A252477 Integer part of 1/(sqrt(prime(n+1))-sqrt(prime(n))).

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

Offset: 1

Author

M. F. Hasler, Dec 31 2014

Keywords

• nonn
• changed

Comments

Andrica's conjecture states that sqrt(prime(n+1))-sqrt(prime(n)) < 1 for all n. Since equality cannot happen, this is equivalent to say that all terms of is sequence are >= 1.

Sequence A074976 is based on the same idea (rounding to the nearest integer instead).

It is a remarkable coincidence(?) that very often, especially around "peaks", a symmetric pattern "x, y, x" occurs: 2, 7, 2,... 10, 5, 10,... 13, 2, 13,... 20, 4, 20, ..., 11, 5, 11, ...

Equal to the integer part of (A000006(n+1)+A000006(n))/(prime(n+1)-prime(n)) for most indices; exceptions are 1, 129, 1667, 2004, 2088, 2334, 3377, 3585, 3695, 3834, 4978, 7057, 7950, 8103, 9525, 9805,...

Examples

			a(1) = floor(1/(sqrt(3) - sqrt(2))) = floor(1/(1.73-1.41)) = floor(1/0.32) = floor(3.15) = 3.
a(2) = floor(1/(sqrt(5) - sqrt(3))) = floor(1/(2.236-1.732)) = floor(1/0.504) = floor(1.98) = 1.
		

Links

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

Crossrefs

Cf. A000040, A074976, A079636.

Programs

Haskell

a252477 n = a252477_list !! (n-1)
a252477_list = map (floor . recip) $ zipWith (-) (tail rs) rs
               where rs = map (sqrt . fromIntegral) a000040_list
-- Reinhard Zumkeller, Jan 04 2015

Maple

a:= n-> ((w, v)-> floor(1/(w-v)))(map(sqrt@ithprime, [n+1, n])[]):
seq(a(n), n=1..92);  # Alois P. Heinz, Aug 23 2025

Mathematica

Floor[1/Subtract @@@ Reverse[Partition[Sqrt[Prime[Range[100]]], 2, 1], 2]] (* Paolo Xausa, Aug 24 2025 *)

PARI

a(n)=1\(sqrt(prime(n+1))-sqrt(prime(n))) \\ M. F. Hasler, Dec 31 2014

Formula

a(n) = A079636(n) - 1. - Alois P. Heinz, Aug 23 2025