le 3 select I[n] else 5*Self(n-1) -3*Self(n-2) -Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 21 2017

Table[LucasL[3*n] - 1, {n,0,50}] (* or *) LinearRecurrence[{5,-3,-1}, {1,3,17}, 30] (* G. C. Greubel, Dec 21 2017 *)

a(n)=if(n==0,1,n*polcoeff(log((1-x)/(1-4*x-x^2)+x*O(x^n)),n))

Vec((1-2*x+5*x^2)/((1-x)*(1-4*x-x^2)) + O(x^40)) \\ Colin Barker, Jun 02 2016

a(n) = A014448(n) - 1.
a(n) = 4*a(n-1) + a(n-2) + 4 for n>1, with a(0)=1, a(1)=3.
G.f.: Sum_{n>=1} a(n)*x^n/n = log((1-x)/(1-4*x-x^2)).
a(n) = [x^n] ( 1 + 2*x + sqrt(1 + 2*x + 5*x^2) )^n. Cf. A016064. - Peter Bala, Jun 23 2015
From Colin Barker, Jun 02 2016: (Start)
a(n) = -1+(2-sqrt(5))^n+(2+sqrt(5))^n.
a(n) = 5*a(n-1)-3*a(n-2)-a(n-3) for n>2.
G.f.: (1-2*x+5*x^2) / ((1-x)*(1-4*x-x^2)).
(End)

New definition from Ralf Stephan, Dec 01 2004

A151961 Semiperimeter of the n-th Heronian triangle.

Original entry on oeis.org

3, 6, 21, 78, 291, 1086, 4053, 15126, 56451, 210678, 786261, 2934366, 10951203, 40870446, 152530581, 569251878, 2124476931, 7928655846, 29590146453, 110431929966, 412137573411, 1538118363678, 5740335881301, 21423225161526, 79952564764803, 298387033897686

Offset: 1

Juri-Stepan Gerasimov, Jul 13 2009

The side lengths are consecutive integers (A016064) and the area is an integer (A011945).
Except for the first term, positive values of x (or y) satisfying  x^2 - 4*x*y + y^2 + 27 = 0. - Colin Barker, Feb 08 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 432 = 0. - Colin Barker, Feb 16 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
G. Jacob Martens, Rational right triangles and the Congruent Number Problem, arXiv:2112.09553 [math.GM], 2021, see section 10.1 The Brahmaguptra triangles, equation (99).
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001075, A003500, A011945, A016064.

I:=[3,6]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 11 2014

CoefficientList[Series[3(1-2x)/(1-4x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Feb 11 2014 *)
3*ChebyshevT[Range[0, 40], 2] (* G. C. Greubel, Oct 10 2022 *)
LinearRecurrence[{4,-1},{3,6},30] (* Harvey P. Dale, Dec 21 2022 *)

Vec(3*x*(1-2*x)/(1-4*x+x^2) + O(x^40)) \\ Colin Barker, Oct 12 2015

[3*chebyshev_T(n, 2) for n in range(41)] # G. C. Greubel, Oct 10 2022

a(n) = 3 * A001075(n-1). - Joerg Arndt, Oct 10 2022
a(n) = 3*(A016064(n-1) + 1)/2 = 3*A003500(n-1)/2. - R. J. Mathar, Jul 27 2009
From Colin Barker, Mar 30 2012: (Start)
a(n) = 4*a(n-1) - a(n-2).
G.f.: 3*x*(1-2*x)/(1-4*x+x^2). (End)
a(n) = 3*( (2+sqrt(3))*(2-sqrt(3))^n + (2-sqrt(3))*(2+sqrt(3))^n )/2. - Colin Barker, Oct 12 2015

More terms from R. J. Mathar, Jul 27 2009

A087601 Primes occurring as lesser side of Heronian triangle (sides are consecutive integers, area and inradius are integers).

Original entry on oeis.org

3, 13, 193, 37633, 7300801, 1416317953

Offset: 1

Zak Seidov, Aug 07 2003

Next term (if it exists) is greater than 10^10000. - Ray Chandler, Jul 04 2015

			a(1)=3 because 3, 4 and 5 are sides of Heronian triangle (area is 6, inradius is 1).

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.

Cf. A001353, A016064, A003500, A087602.

Prime elements of A016064. Alternatively, primes p such that 3*(p+1)^2-12 is a square, i.e., p+1 belongs to A003500. - Max Alekseyev, May 14 2010

A087602 Primes occurring as larger side of Heronian triangle (sides are consecutive integers, area and inradius are integers).

Original entry on oeis.org

3, 5, 53, 140453

Offset: 1

Zak Seidov, Aug 07 2003

Next term (if it exists) is greater than 10^10000. - Ray Chandler, Jul 04 2015

			a(3)=53 because 53, 52 and 51 are sides of Heronian triangle (area is 1170, inradius is 15).

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.

Cf. A001353, A016064, A003500, A087601.

p, p-1 and p-2 are sides of Heronian triangle.

Primes p such that 3*(p-1)^2-12 is a square, i.e., p-1 belongs to A003500. - Max Alekseyev, May 14 2010

A131515 a(n+2) = 34*a(n+1) - a(n) + 2; a(1) = 1 and a(2) = 16.

Original entry on oeis.org

1, 16, 545, 18516, 629001, 21367520, 725866681, 24658099636, 837649520945, 28455425612496, 966646821303921, 32837536498720820, 1115509594135203961, 37894488664098213856, 1287297104985204067145, 43730207080832840069076, 1485539743643331358281441

Offset: 1

Parthasarathy Nambi, Aug 14 2007

			If n=3 then a(3) = 34*a(2) - a(1) + 2 = 545 which is the third term in the sequence.

Colin Barker, Table of n, a(n) for n = 1..650
Chip Curtis, Generating Heronian Triangles, The College Mathematics Journal, vol. 38 (2007), pp. 315-316. See page 316.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A016064, A001110.

RecurrenceTable[{a[1]==1,a[2]==16,a[n]==34a[n-1]-a[n-2]+2},a,{n,20}] (* or *) LinearRecurrence[{35,-35,1},{1,16,545},20] (* Harvey P. Dale, Feb 10 2015 *)

Vec(x*(1-19*x+20*x^2)/((1-x)*(1-34*x+x^2)) + O(x^20)) \\ Colin Barker, Mar 02 2016

a(n)=([0,1,0;0,0,1;1,-35,35]^n*[20;1;16])[1,1] \\ Charles R Greathouse IV, Aug 05 2016

G.f.: -x*(1-19*x+20*x^2)/(-1+x)/(1-34*x+x^2). - R. J. Mathar, Nov 14 2007
a(1)=1, a(2)=16, a(3)=545, a(n) = 35*a(n-1)-35*a(n-2)+a(n-3). - Harvey P. Dale, Feb 10 2015
a(n) = (-6 + (963-680*sqrt(2))*(17+12*sqrt(2))^n + (17+12*sqrt(2))^(-n)*(963+680*sqrt(2)))/96. - Colin Barker, Mar 02 2016

Definition corrected by Rick L. Shepherd, Aug 17 2007

More terms from Harvey P. Dale, Feb 10 2015

A242497 Sides of (Heronian) triangles where sides are consecutive integers and area is an integer.

Original entry on oeis.org

3, 4, 5, 13, 14, 15, 51, 52, 53, 193, 194, 195, 723, 724, 725, 2701, 2702, 2703, 10083, 10084, 10085, 37633, 37634, 37635, 140451, 140452, 140453, 524173, 524174, 524175, 1956243, 1956244, 1956245, 7300801, 7300802, 7300803, 27246963, 27246964, 27246965

Offset: 1

R. K. Guy and Charles R Greathouse IV, May 16 2014

Let the edge lengths of the triangle be 2x-1, 2x, 2x+1 so that area = sqrt{3x * x * (x-1) * (x+1)} and we need x^2 - 1 to be of shape 3y^2. That is, x/y is an even rank convergent to the continued fraction of sqrt(3) and x is A001075.

The intermediate length sides are given by A003500(n), n >= 1. Note that A003500(0) = 2 corresponds to the degenerate (Heronian) triangle with sides {1, 2, 3} and area 0. - Daniel Forgues, May 28 2014

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

Harvey P. Dale, Table of n, a(n) for n = 1..1000
David Eugene Smith and Yoshio Mikami, A history of Japanese mathematics, Dover, 2004, p. 168.
Index entries for linear recurrences with constant coefficients, signature (-1,-1,4,4,4,-1,-1,-1).

A016064 is the main entry for this sequence.

LinearRecurrence[{-1,-1,4,4,4,-1,-1,-1},{3,4,5,13,14,15,51,52},40] (* Harvey P. Dale, May 04 2021 *)

Vec((-3*x^7 - 5*x^6 - 6*x^5 + 4*x^4 + 10*x^3 + 12*x^2 + 7*x + 3)/(x^8 + x^7+ x^6 - 4*x^5 - 4*x^4 - 4*x^3 + x^2 + x + 1)+O(x^99))

G.f.: (-3*x^7 - 5*x^6 - 6*x^5 + 4*x^4 + 10*x^3 + 12*x^2 + 7*x + 3)/ ((1+x+x^2)*(1-4*x^3+x^6)). - R. J. Mathar, May 30 2023

A240240 Consider primitive Heronian triangles with integer area and with sides {m, m+1, c}, where c > m+1. The sequence gives the possible values of m.

Original entry on oeis.org

3, 9, 13, 19, 20, 33, 51, 65, 73, 99, 119, 129, 163, 170, 174, 193, 201, 203, 220, 243, 260, 269, 287, 289, 339, 362, 377, 393, 450, 451, 513, 532, 559, 579, 615, 649, 696, 702, 714, 723, 740, 771, 801, 883, 909, 940, 969, 975, 1059, 1112, 1153, 1155, 1156, 1164, 1251, 1299, 1325, 1332, 1353, 1424, 1455, 1459, 1569, 1605, 1615, 1683, 1690, 1716, 1801, 1869, 1919, 1923

Offset: 1

Zak Seidov, Apr 03 2014

nonn

Corresponding values of c are 5, 17, 15, 37, 29, 65, 101, 109, 145.
And corresponding values of area/6 are 1, 6, 14, 19, 35, 44, 85, 330, 146, 231, 1190.
The sequence includes all terms of A016064 (where c = m+2) except for the first term, 1 (case with zero area).
Note that in all cases c is odd and m+2 <= c < 2m+1.

			First triangle has sides (3,4,5) and area 6.
2nd triangle has sides (9,10,17) and area 36.
3rd triangle has sides (13,14,15) and area 84.

Cf. A083875, A016064, A224301, A227003, A239978, A227166.

re=Reap[Do[a=m;b=m+1;Do[s=(a+b+c)/2;area=Sqrt[s(s-a)(s-b)(s-c)];If[IntegerQ[area],Sow[{a,b,c,area}];Break[]],{c,2m-1,m+2,-2 }],{m,3,2000}]][[2,1]];#[[1]]&/@ re

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A335025 Largest side lengths of almost-equilateral Heronian triangles.

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A100233 a(n) = Lucas(3*n) - 1.

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A151961 Semiperimeter of the n-th Heronian triangle.

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A087601 Primes occurring as lesser side of Heronian triangle (sides are consecutive integers, area and inradius are integers).

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A087602 Primes occurring as larger side of Heronian triangle (sides are consecutive integers, area and inradius are integers).

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A131515 a(n+2) = 34*a(n+1) - a(n) + 2; a(1) = 1 and a(2) = 16.

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A242497 Sides of (Heronian) triangles where sides are consecutive integers and area is an integer.

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