A272365 - OEIS

%I #19 Apr 29 2016 13:01:09

%S 10,34,250,1954,15370,120994,952570,7499554,59043850,464851234,

%T 3659766010,28813276834,226846448650,1785958312354,14060820050170,

%U 110700602088994,871543996661770,6861651371205154,54021666972979450,425311684412630434,3348471808328064010,26362462782211881634

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

%C 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.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-9,1).

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

%F From _Ilya Gutkovskiy_, Apr 27 2016: (Start)

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

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

%e 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.

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

%o (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

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

%Y Cf. A000005, A000010, A003500, A268281

%K nonn,easy

%O 1,1

%A _Frank M Jackson_, Apr 27 2016