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

0,4

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.

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

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 *)

Prog

(PARI) vector(10001, n, f=fibonacci(n-1); f\10^(#Str(f)-1))
(Haskell)
a008963 = a000030 . a000045  -- Reinhard Zumkeller, Sep 09 2015

Crossrefs

Cf. A000045, A003893 (final digit).

Cf. A000030, A261607, A213201.

Sequence in context: A105994 A120496 A105150 * A031324 A226251 A093086

Adjacent sequences: A008960 A008961 A008962 * A008964 A008965 A008966

Keyword

nonn,base,easy

Author

N. J. A. Sloane.

Status

approved