%I #20 Jun 30 2024 19:31:54
%S 5,23,499,13093,501343,162598021,25296334003
%N Least starting prime of exactly n consecutive primes p_i (i = 1..n) such that omega(p_i + 1) = 1 + i.
%e a(1) = 5, because omega(5+1) = 2, and no lesser number has this property.
%e For n=3, the primes starting at a(3) = 499 are as follows and are a run of exactly 3 omega = i+1,
%e i = 1 2 3
%e p = 499, 503, 509, 521
%e omega(p+1) = 2, 3, 4, 3
%e \-----------/
%t a[n_]:=Module[{k=1},While[Product[Boole[PrimeNu[Prime[k+i-1]+1]==1+i],{i,n}]==0, k++]; Prime[k]]; Array[a,5] (* _Stefano Spezia_, Jun 10 2024 *)
%Y Cf. A001221, A086560, A369097.
%K nonn,more
%O 1,1
%A _Jean-Marc Rebert_, Jun 08 2024