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A351545
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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).
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4
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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, 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, 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
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OFFSET
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1,2
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LINKS
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Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences computed from indices in prime factorization
Index entries for sequences related to sigma(n)
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FORMULA
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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).
a(n) = A000203(n) / A351547(n).
For all n >= 1, a(n) is a divisor of A351544(n), which is a divisor of A000203(n).
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PROG
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(PARI)
A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
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)); };
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CROSSREFS
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Cf. A000203, A003961, A351544, A351547.
Sequence in context: A091842 A306346 A060901 * A087612 A260626 A155828
Adjacent sequences: A351542 A351543 A351544 * A351546 A351547 A351548
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KEYWORD
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nonn
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AUTHOR
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Antti Karttunen, Feb 16 2022
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STATUS
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approved
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