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%I #11 Feb 17 2022 00:02:10
%S 1,3,1,1,1,3,1,3,1,9,1,1,1,3,1,1,1,3,1,21,1,9,1,15,1,3,5,1,1,9,1,9,1,
%T 27,1,1,1,3,1,9,1,3,1,3,1,9,1,1,1,3,1,1,1,15,1,3,5,9,1,21,1,3,1,1,7,9,
%U 1,9,1,9,1,15,1,3,1,1,1,3,1,3,1,9,1,1,1,3,5,9,1,9,1,3,1,9,1,9,1,9,13,7,1,27
%N a(n) is the largest unitary divisor of sigma(n) such that its every prime factor also divides A003961(n), where A003961 is fully multiplicative with a(p) = nextprime(p), and sigma is the sum of divisors function.
%H Antti Karttunen, <a href="/A351544/b351544.txt">Table of n, a(n) for n = 1..65537</a>
%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>
%F a(n) = Product_{p^e || A000203(n)} p^(e*[p divides A003961(n)]), where [ ] is the Iverson bracket, returning 1 if p is a divisor of A003961(n), and 0 otherwise. Here p^e is the largest power of prime p dividing sigma(n).
%F a(n) = A000203(n) / A351546(n).
%F For all n >= 1, a(n) is a multiple of A351545(n).
%o (PARI)
%o A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
%o A351544(n) = { my(s=sigma(n),f=factor(s),u=A003961(n)); prod(k=1,#f~,if(!(u%f[k,1]), f[k,1]^f[k,2], 1)); };
%Y Cf. A000203, A003961, A351545, A351546.
%Y Cf. A342671, A349161, A349162.
%K nonn
%O 1,2
%A _Antti Karttunen_, Feb 16 2022
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