cp's OEIS Frontend

From Vladimir Pletser, Mar 16 2013: (Start)

As Li(2) = 1.04516..., a(n) = A057752(n) - 1.

This sequence gives the exact values of the difference between Gauss's Li (defined as integral(2..10^n, dt/log(t)) or Li(10^n)-Li(2)) and the number of primes <= 10^n (A006880). For large values of x=10^n, Li(2) can be neglected but for small values of x=10^n, the value of Li(2) cannot be neglected.

This sequence yields a better average relative difference, i.e., average(a(n)/pi(10^n)) = 2.0116...x10^-2 for 1<=n<=24, compared to average(A057752(n)/pi(10^n)) = 3.2486...x10^-2. However see also Li(10^n)-Li(3) in A223166 and A223167.

Note that most of the Tables in the literature giving the difference of Li(10^n) - pi(10^n) use the values of A057752 as the difference between Gauss's Li values and pi(10^n). This is incorrect and the values above should be used instead. For example (certainly not exhaustive):

- John H. Conway and R. K. Guy in "The Book of Numbers" show in Fig. 5.2, p. 144, Li(N) as integral(2..10^n, dt/log(t)) but reports values of A057752 (the difference of integral(0..10^n, dt/log(t)) and pi(10^n)) in Table 5.2, p. 146;

- Eric Weisstein in "Prime Counting Function" gives also values of -(A057752) for pi(10^n)-Li(10^n)

- Wikipedia gives a Table with Li(10^n)-pi(10^n) (A057752);

- C. K. Caldwell in Table 3 in the link below give values of Li(10^n) while values of Li(10^n) - Li(2) would be more suited. (End)

"Adrien-Marie Legendre in 1778 published his work 'Essai sur la théorie des nombres' where he proposed a modified form of the first approximation, pi(n) ~ n/ln n." (Gullberg)

From Mats Granvik, Jun 14 2013: (Start)

The logarithmic integral li(x) = exponential integral Ei(log(x)).

The generating function for tau A000005, the number of divisors of n is: Sum_{n >= 1} a(n) x^n = Sum_{k > 0} x^k/(1 - x^k). Another way to write the generating function for tau A000005 is Sum_{n>=1} A000005(n) x^n = Sum_{a=1..Infinity} Sum_{b>=1} x^(a*b).

If we instead think of the integral with the same form, evaluate at x = exp(1) = 2.7182818284... = A001113 and set the integration limits to zero and sqrt(log(n)), we get for n >= 0:

Logarithmic integral li(n) = Integral_{a = 0..sqrt(log(n))} Integral_{b=0..sqrt(log(n))} exp(1)^(a*b) + EulerGamma + log(log(n)). (End)

li(2)-1 is the minimum [known to date, for n>1] of |li(n) - PrimePi(n)|. - Jean-François Alcover, Jul 10 2013

The modern logarithmic integral function li(x) = Integral_{t=0..x} (1/log(t)) replaced the Li(x) = Integral_{t=2..x} (1/log(t)) which was sometimes used because it avoids the singularity at x=1. This constant is the offset between the two functions: li(2) = li(x) - Li(x) = Integral_{t=0..2} (1/log(t)). - Stanislav Sykora, May 09 2015

As Li(3)= 2.163588..., A057752(n)-a(n) = 2, except for n =3, 6, 10, 11, 15, 20 where A057752(n)-a(n)= 3.

This sequence yields an even better average relative difference than Gauss's approximation (A106313), i.e., Average(a(n)/pi(10^n)) = 7.4969...*10^-3 for 1<=n<=24, compared to Average(A057752(n)/pi(10^n)) = 3.2486...*10^-2 and Average(A106313(n)/pi(10^n)) = 2.0116...*10^-2, showing that, when using the logarithmic integral, Li(10^n)-Li(3) (A223166) gives a better approximation to pi(10^n) than Li(10^n)-Li(2) (A190802) and than Li(10^n) (A057754).

The sequence gives exactly the values of pi(10^n) for n = 1 to 3.

A228724 gives the difference between A006880 and this sequence.

In computing Li(x) we can limit the iterations to 2*log(x) + m where m is suitably large to allow convergence to the precision desired. If we let m = floor(log(log(x))) we get a better approximation of Pi(x) than the full Li(x) expansion. With this m we get Li(x) < Pi(x) often but still closer in absolute value to Pi(x). Note the use of the gamma function to quickly compute factorials in the precision range i.e. gamma(x+1) = x!.

See A057754 for the round() variant. [From R. J. Mathar, Oct 09 2010]

a(n) closely approximates the number of primes < n^2, that is, A038107(n) = Pi(n^2).

In fact, the sum is as good as Li(n^2). For n = 10^9,

a(n) = 24739954333817884.

Pi(n^2) = 24739954287740860 = A006880(18).

Li(n^2) = 24739954309690415 = A057754(18) = A089896(18).

R(n^2) = 24739954284239494 = A057793(18).

Now x/(log(x)-1) is a much better approximation of Pi(x) than x/log(x).

10^18/(log(10^18)-1) = 24723998785919976 and

10^18/log(10^18) = 24127471216847323.

Ironically though, a(n) = Sum_{x=2..n} x/(log(x)-1) is far from Pi(n^2), see A058290.