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A125168
a(n) = gcd(n, A032741(n)) where A032741(n) is the number of proper divisors of n.
3
1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 3, 4, 1, 1, 1, 5, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 4, 1, 1, 3, 1, 1, 7, 1, 1, 5, 1, 1, 3, 1, 5, 3, 1, 1, 1, 1, 7, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 7, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 7
OFFSET
1,4
COMMENTS
First occurrence of k: 1, 4, 6, 16, 20, 3240000, 42, 256, 162, 18662400, 132, 5308416, 832, 784, 120, 65536, 612, 2985984, 912, 1600, 9240, 98010000, 1380, 1296, 100800, ..., (10^7). - Robert G. Wilson v, Jan 23 2007
Do all values appear? - Robert G. Wilson v, Jan 23 2007
From Bernard Schott, Oct 19 2019: (Start)
a(n) = 1 if n = p^k, p prime, k >= 0 and k <> p or,
n = p*q, p<q primes <> 3 or
n = p*q*r, p<q<r primes <> 7 or,
n = p^2*q, p<q primes <> 5 or
n = p^3*q, p<q primes <> 7.
a(n) = 2 if n = 2^2 or n = 2^(2*p), p prime <> 2,
a(n) = 3 if n = 3*p, p prime <> 3 or n = 3^3,
a(n) = 4 if n = 4*p^2, p prime,
a(n) = 5 if n = 5*p^2, p prime <> 5, or n = 25*p, p prime <> 5, or n = 5^5,
a(n) = 7 if n = 7*p*q with p<q primes <> 7 or n = 7*p^3, p prime <> 7, or n = 7^7,
a(n) = p if n = p^p, p prime. (End)
LINKS
FORMULA
a(n) = gcd(n, A032741(n)) = gcd(n, A062968(n)).
EXAMPLE
a(6)=3 because 6 has 3 proper divisors {1,2,3} and gcd(6,3) is 3.
MATHEMATICA
f[n_] := GCD[n, DivisorSigma[0, n] - 1]; Array[f, 105] (* Robert G. Wilson v *).
PROG
(PARI) A125168(n) = gcd(n, numdiv(n)-1); \\ Antti Karttunen, Sep 25 2018
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Mitch Cervinka (puritan(AT)toast.net), Jan 12 2007
EXTENSIONS
More terms from Robert G. Wilson v, Jan 23 2007
STATUS
approved