OFFSET
2,2
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
LINKS
Ivan Neretin, Table of n, a(n) for n = 2..10000
EXAMPLE
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.
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}]
PROG
(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
(Haskell)
a258409 n = foldl1 gcd $ map (subtract 1) $ tail $ a027750_row' n
-- Reinhard Zumkeller, Jun 25 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Ivan Neretin, May 29 2015
STATUS
approved