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%I #23 Sep 08 2022 08:45:07
%S 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,
%T 18,37,140,43,67,23,120,25,77,80,55,85,88,30,155,32,65,33,205,217,75,
%U 38,77,118,40,205,127,130,133,45,137,93,47,240,350,103,52,105,378,167,285
%N Difference of successive primes squared divided by 24, (prime(n+1)^2-prime(n)^2)/24.
%C For n>=3, prime(n+1)^2-prime(n)^2 is always divisible by 24.
%C 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]
%H G. C. Greubel, <a href="/A075888/b075888.txt">Table of n, a(n) for n = 3..1000</a>
%F a(n) = (prime(n+1)^2 - prime(n)^2)/24.
%e a(4)=3 because (prime(5)^2-prime(4)^2)/24=(11^2-7^2)/24=3.
%t (#[[2]]-#[[1]])/24&/@(Partition[Prime[Range[3,70]],2,1]^2) (* _Harvey P. Dale_, Apr 06 2013 *)
%t Table[(Prime[n + 1]^2 - Prime[n]^2)/24, {n,3,50}] (* _G. C. Greubel_, Feb 18 2017 *)
%o (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
%o (Magma) [(NthPrime(n+1)^2 - NthPrime(n)^2)/24: n in [3..100]]; // _Vincenzo Librandi_, Mar 07 2015
%Y Cf. A024702.
%K easy,nonn
%O 3,2
%A _Zak Seidov_, Oct 17 2002
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