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A008963
Initial digit of Fibonacci number F(n).
26
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.
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).
Sequence in context: A105994 A120496 A105150 * A031324 A226251 A093086
KEYWORD
nonn,base,easy
AUTHOR
STATUS
approved