A072708 Last digit of F(n) is 6 where F(n) is the n-th Fibonacci number.

21, 39, 42, 48, 81, 99, 102, 108, 141, 159, 162, 168, 201, 219, 222, 228, 261, 279, 282, 288, 321, 339, 342, 348, 381, 399, 402, 408, 441, 459, 462, 468, 501, 519, 522, 528, 561, 579, 582, 588, 621, 639, 642, 648, 681, 699, 702, 708, 741, 759, 762, 768, 801

Offset

1,1

Comments

Sequence contains numbers of form (21+60k), (39+60k), (42+60k), (48+60k), with k>=0.

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Formula

G.f.: x*(12*x^4+6*x^3+3*x^2+18*x+21) / (x^5-x^4-x+1). - Colin Barker, Jun 16 2013

a(n) = (-3/4+(3*i)/4)*((1+i)*(-1)^n + (5+2*i)*(-i)^n + (2+5*i)*i^n - (10+10*i)*n) where i=sqrt(-1). - Colin Barker, Oct 16 2015

Mathematica

LinearRecurrence[{1, 0, 0, 1, -1}, {21, 39, 42, 48, 81}, 60] (* Harvey P. Dale, Aug 28 2017 *)

Prog

(PARI) a(n) = (-3/4+(3*I)/4)*((1+I)*(-1)^n + (5+2*I)*(-I)^n + (2+5*I)*I^n - (10+10*I)*n) \\ Colin Barker, Oct 16 2015
(PARI) Vec(x*(12*x^4+6*x^3+3*x^2+18*x+21)/(x^5-x^4-x+1) + O(x^100)) \\ Colin Barker, Oct 16 2015

Crossrefs

Cf. A000045, A003893.

Sequence in context: A139768 A307278 A176071 * A338330 A102478 A221048

Adjacent sequences: A072705 A072706 A072707 * A072709 A072710 A072711

Keyword

nonn,base,easy

Author

Benoit Cloitre, Aug 07 2002

Status

approved