OFFSET
0,1
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)
LINKS
Kevin Brown, Benford's Law.
Eric Weisstein's World of Mathematics, Benford's Law.
Wikipedia, Benford's Law.
FORMULA
a(n) = floor(F(5*n+3)/10^(floor(log(F(5*n+3))/log(10)))). - Johannes W. Meijer, Jul 06 2011
EXAMPLE
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)
MAPLE
A134490 := proc(n) combinat[fibonacci](5*n+3) ; end proc:
seq(A141053(n), n=0..70) ; # R. J. Mathar, Jul 04 2011
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
KEYWORD
nonn,base,less
AUTHOR
Paul Curtz, Aug 01 2008
EXTENSIONS
Edited by Johannes W. Meijer, Jul 06 2011
STATUS
approved