OFFSET
0,2
COMMENTS
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
REFERENCES
Nakane Genkei (Nakane the Elder), Shichijo Beki Yenshiki, 1691.
LINKS
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).
FORMULA
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) = [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
EXAMPLE
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
MATHEMATICA
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 *)
PROG
(PARI) 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
(PARI) a(n)=if(n<1, 1, -1+ceil((2+sqrt(3))^(n))) \\ Ralf Stephan
(PARI) is(n)=issquare(3*n^2+6*n-9) \\ Charles R Greathouse IV, May 16 2014
(PARI) {a(n) = 2 * polchebyshev(n, 1, 2) - 1}; /* Michael Somos, May 11 2019 */
(Magma) 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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
More terms from Rick L. Shepherd, Feb 18 2007
Defition revised by N. J. A. Sloane, Feb 04 2021
STATUS
approved