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%I #16 Mar 16 2023 09:17:58
%S 0,1,2,2,2,4,2,1,2,2,4,2,2,4,1,6,2,1,2,2,4,1,3,5,2,2,4,2,1,14,2,1,2,1,
%T 2,6,4,4,3,1,2,8,2,4,2,6,4,4,2,1,2,2,7,3,1,6,2,1,2,2,3,14,2,2,4,2,2,1,
%U 2,1,4,5,4,2,1,1,2,2,8,2,2,4,2,3,1,2,5,2,2,4,9,2,2,8,1,3,2
%N q + p - 1 mod q - p + 1 where p, q are consecutive primes.
%C p(n + 1) + p(n) - 1 mod p(n + 1) - p(n) + 1 where p(n) is n-th prime.
%H Harvey P. Dale, <a href="/A226534/b226534.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = A076273(n + 1) mod A076368(n + 1).
%e 0 is in the sequence because (3 + 2 - 1 mod 3 - 2 + 1) = (4 mod 2) = 0 where 2, 3 are consecutive primes.
%t Table[Mod[Prime[p + 1] + Prime[p] - 1, Prime[p + 1] - Prime[p] + 1], {p, 100}] (* _Alonso del Arte_, Jan 18 2014 *)
%t Mod[Total[#]-1,#[[2]]-#[[1]]+1]&/@Partition[Prime[Range[100]],2,1] (* _Harvey P. Dale_, Mar 16 2023 *)
%o (PARI) a(n)=lift(Mod(prime(n+1)+prime(n)-1,prime(n+1)-prime(n)+1)) /* _Ralf Stephan_, Sep 03 2013 */
%K nonn,easy
%O 1,3
%A _Juri-Stepan Gerasimov_, Aug 31 2013
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