A074976 a(n) = round(1/(sqrt(prime(n+1)) - sqrt(prime(n)))).

3, 2, 2, 1, 3, 2, 4, 2, 2, 5, 2, 3, 6, 3, 2, 2, 8, 3, 4, 8, 3, 4, 3, 2, 5, 10, 5, 10, 5, 2, 6, 4, 12, 2, 12, 4, 4, 6, 4, 4, 13, 3, 14, 7, 14, 2, 2, 7, 15, 8, 5, 15, 3, 5, 5, 5, 16, 6, 8, 17, 3, 2, 9, 18, 9, 3, 6, 4, 19, 9, 6, 5, 6, 6, 10, 7, 5, 10, 5, 4, 20, 4, 21, 7, 10, 7, 5, 11, 21, 11, 4, 5, 11, 6, 11

Offset: 1

Author

Werner D. Sand and Benoit Cloitre, Oct 06 2002

Keywords

• nonn

Comments

If Andrica's conjecture is true, each term in the sequence is >=1.

Since Andrica's conjecture is bounded below only by zero, a(n) is not bounded above.

Examples

			a(1) = round(1/(sqrt(3) - sqrt(2))) = round(1/(1.7320.. - 1.4142..)) = round(1/0.3178..) = round(3.1462..) = 3.
		

References

• C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 482.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Andrica's conjecture

Crossrefs

Cf. A000040, A252477.

Programs

Haskell

a074976 n = a074976_list !! (n-1)
a074976_list = map (round . recip) $ zipWith (-) (tail rs) rs
               where rs = map (sqrt . fromIntegral) a000040_list
-- Reinhard Zumkeller, Jan 04 2015

Mathematica

Round[1/(Sqrt[#[[2]]]-Sqrt[#[[1]]])]&/@Partition[Prime[Range[100]],2,1] (* Harvey P. Dale, May 30 2022 *)

PARI

a(n) = round(1/(sqrt(prime(n+1))-sqrt(prime(n)))) \\ Michel Marcus, May 22 2013

Formula

Conjecture: for n>=4, a(n)>=2. More generally, for any m >=1, the set of k such that a(k)=m is finite. I.e., if n>=217, a(n)>=3; if n>=263, a(n)>=4; if n>=590, a(n)>=5; if n>=3385, a(n)>=6; ...