A272365 a(n) = 9a(n-1) - 9a(n-2) + a(n-3).

10, 34, 250, 1954, 15370, 120994, 952570, 7499554, 59043850, 464851234, 3659766010, 28813276834, 226846448650, 1785958312354, 14060820050170, 110700602088994, 871543996661770, 6861651371205154, 54021666972979450, 425311684412630434, 3348471808328064010, 26362462782211881634

Offset

1,1

Comments

For n>1 this linear recurrence generates Heronian triangles whose sides are a(n), a(n)-4, a(n)/2-1 and whose area K = (a(n)-2)*sqrt(15(a(n)-10)(a(n)+6))/16. When a(n) = 2p and p is an odd prime then 2p-tau(2p) = a(n)-4 and phi(2p) = a(n)/2-1, where tau=A000005 is the number of divisors and phi=A000010 the totient. Hence when a(n) = 2p for some odd prime p, it is a member of A268281.

Links

Table of n, a(n) for n=1..22.

Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Formula

a(n) = 9a(n-1) - 9a(n-2) + a(n-3).

From Ilya Gutkovskiy, Apr 27 2016: (Start)

G.f.: -2*x*(5-28*x+17*x^2) / ( (x-1)*(x^2-8*x+1) )

a(n) = 2*(2*(4 + sqrt(15))*(4 - sqrt(15))^n - 2*(sqrt(15) - 4)*(4 + sqrt(15))^n + 1). (End)

Example

a(2) = 34 because the triangle so formed has sides 34, 30, 16. It is Heronian with integer area 240 and is also Pythagorean. Because 34 = 2*17 and 17 is prime, it is also a member of A268281.

Mathematica

LinearRecurrence[{9, -9, 1}, {10, 34, 250}, 24]

Prog

(PARI) Vec(2*x*(5-28*x+17*x^2)/(1-9*x+9*x^2-x^3) + O(x^99)) \\ Altug Alkan, Apr 28 2016

Crossrefs

A268281 is a member of this sequence iff A268281/2 is prime.

Cf. A000005, A000010, A003500, A268281

Sequence in context: A001890 A221810 A219591 * A230895 A254674 A260336

Adjacent sequences: A272362 A272363 A272364 * A272366 A272367 A272368

Keyword

nonn,easy

Author

Frank M Jackson, Apr 27 2016

Status

approved