A075888 Difference of successive primes squared divided by 24, (prime(n+1)^2-prime(n)^2)/24.

1, 3, 2, 5, 3, 7, 13, 5, 17, 13, 7, 15, 25, 28, 10, 32, 23, 12, 38, 27, 43, 62, 33, 17, 35, 18, 37, 140, 43, 67, 23, 120, 25, 77, 80, 55, 85, 88, 30, 155, 32, 65, 33, 205, 217, 75, 38, 77, 118, 40, 205, 127, 130, 133, 45, 137, 93, 47, 240, 350, 103, 52, 105, 378, 167, 285

Offset: 3

Author

Zak Seidov, Oct 17 2002

Keywords

• easy
• nonn

Comments

For n>=3, prime(n+1)^2-prime(n)^2 is always divisible by 24.

It follows from the previous comment that for n>=3, prime(n)= sqrt(5^2 + k*24) where integer k>= 0 Then it follows from above that for n>=3, ((prime(n))^2 - 1)/24 always gives integral values - see A024702. [From Alexander R. Povolotsky, Sep 20 2008]

Examples

			a(4)=3 because (prime(5)^2-prime(4)^2)/24=(11^2-7^2)/24=3.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 3..1000

Crossrefs

Cf. A024702.

Programs

Magma

[(NthPrime(n+1)^2 - NthPrime(n)^2)/24: n in [3..100]]; // Vincenzo Librandi, Mar 07 2015

Mathematica

(#[[2]]-#[[1]])/24&/@(Partition[Prime[Range[3,70]],2,1]^2) (* Harvey P. Dale, Apr 06 2013 *)
Table[(Prime[n + 1]^2 - Prime[n]^2)/24, {n,3,50}] (* G. C. Greubel, Feb 18 2017 *)

PARI

j=[];for(n=3, 300, if(((floor((((prime(n+1))^2)-((prime(n))^2))/24))==(ceil(((((prime(n+1))^2)-((prime(n))^2))/24)))), j=concat(j, ((((prime(n+1))^2) - ((prime(n))^2))/24)), j=concat(j,-1)));j \\ Alexander R. Povolotsky, Sep 08 2008

Formula

a(n) = (prime(n+1)^2 - prime(n)^2)/24.