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A016064 Shortest legs of Heronian triangles (sides are consecutive integers, area is an integer). 15
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

Least side in (m, m + 1, m + 2) integer-sided triangle with integer area.

Also describes triangles whose sides are consecutive integers and in which an inscribed circle has an integer radius - Harvey P. Dale, Dec 28 2000

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

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.

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) = 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

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

Corresponding areas are in A011945.

Cf. A001353, A019973 (2 + sqrt(3)), A102341, A103974, A103975.

Cf. A072221.

Sequence in context: A026529 A286182 A101052 * A163774 A304629 A301458

Adjacent sequences:  A016061 A016062 A016063 * A016065 A016066 A016067

KEYWORD

nonn,easy

AUTHOR

Robert G. Wilson v

EXTENSIONS

More terms from Rick L. Shepherd, Feb 18 2007

STATUS

approved

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Last modified February 19 22:36 EST 2020. Contains 332061 sequences. (Running on oeis4.)