A008963 Initial digit of Fibonacci number F(n).

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, 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, 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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• base
• easy
• changed

Comments

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

Wlodarski observed and Webb proved that the distribution of terms of this sequence follows Benford's law. - Amiram Eldar, Sep 23 2019

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from T. D. Noe)
• William Webb, Distribution of the first digits of Fibonacci numbers, The Fibonacci Quarterly, Vol. 13, No. 4 (1975), pp. 334-336.
• Wikipedia, Benford's law.
• J. Wlodarski, Fibonacci and Lucas Numbers Tend to Obey Benford's Law, The Fibonacci Quarterly, Vol. 9, No. 1 (1971), pp. 87-88.
• Index entries for sequences related to Benford's law.

Crossrefs

Cf. A000045, A003893 (final digit).

Cf. A000030, A261607, A213201.

Programs

Haskell

a008963 = a000030 . a000045  -- Reinhard Zumkeller, Sep 09 2015

Maple

F:= combinat[fibonacci]:
a:= n-> parse(""||(F(n))[1]):
seq(a(n), n=0..100);  # Alois P. Heinz, Nov 22 2023

Mathematica

Table[IntegerDigits[Fibonacci[n]][[1]], {n, 0, 100}] (* T. D. Noe, Sep 23 2011 *)

PARI

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

Formula

a(n) = A000030(A000045(n)). - Amiram Eldar, Sep 23 2019

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

For n>5, a(n) = floor(10^{alpha*n-beta}), where alpha=log_10(phi), beta=log_10(5)/2, {x}=x-floor(x) denotes the fractional part of x, log_10(phi) = A097348, and phi = (1+sqrt(5))/2 = A001622. - Hans J. H. Tuenter, Aug 20 2025