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A389160
a(n) = A350388(A228058(n)), where A350388(x) is the largest unitary divisor of x that is a square, and A228058 gives the numbers satisfying Euler's condition for odd perfect numbers.
4
9, 9, 9, 49, 9, 25, 9, 9, 81, 25, 9, 9, 121, 49, 9, 25, 9, 49, 169, 9, 9, 25, 9, 9, 25, 81, 9, 25, 9, 81, 9, 49, 289, 25, 9, 121, 9, 9, 9, 361, 49, 25, 49, 121, 9, 9, 9, 441, 25, 9, 81, 9, 25, 9, 25, 9, 49, 9, 529, 25, 9, 25, 9, 169, 225, 49, 81, 9, 9, 9, 81, 9, 25, 9, 121, 9, 49, 9, 729, 9, 25, 289, 9, 225, 9, 25
OFFSET
1,1
COMMENTS
For all odd numbers of the form p^(1+4k) * r^2, where p is prime of the form 1+4m (so called "special prime"), r > 1, and gcd(p,r) = 1, a(n) = r^2.
LINKS
FORMULA
a(n) = A228058(n) / A389161(n).
PROG
(PARI)
up_to = 20000;
isA228058(n) = if(!(n%2)||(omega(n)<2), 0, my(f=factor(n), y=0); for(i=1, #f~, if(1==(f[i, 2]%4), if((1==y)||(1!=(f[i, 1]%4)), return(0), y=1), if(f[i, 2]%2, return(0)))); (y));
A228058list(up_to) = { my(v=vector(up_to), k=0, n=0); while(k<up_to, n++; if(isA228058(n), k++; v[k] = n)); (v); };
v228058 = A228058list(up_to);
A228058(n) = v228058[n];
A350388(n) = { my(m=1, f=factor(n)); for(k=1, #f~, if(0==(f[k, 2]%2), m *= (f[k, 1]^f[k, 2]))); (m); };
CROSSREFS
Cf. A228058, A350388, A389161, A389200 [= sigma(a(n))].
Differs from A325320 at n = 9, 26, 30, 48, 51, 65, 67, 71, 79, 84, 91, 105, ...
Differs from A389166 first at n=520, where a(520) = 9, while A389166(520) = 5625.
Sequence in context: A160761 A082049 A282242 * A389166 A325320 A282910
KEYWORD
nonn
AUTHOR
Antti Karttunen, Sep 28 2025
STATUS
approved