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Numbers k for which bigomega(k) == 1 (mod 4).
5

%I #11 Nov 30 2022 16:11:28

%S 2,3,5,7,11,13,17,19,23,29,31,32,37,41,43,47,48,53,59,61,67,71,72,73,

%T 79,80,83,89,97,101,103,107,108,109,112,113,120,127,131,137,139,149,

%U 151,157,162,163,167,168,173,176,179,180,181,191,193,197,199,200,208,211,223,227,229,233,239,241,243

%N Numbers k for which bigomega(k) == 1 (mod 4).

%C Numbers k such that number of their prime factors (when counted with multiplicity, with A001222) is of the form 4n+1: 1, 5, 9, 13, 17, ..., A016813.

%C Equally, numbers k for which A349905(k) == 1 (mod 4).

%F {k | A010873(A001222(k)) = 1}.

%e 48 = 2^4 * 3 has 5 prime factors in total, and 5 is a number of the form 5n+1 (in A016813), therefore 48 is included in this sequence. Or equivalently, because A349905(48) = 621 = 4*155 + 1.

%o (PARI) isA358761(n) = A358751(n);

%Y Cf. A000040 (subsequence), A001222, A003415, A003961, A010051, A010873, A016813, A349905, A358751 (characteristic function).

%Y Setwise difference A026424 \ A358763.

%Y Cf. also A358760, A358762.

%K nonn

%O 1,1

%A _Antti Karttunen_, Nov 29 2022