%I #44 Dec 07 2022 16:40:05
%S 2,1,2,1,2,1,2,3,1,3,2,1,2,3,1,1,3,2,1,3,2,3,4,1,1,2,1,2,7,2,3,1,5,1,
%T 3,1,2,3,1,1,5,1,2,1,6,1,1,1,2,3,1,5,3,1,1,1,3,2,1,5,7,2,1,2,7,3,5,1,
%U 2,3,4,3,1,2,3,4,1,2,5,1,5,1,3,2,3,4,1,1,2,3,2,1,2,1,3,2
%N Numerator of the ratio of consecutive prime gaps.
%H Vincenzo Librandi, <a href="/A272863/b272863.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = numerator((prime(n+2)-prime(n+1))/(prime(n+1)-prime(n))).
%F A001223(n) = Product_{k=1..n-1} a(k)/A274225(k). - _Andres Cicuttin_, Apr 26 2017
%F A000040(n) = 3 + Sum_{j=1..n-1} Product_{k=1..j} a(k)/A274225(k), for n>1. - _Andres Cicuttin_, Apr 26 2017
%t Table[(Prime[j+2]-Prime[j+1])/(Prime[j+1]-Prime[j]),{j,1,120}]//Numerator
%t Numerator[#[[2]]/#[[1]]]&/@Partition[Differences[Prime[Range[100]]],2,1] (* _Harvey P. Dale_, Dec 07 2022 *)
%o (Magma) [Numerator((NthPrime(n+2)-NthPrime(n+1))/(NthPrime(n+1)-NthPrime(n))): n in [1..100]]; // _Vincenzo Librandi_, Apr 27 2017
%o (PARI) a(n) = numerator((prime(n+2)-prime(n+1))/(prime(n+1)-prime(n))); \\ _Michel Marcus_, Apr 29 2017
%Y Cf. A000040, A001223, A274225 (denominators), A274263.
%Y Cf. A184247 (primes of the form prime(k+1) when a(k)=1).
%K nonn,frac
%O 1,1
%A _Andres Cicuttin_, Jun 21 2016
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