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A125168 a(n) = gcd(n, A032741(n)) where A032741(n) is the number of proper divisors of n. 3

%I

%S 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,

%T 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,

%U 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

%N a(n) = gcd(n, A032741(n)) where A032741(n) is the number of proper divisors of n.

%C 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

%C Do all values appear? - _Robert G. Wilson v_, Jan 23 2007

%C From _Bernard Schott_, Oct 19 2019: (Start)

%C a(n) = 1 if n = p^k, p prime, k >= 0 and k <> p or,

%C n = p*q, p<q primes <> 3 or

%C n = p*q*r, p<q<r primes <> 7 or,

%C n = p^2*q, p<q primes <> 5 or

%C n = p^3*q, p<q primes <> 7.

%C a(n) = 2 if n = 2^2 or n = 2^(2*p), p prime <> 2,

%C a(n) = 3 if n = 3*p, p prime <> 3 or n = 3^3,

%C a(n) = 4 if n = 4*p^2, p prime,

%C a(n) = 5 if n = 5*p^2, p prime <> 5, or n = 25*p, p prime <> 5, or n = 5^5,

%C 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,

%C a(n) = p if n = p^p, p prime. (End)

%H Antti Karttunen, <a href="/A125168/b125168.txt">Table of n, a(n) for n = 1..65537</a>

%F a(n) = gcd(n, A032741(n)) = gcd(n, A062968(n)).

%e a(6)=3 because 6 has 3 proper divisors {1,2,3} and gcd(6,3) is 3.

%t f[n_] := GCD[n, DivisorSigma[0, n] - 1]; Array[f, 105] (* _Robert G. Wilson v_ *).

%o (PARI) A125168(n) = gcd(n,numdiv(n)-1); \\ _Antti Karttunen_, Sep 25 2018

%Y Cf. A032741, A009191, A062968.

%K easy,nonn

%O 1,4

%A Mitch Cervinka (puritan(AT)toast.net), Jan 12 2007

%E More terms from _Robert G. Wilson v_, Jan 23 2007

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Last modified January 28 00:31 EST 2020. Contains 331313 sequences. (Running on oeis4.)