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A236411 Let p(k) denote the k-th prime; a(n) = smallest p(m) > p(n) such that the n-2 differences between [p(n), p(n+1), ..., p(2n-2)] are the same as the n-2 differences between [p(m), p(m+1), ..., p(m+n-2)].

%I A236411 #32 Oct 09 2023 19:57:53
%S A236411 5,11,13,101,37,1277,1279,1616603,57405419,51448351,76623356077,
%T A236411 115438255651991,433241801791933
%N A236411 Let p(k) denote the k-th prime; a(n) = smallest p(m) > p(n) such that the n-2 differences between [p(n), p(n+1), ..., p(2n-2)] are the same as the n-2 differences between [p(m), p(m+1), ..., p(m+n-2)].
%e A236411 n=5: We take the four primes [p(5)=11, 13, 17, 19], whose successive differences are 2, 4, 2. The next time we see this sequence of differences is at [101, 103, 107, 109], so a(5) = 101.
%t A236411 (* This program generates the first ten terms of the sequence.  To generate more would require significantly greater computing resources *) dbp[n_]:=Differences[ Prime[ Range[ n,2n-2]]]; With[{prs=Prime[Range[ 3500000]]}, First/@ Flatten[ Table[Select[Partition[Drop[prs,n],n-1,1], Differences[#]==dbp[n]&,1],{n,2,11}],1]] (* _Harvey P. Dale_, Feb 05 2014 *)
%o A236411 (PARI) A236411 = n->{d=vector(n-2,i,prime(n+i)-prime(n));
%o A236411   forprime(p=prime(n+1),,
%o A236411     for(k=1,#d, isprime(p+d[k])||next(2));
%o A236411     for(k=1,#d, p+d[k]==nextprime(p+if(k>1,d[k-1])+1)||next(2));
%o A236411   return(p))} \\ The second k-loop would suffice, but the first makes it 5x faster. Yields a(10), a(11) in ca. 3 sec (i7, 1.9Ghz). - _M. F. Hasler_, Feb 05 2014. [Erroneous ')' removed, Oct 09 2023]
%Y A236411 See A073615 for a very similar sequence.
%K A236411 nonn,more
%O A236411 2,1
%A A236411 _Don Reble_, Feb 05 2014
%E A236411 Edited by _N. J. A. Sloane_, Feb 05 2014