A141053 Most-significant decimal digit of Fibonacci(5n+3).

2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 7, 8, 8, 9, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2

Offset: 0

Author

Paul Curtz, Aug 01 2008

Keywords

• nonn
• base
• less

Comments

Leading digit of A134490(n).

From Johannes W. Meijer, Jul 06 2011: (Start)

The leading digit d, 1 <= d <= 9, of A141053 follows Benford’s Law. This law states that the probability for the leading digit is p(d) = log_10(1+1/d), see the examples.

We observe that the last digit of A134490(n), i.e. F(5*n+3) mod 10, leads to the Lucas sequence A000032(n) (mod 10), i.e. a repetitive sequence of 12 digits [2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9] with p(0) = p(5) = 0, p(1) = p(3) = p(7) = p(9) = 1/6 and p(2) = p(4) = p(6) = p(8) = 1/12. This does not obey Benford’s Law, which would predict that the last digit would satisfy p(d) = 1/10, see the links. (End)

Examples

			From _Johannes W. Meijer_, Jul 06 2011: (Start)
d     p(N=2000) p(N=4000) p(N=6000) p(Benford)
1      0.29900   0.29950   0.30033   0.30103
2      0.17700   0.17675   0.17650   0.17609
3      0.12550   0.12525   0.12517   0.12494
4      0.09650   0.09675   0.09700   0.09691
5      0.07950   0.07950   0.07933   0.07918
6      0.06700   0.06675   0.06700   0.06695
7      0.05800   0.05825   0.05800   0.05799
8      0.05150   0.05125   0.05100   0.05115
9      0.04600   0.04600   0.04567   0.04576
Total  1.00000   1.00000   1.00000   1.00000 (End)
		

Links

• Kevin Brown, Benford's Law.
• Eric Weisstein's World of Mathematics, Benford's Law.
• Wikipedia, Benford's Law.
• Index entries for sequences related to Benford's law

Crossrefs

Cf. A000045 (F(n)), A008963 (Initial digit F(n)), A105511-A105519, A003893 (F(n) mod 10), A130893, A186190 (First digit tribonacci), A008952 (Leading digit 2^n), A008905 (Leading digit n!), A045510, A112420 (Leading digit Collatz 3*n+1 starting with 1117065), A007524 (log_10(2)), A104140 (1-log_10(9)). - Johannes W. Meijer, Jul 06 2011

Programs

Maple

A134490 := proc(n) combinat[fibonacci](5*n+3) ; end proc:
A141053 := proc(n) convert(A134490(n),base,10) ; op(-1,%) ; end proc:
seq(A141053(n),n=0..70) ; # R. J. Mathar, Jul 04 2011

Mathematica

Table[IntegerDigits[Fibonacci[5n+3]][[1]],{n,0,70}] (* Harvey P. Dale, Jun 22 2025 *)

Formula

a(n) = floor(F(5*n+3)/10^(floor(log(F(5*n+3))/log(10)))). - Johannes W. Meijer, Jul 06 2011

Extensions

Edited by Johannes W. Meijer, Jul 06 2011