|
|
A111497
|
|
Difference between successive terms of floor(10^n/Li(10^n) - 1).
|
|
0
|
|
|
2, 3, 2, 2, 2, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 2, 2, 2
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
10^n/Li(10^n) - 1) is the ratio of estimated composite numbers less than 10^n to the estimated prime numbers less than 10^n. Conjecture: 2 and 3 are the only numbers in this sequence.
|
|
LINKS
|
|
|
FORMULA
|
Li(n) is the logarithmic integral which approximates the number of primes less than n. n Li(n) = Int dt/log(t) 2
|
|
PROG
|
(PARI) LiRatioDiff(m, n) = { local(x, p1, p2, a, b); forstep(x=m, n, 2, p1=10.^x; p2=10^(x+1); a=floor(p1/Li(p1)-1); b=floor(p2/Li(p2)-1); print1(b-a, ", ") ) } Li(x) = \ Logarithmic integral { -eint1(log(1/x)) }
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|