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A351545 a(n) is the largest unitary divisor of sigma(n) such that its every prime factor p also divides A003961(n), and valuation(sigma(n),p) >= valuation(A003961(n),p).

%I A351545 #11 Feb 17 2022 00:02:05
%S A351545 1,3,1,1,1,3,1,1,1,9,1,1,1,3,1,1,1,3,1,7,1,9,1,5,1,3,1,1,1,9,1,1,1,27,
%T A351545 1,1,1,3,1,1,1,3,1,1,1,9,1,1,1,3,1,1,1,3,1,1,5,9,1,7,1,3,1,1,7,9,1,9,
%U A351545 1,9,1,1,1,3,1,1,1,3,1,1,1,9,1,1,1,3,5,1,1,9,1,1,1,9,1,1,1,9,13,1,1,27
%N A351545 a(n) is the largest unitary divisor of sigma(n) such that its every prime factor p also divides A003961(n), and valuation(sigma(n),p) >= valuation(A003961(n),p).
%H A351545 Antti Karttunen, <a href="/A351545/b351545.txt">Table of n, a(n) for n = 1..65537</a>
%H A351545 <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%H A351545 <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>
%F A351545 a(n) = Product_{p^e || A000203(n)} p^(e*[p divides A003961(n) but p^(1+e) does not divide A003961(n)]), where [ ] is the Iverson bracket, returning 1 if the condition holds, and 0 otherwise. Here p^e is the largest power of prime p dividing sigma(n).
%F A351545 a(n) = A000203(n) / A351547(n).
%F A351545 For all n >= 1, a(n) is a divisor of A351544(n), which is a divisor of A000203(n).
%o A351545 (PARI)
%o A351545 A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
%o A351545 A351545(n) = { my(s=sigma(n),f=factor(s),u=A003961(n)); prod(k=1,#f~,if(!(u%f[k,1]) && (f[k,2]>=valuation(u,f[k,1])), f[k,1]^f[k,2], 1)); };
%Y A351545 Cf. A000203, A003961, A351544, A351547.
%K A351545 nonn
%O A351545 1,2
%A A351545 _Antti Karttunen_, Feb 16 2022