A008963 - OEIS

%I #36 Nov 22 2023 11:30:36

%S 0,1,1,2,3,5,8,1,2,3,5,8,1,2,3,6,9,1,2,4,6,1,1,2,4,7,1,1,3,5,8,1,2,3,

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

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

%N Initial digit of Fibonacci number F(n).

%C Benford's law applies since the Fibonacci sequence is of exponential growth: P(d)=log_10(1+1/d), in fact among first 5000 values the digit d=1 appears 1505 times, while 5000*P(1) is about 1505.15. - _Carmine Suriano_, Feb 14 2011

%C Wlodarski observed and Webb proved that the distribution of terms of this sequence follows Benford's law. - _Amiram Eldar_, Sep 23 2019

%H Alois P. Heinz, <a href="/A008963/b008963.txt">Table of n, a(n) for n = 0..10000</a> (first 1001 terms from T. D. Noe)

%H William Webb, <a href="https://www.fq.math.ca/Scanned/13-4/webb.pdf">Distribution of the first digits of Fibonacci numbers</a>, The Fibonacci Quarterly, Vol. 13, No. 4 (1975), pp. 334-336.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Benford&#39;s_law">Benford's law</a>.

%H J. Wlodarski, <a href="https://www.fq.math.ca/Scanned/9-1/wlodarski2.pdf">Fibonacci and Lucas Numbers Tend to Obey Benford's Law</a>, The Fibonacci Quarterly, Vol. 9, No. 1 (1971), pp. 87-88.

%H <a href="/index/Be#Benford">Index entries for sequences related to Benford's law</a>.

%F a(n) = A000030(A000045(n)). - _Amiram Eldar_, Sep 23 2019

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{d=1..9} d*log(1+1/d)/log(10) = 3.440236... (A213201). - _Amiram Eldar_, Jan 12 2023

%p F:= combinat[fibonacci]:

%p a:= n-> parse(""||(F(n))[1]):

%p seq(a(n), n=0..100);  # _Alois P. Heinz_, Nov 22 2023

%t Table[IntegerDigits[Fibonacci[n]][[1]], {n, 0, 100}] (* _T. D. Noe_, Sep 23 2011 *)

%o (PARI) vector(10001,n,f=fibonacci(n-1);f\10^(#Str(f)-1))

%o (Haskell)

%o a008963 = a000030 . a000045  -- _Reinhard Zumkeller_, Sep 09 2015

%Y Cf. A000045, A003893 (final digit).

%Y Cf. A000030, A261607, A213201.

%K nonn,base,easy

%O 0,4

%A _N. J. A. Sloane_.