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A103974 Smaller sides (a) in (a,a,a+1)-integer triangle with integer area. 9
1, 5, 65, 901, 12545, 174725, 2433601, 33895685, 472105985, 6575588101, 91586127425, 1275630195845, 17767236614401, 247465682405765, 3446752317066305, 48007066756522501, 668652182274248705 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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

LINKS

Table of n, a(n) for n=1..17.

Project Euler, Problem 94: Almost Equilateral Triangles.

FORMULA

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, x^2-3*y^2=1, 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.

"Formula: 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 (conjectured). - Rick L. Shepherd, Sep 04 2005

a(n)= [1+14*A007655(n+2)-194*A007655(n+1)]/3. - R. J. Mathar, Nov 16 2007

MAPLE

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

MATHEMATICA

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 *)

PROG

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 ]

(PARI) 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

CROSSREFS

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

Sequence in context: A199024 A155653 A087453 * A233093 A208588 A249930

Adjacent sequences:  A103971 A103972 A103973 * A103975 A103976 A103977

KEYWORD

nonn

AUTHOR

Zak Seidov, Feb 23 2005

EXTENSIONS

More terms from Creighton Dement, Apr 18 2005

STATUS

approved

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Last modified November 21 13:53 EST 2017. Contains 295001 sequences.