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

Author

Benoit Cloitre, Aug 07 2002

Keywords

• nonn
• base
• easy

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).

Crossrefs

Cf. A000045, A003893.

Programs

Mathematica

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

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

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