A052011 Number of primes between successive Fibonacci numbers exclusive.

0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 10, 16, 23, 37, 55, 84, 125, 198, 297, 458, 704, 1087, 1673, 2602, 4029, 6263, 9738, 15186, 23704, 36981, 57909, 90550, 142033, 222855, 349862, 549903, 865019, 1361581, 2145191, 3381318, 5334509, 8419527, 13298630

Offset: 1

Author

Patrick De Geest, Nov 15 1999

Keywords

• nonn
• nice

Comments

With the given sequence data, we see that neither endpoint is included, so we count primes p in the open interval F(n)Jeppe Stig Nielsen, Jun 06 2015

Examples

			Between Fib(9)=34 and Fib(10)=55 we find the following primes: 37, 41, 43, 47 and 53 hence a(9)=5.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..122 (calculated from the data at A054782 and A001605)

Crossrefs

Cf. A000040, A001605, A005478 (endpoint primes), A010051, A052012, A054782.

Programs

Haskell

a052011 n = a052011_list !! (n-1)
a052011_list = c 0 0 $ drop 2 a000045_list where
  c x y fs'@(f:fs) | x < f     = c (x+1) (y + a010051 x) fs'
                   | otherwise = y : c (x+1) 0 fs
-- Reinhard Zumkeller, Dec 18 2011

Maple

for n from 1 to 43 do T[n]:= numtheory:-pi(combinat:-fibonacci(n)) od:
seq(T[n]-T[n-1]-`if`(isprime(combinat:-fibonacci(n)),1,0), n=2..43); # Robert Israel, Jun 08 2015

Mathematica

lst={};Do[p=0;Do[If[PrimeQ[a],p++ ],{a,Fibonacci[n]+1,Fibonacci[n+1]-1}];AppendTo[lst,p],{n,50}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
pbf[n_]:=Module[{fib1=If[PrimeQ[Fibonacci[n+1]],PrimePi[Fibonacci[n+1]-1], PrimePi[ Fibonacci[n+1]]], fib0=If[PrimeQ[Fibonacci[n]], PrimePi[ Fibonacci[n]+1],PrimePi[Fibonacci[n]]]},Max[0,fib1-fib0]]; Array[pbf,50] (* Harvey P. Dale, Mar 01 2012 *)

PARI

a(n)=my(s); forprime(p=fibonacci(n)+1,fibonacci(n+1)-1,s++); s \\ Charles R Greathouse IV, Jun 08 2015

Formula

a(n) = pi(F(n+1)-1) - pi(F(n)) = A000720(A000045(n+1)-1) - A000720(A000045(n)). - Jonathan Vos Post, Mar 08 2010; corrected by Jeppe Stig Nielsen, Jun 06 2015

a(n) ~ phi^(n-1)/(n*sqrt(5)*log(phi)), where phi = (1+sqrt(5))/2 is the golden ratio. - Charles R Greathouse IV, Jun 08 2015

a(n) = A054782(n+1) - A054782(n) - [n+1 in A001605], where [] denotes the Iverson bracket. - Amiram Eldar, Jun 10 2024