A258409 Greatest common divisor of all (d-1)'s, where the d's are the positive divisors of n.

1, 2, 1, 4, 1, 6, 1, 2, 1, 10, 1, 12, 1, 2, 1, 16, 1, 18, 1, 2, 1, 22, 1, 4, 1, 2, 1, 28, 1, 30, 1, 2, 1, 2, 1, 36, 1, 2, 1, 40, 1, 42, 1, 2, 1, 46, 1, 6, 1, 2, 1, 52, 1, 2, 1, 2, 1, 58, 1, 60, 1, 2, 1, 4, 1, 66, 1, 2, 1, 70, 1, 72, 1, 2, 1, 2, 1, 78, 1

Offset: 2

Author

Ivan Neretin, May 29 2015

Keywords

• nonn

Comments

a(n) = 1 for even n; a(p) = p-1 for prime p.

a(n) is even for odd n (since all divisors of n are odd).

It appears that a(n) = A052409(A005179(n)), i.e., it is the largest integer power of the smallest number with exactly n divisors. - Michel Marcus, Nov 10 2015

Conjecture: GCD of all (p-1) for prime p|n. - Thomas Ordowski, Sep 14 2016

Conjecture is true, because the set of numbers == 1 (mod g) is closed under multiplication. - Robert Israel, Sep 14 2016

Conjecture: a(n) = A289508(A328023(n)) = GCD of the differences between consecutive divisors of n. See A328163 and A328164. - Gus Wiseman, Oct 16 2019

Examples

			65 has divisors 1, 5, 13, and 65, hence a(65) = gcd(1-1,5-1,13-1,65-1) = gcd(0,4,12,64) = 4.
		

Links

• Ivan Neretin, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A049559, A057237, A060680, A063994, A187730, A027750.

Cf. A084190 (similar but with LCM).

Looking at prime indices instead of divisors gives A328167.

Partitions whose parts minus 1 are relatively prime are A328170.

Cf. A000005, A060681, A060683, A193829, A289508.

Programs

Haskell

a258409 n = foldl1 gcd $ map (subtract 1) $ tail $ a027750_row' n
-- Reinhard Zumkeller, Jun 25 2015

Maple

f:= n -> igcd(op(map(`-`,numtheory:-factorset(n),-1))):
map(f, [$2..100]); # Robert Israel, Sep 14 2016

Mathematica

Table[GCD @@ (Divisors[n] - 1), {n, 2, 100}]

PARI

a(n) = my(g=0); fordiv(n, d, g = gcd(g, d-1)); g; \\ Michel Marcus, May 29 2015

PARI

a(n) = gcd(apply(x->x-1, divisors(n))); \\ Michel Marcus, Nov 10 2015

PARI

a(n)=if(n%2==0, return(1)); if(n%3==0, return(2)); if(n%5==0 && n%4 != 1, return(2)); gcd(apply(p->p-1, factor(n)[,1])) \\ Charles R Greathouse IV, Sep 19 2016