0,3

a(n) / a(n-1) converges to 10^1/2 + 2 as n approaches infinity. 10^1/2 + 2 can also be written as 2^1/2 * (2^1/2 + 5^1/2), ((2 * 2^1/2) * Phi) - 2^1/2 + 2 and 2^1/2 * (2^1/2 + (L(n) / F(n))), where L(n) is the n-th Lucas number and F(n) is the n-th Fibonacci number as n approaches infinity.

Eric Weisstein, Lucas Number

Eric Weisstein, Lucas Sequence

Eric Weisstein, Horadam Sequence

Eric Weisstein, Fibonacci Number

Eric Weisstein, Pell Number

a(n) = s*a(n-1) + r*a(n-2); for n > 1, where a(0) = 0, a(1) = 1, s = 4, r = 6

a(n)=((2+sqrt(10))^n-(2-sqrt(10))^n)/(2*sqrt(10)) [From Rolf Pleisch (r_pleisch(AT)gmx.ch), Jul 06 2009]

a(4) = 112 because a(3) = 22, a(2) = 4, s = 4, r = 6 and (4 * 22) + (6 * 4) = 112.

(Other) sage: [lucas_number1(n, 4, -6) for n in xrange(0, 21)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 23 2009]

Cf. A024318, A000032, A000129.

Sequence in context: A144047 A077543 A084157 this_sequence A106835 A155596 A025569

Adjacent sequences: A085936 A085937 A085938 this_sequence A085940 A085941 A085942

easy,nonn

Ross La Haye (rlahaye(AT)new.rr.com), Aug 16 2003