%I #40 Oct 16 2019 20:22:36
%S 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,
%T 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,
%U 1,4,1,66,1,2,1,70,1,72,1,2,1,2,1,78,1
%N Greatest common divisor of all (d-1)'s, where the d's are the positive divisors of n.
%C a(n) = 1 for even n; a(p) = p-1 for prime p.
%C a(n) is even for odd n (since all divisors of n are odd).
%C 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
%C Conjecture: GCD of all (p-1) for prime p|n. - _Thomas Ordowski_, Sep 14 2016
%C Conjecture is true, because the set of numbers == 1 (mod g) is closed under multiplication. - _Robert Israel_, Sep 14 2016
%C Conjecture: a(n) = A289508(A328023(n)) = GCD of the differences between consecutive divisors of n. See A328163 and A328164. - _Gus Wiseman_, Oct 16 2019
%H Ivan Neretin, <a href="/A258409/b258409.txt">Table of n, a(n) for n = 2..10000</a>
%e 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.
%p f:= n -> igcd(op(map(`-`,numtheory:-factorset(n),-1))):
%p map(f, [$2..100]); # _Robert Israel_, Sep 14 2016
%t Table[GCD @@ (Divisors[n] - 1), {n, 2, 100}]
%o (PARI) a(n) = my(g=0); fordiv(n, d, g = gcd(g, d-1)); g; \\ _Michel Marcus_, May 29 2015
%o (PARI) a(n) = gcd(apply(x->x-1, divisors(n))); \\ _Michel Marcus_, Nov 10 2015
%o (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
%o (Haskell)
%o a258409 n = foldl1 gcd $ map (subtract 1) $ tail $ a027750_row' n
%o -- _Reinhard Zumkeller_, Jun 25 2015
%Y Cf. A049559, A057237, A060680, A063994, A187730, A027750.
%Y Cf. A084190 (similar but with LCM).
%Y Looking at prime indices instead of divisors gives A328167.
%Y Partitions whose parts minus 1 are relatively prime are A328170.
%Y Cf. A000005, A060681, A060683, A193829, A289508.
%K nonn
%O 2,2
%A _Ivan Neretin_, May 29 2015