A099000
Indices k such that the k-th prime is a Fibonacci number.
Original entry on oeis.org
1, 2, 3, 6, 24, 51, 251, 3121, 42613, 23023556, 143130479, 2602986161967491
Offset: 1
Cf.
A001605 (n-th Fibonacci number is prime),
A005478 (Prime Fibonacci numbers).
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PrimePi[Select[Fibonacci[Range[80]], PrimeQ]]
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print1("1, 2");forprime(p=5,47,if(isprime(fibonacci(p)),print1(", "primepi(fibonacci(p))))) \\ Charles R Greathouse IV, Aug 21 2011
A095124
a(n) = prime(2^(2^n)).
Original entry on oeis.org
3, 7, 53, 1619, 821641, 104484802057, 870566678511500413493
Offset: 0
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Table[Prime[2^(2^n)],{n,0,5}] (* Harvey P. Dale, Oct 13 2013 *)
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f(n) = for(x=0,n,print1(prime(2^2^x)","))
A123233
Difference between the (10^n)-th prime and the Riemann-Gram approximation of the (10^n)-th prime.
Original entry on oeis.org
1, 0, 5, -4, -39, -24, 1823, -6566, -1844, -34087, 84846, -449836, -1117632, -3465179, -1766196, -11290074, 105510354, -208774399, 704933861
Offset: 0
a(1) = prime(10) - primeGR(10) = 29 - 29 = 0.
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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) }
Showing 1-3 of 3 results.
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