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Least starting prime of exactly n consecutive primes p_i (i = 1..n) such that omega(p_i + 1) = 1 + i.
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%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