A277369 a(0) = 5, a(1) = 8; for n>1, a(n) = 2*a(n-1) + a(n-2).

5, 8, 21, 50, 121, 292, 705, 1702, 4109, 9920, 23949, 57818, 139585, 336988, 813561, 1964110, 4741781, 11447672, 27637125, 66721922, 161080969, 388883860, 938848689, 2266581238, 5472011165, 13210603568, 31893218301, 76997040170, 185887298641, 448771637452, 1083430573545

Offset

0,1

Comments

After the first term, there are no primes in this sequence. In fact:

a(12*k) is divisible by 5,

a(12*k+1) is divisible by 2,

a(12*k+2) is divisible by 3,

a(12*k+3) is divisible by 2,

a(12*k+4) is divisible by 11,

a(12*k+5) is divisible by 2,

a(12*k+6) is divisible by 3,

a(12*k+7) is divisible by 2,

a(12*k+8) is divisible by 7,

a(12*k+9) is divisible by 2,

a(12*k+10) is divisible by 3,

a(12*k+11) is divisible by 2.

Therefore, every term is divisible by 2, 3, 5, 7, or 11.

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2,1).

Formula

From Colin Barker, Oct 11 2016: (Start)

a(n) = (((1-sqrt(2))^n*(-3+5*sqrt(2))+(1+sqrt(2))^n*(3+5*sqrt(2))))/(2*sqrt(2)).

G.f.: (5-2*x) / (1-2*x-x^2).

(End)

Mathematica

LinearRecurrence[{2, 1}, {5, 8}, 40] (* Alonso del Arte, Oct 11 2016 *)

Prog

(PARI) lista(n) = n++; my(v=vector(max(2, n))); v[1]=5; v[2]=8; for(i=3, n, v[i]=2*v[i-1] + v[i-2]); v \\ David A. Corneth, Oct 11 2016
(PARI) Vec((5-2*x)/(1-2*x-x^2) + O(x^40)) \\ Colin Barker, Oct 11 2016

Crossrefs

Cf. A276849.

Sequence in context: A120036 A036381 A363224 * A140419 A292851 A138023

Adjacent sequences: A277366 A277367 A277368 * A277370 A277371 A277372

Keyword

nonn,easy

Author

Bobby Jacobs, Oct 11 2016

Extensions

More terms from David A. Corneth, Oct 11 2016

Status

approved