A072675 Integers m such that the last digit of Fibonacci(m) is 1.

1, 2, 8, 19, 22, 28, 41, 59, 61, 62, 68, 79, 82, 88, 101, 119, 121, 122, 128, 139, 142, 148, 161, 179, 181, 182, 188, 199, 202, 208, 221, 239, 241, 242, 248, 259, 262, 268, 281, 299, 301, 302, 308, 319, 322, 328, 341, 359, 361, 362, 368, 379, 382, 388, 401

Offset: 1

Author

Benoit Cloitre, Aug 07 2002

Keywords

• nonn
• base
• easy

Examples

			Fibonacci(28) = 317811, so 28 is a term.
		

Links

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

Crossrefs

Cf. A000045, A003893.

Programs

Mathematica

Position[Fibonacci[Range[500]],?(Mod[#,10]==1&)]//Flatten (* or *) LinearRecurrence[{1,0,0,0,0,0,0,1,-1},{1,2,8,19,22,28,41,59,61},70] (* _Harvey P. Dale, Sep 17 2018 *)

Formula

Sequence contains numbers of form (1+60k) (2+60k) (8+60k) (19+60k) (22+60k) (28+60k) (41+60k) (59+60k) k>=0.

G.f.: x*(x^8+18*x^7+13*x^6+6*x^5+3*x^4+11*x^3+6*x^2+x+1) / (x^9-x^8-x+1). - Colin Barker, Jun 16 2013