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%I #65 Jun 11 2024 03:24:53
%S 3,5,541,997,328753,5385217,1287133,9483302497,107887226353
%N Least starting prime of exactly n consecutive primes p_i (i = 1..n) such that bigomega(p_i + 1) = 1 + i.
%e a(1) = 3, because bigomega(3+1) = 2 and no lesser number has this property.
%e a(2) = 5, because bigomega(5+1) = 2 and bigomega(7+1) = 3, and no lesser number has this property.
%e a(3) = 541, because bigomega(541+1) = 2, bigomega(547+1) = 3, bigomega(557+1) = 4 and no lesser number has this property.
%o (PARI) isok(p, n) = if (bigomega(p+1) != 2, return(0)); for (i=1, n-1, p = nextprime(p+1); if (bigomega(p+1) != i+2, return(0))); if (bigomega(nextprime(p+1)+1) == n+2, return(0)); return(1);
%o a(n) = my(p=2); while (!isok(p, n), p = nextprime(p+1)); p; \\ _Michel Marcus_, Jun 07 2024
%Y Cf. A001222, A072875.
%K nonn,more
%O 1,1
%A _Jean-Marc Rebert_, Jun 07 2024
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