A123233 Difference between the (10^n)-th prime and the Riemann-Gram approximation of the (10^n)-th prime.

1, 0, 5, -4, -39, -24, 1823, -6566, -1844, -34087, 84846, -449836, -1117632, -3465179, -1766196, -11290074, 105510354

Offset

0,3

Comments

The algorithm in the PARI script below produces the 10^n-th prime accurate to first n/2 places. Conjecture: The sign of the terms in this sequence changes infinitely often. Based on the small sample presented here, it appears the negative terms occur much more often.

Links

Table of n, a(n) for n=0..16.

Formula

prime(10^x)-primeRG(10^x), where prime(n) is the n-th prime and primeRG(n)is an approximation of the n-th prime number based on an exponential bisection routine that uses the Riemann-Gram approximation of Pi(x). The flow of the routine is evident in the PARI script below.

Example

prime(10) - primeGR(10) = 29 - 29 = 0, the second term.

Prog

(PARI)
primeGR(n) =
\\ A good approximation for the n-th prime number using
\\ the Gram-Riemann approximation of Pi(x)
{ local(x, px, r1, r2, r, p10, b, e); b=10; p10=log(n)/log(10); if(Rg(b^p10*log(b^(p10+1)))< b^p10, m=p10+1, m=p10); r1 = 0; r2 = 7.18281828; for(x=1, 400, r=(r1+r2)/2; px = Rg(b^p10*log(b^(m+r))); if(px <= b^p10, r1=r, r2=r); r=(r1+r2)/2; ); floor(b^p10*log(b^(m+r))+.5); }
Rg(x) =
\\ Gram's Riemann's Approx of Pi(x)
{ local(n=1, L, s=1, r); L=r=log(x); while(s<10^40*r, s=s+r/zeta(n+1)/n; n=n+1; r=r*L/n); (s) }

Crossrefs

Sequence in context: A298548 A078811 A093399 * A189748 A304151 A223527

Adjacent sequences: A123230 A123231 A123232 * A123234 A123235 A123236

Keyword

sign,hard,more

Author

Cino Hilliard, Oct 06 2006

Status

approved