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a(n) is the least k>1 such that omega(k) is equal to (omega(n*k + 1) - 1)/n.
1

%I #64 Mar 15 2022 02:59:25

%S 5,97,443,5801,42697,7813639,10303967,1225192093

%N a(n) is the least k>1 such that omega(k) is equal to (omega(n*k + 1) - 1)/n.

%C Are all terms prime numbers?

%C a(9) <= 14567138141, a(10) <= 5509396663871, a(11) <= 4128894057139, a(12) <= 13264466350165447, a(13) <= 6115610326638653. - _Daniel Suteu_, Mar 14 2022

%e a(2) = 97 because omega(97) = (omega(2*97 + 1) - 1)/2 = (omega(3*5*13) - 1)/2 = 1.

%t a[n_] := Module[{k = 2}, While[PrimeNu[k] != (PrimeNu[n*k + 1] - 1)/n, k++]; k]; Array[a, 5] (* _Amiram Eldar_, Mar 09 2022 *)

%o (PARI) a(n) = my(k=2); while (omega(k) != (omega(n*k + 1) - 1)/n, k++); k; \\ _Michel Marcus_, Mar 09 2022

%o (Python)

%o from sympy import factorint

%o for n in range(1, 8):

%o for k in range(2, 10**10):

%o if len(factorint(k).keys())*n+1==len(factorint(k*n+1).keys()):

%o print(n, k)

%o break # _Martin Ehrenstein_, Mar 14 2022

%Y Cf. A001221 (omega).

%K nonn,more

%O 1,1

%A _Juri-Stepan Gerasimov_, Mar 09 2022

%E a(8) from _Martin Ehrenstein_, Mar 14 2022