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A065680
Number of primes <= prime(n) which begin with a 1.
8
0, 0, 0, 0, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25
OFFSET
1,6
COMMENTS
Considering the frequency of all decimal digits in leading position of prime numbers (A065681 - A065687), we cannot apply Benford's Law. But we observe at 10^e - levels that the frequency for 0 to 9 decreases monotonically, at least in the small range until 10^7.
The "begins with 9" sequence is too dull to include. - N. J. A. Sloane
Note that the primes do not satisfy Benford's law (see A000040). - N. J. A. Sloane, Feb 08 2017
LINKS
EXAMPLE
13 is the second prime beginning with 1: A000040(6) = 13, therefore a(6) = 2. a(664579) = 80020 (A000040(664579) = 9999991 is the largest prime < 10^7).
MATHEMATICA
Accumulate[If[First[IntegerDigits[#]]==1, 1, 0]&/@Prime[Range[80]]] (* Harvey P. Dale, Jan 22 2013 *)
PROG
(PARI) lista(n) = { my(a=[p\10^logint(p, 10)==1 | p<-primes(n)]); for(i=2, #a, a[i]+=a[i-1]); a} \\ Harry J. Smith, Oct 26 2009
KEYWORD
base,nonn
AUTHOR
Reinhard Zumkeller, Nov 13 2001
STATUS
approved