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%I #20 Jul 09 2022 18:31:32
%S 1,1,4,7,6,4,8,5,13,2,12,28,14,8,24,31,18,13,20,2,32,4,24,4,31,14,8,
%T 56,30,8,32,7,48,2,48,91,38,20,56,10,42,32,44,28,78,8,48,124,57,31,72,
%U 98,54,8,72,40,16,10,60,8,62,32,104,127,12,16,68,14,96,16,72,13,74,38,124,140,96,56,80,62,121,14,84
%N a(n) is the largest unitary divisor of sigma(n) coprime with 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="/A351546/b351546.txt">Table of n, a(n) for n = 1..20000</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 does not divide A003961(n)]), where [ ] is the Iverson bracket, returning 0 if p is a divisor of A003961(n), and 1 otherwise. Here p^e is the largest power of each prime p dividing sigma(n).
%F a(n) = A000203(n) / A351544(n).
%F a(n) = A353666(n) * A353668(n) = A351547(n) / A354997(n). - _Antti Karttunen_, Jul 09 2022
%e For n = 672 = 2^5 * 3^1 * 7^1, and the largest unitary divisor of the sigma(672) [= 2^5 * 3^2 * 7^1] coprime with A003961(672) = 13365 = 3^5 * 5^1 * 11^1 is 2^5 * 7^1 = 224, therefore a(672) = 224.
%o (PARI)
%o A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
%o A351546(n) = { my(f=factor(sigma(n)),u=A003961(n)); prod(k=1,#f~,f[k,1]^((0!=(u%f[k,1]))*f[k,2])); };
%Y Cf. A000203, A003961, A351544, A351547, A351551, A351552, A353666 [gcd(a(n),n)], A353667, A353668, A353633, A354997.
%Y Cf. A342671, A349161, A349162.
%K nonn
%O 1,3
%A _Antti Karttunen_, Feb 16 2022