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A258409
Greatest common divisor of all (d-1)'s, where the d's are the positive divisors of n.
17
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
2,2
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
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
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.
Sequence in context: A281071 A256908 A346466 * A060680 A057237 A187730
KEYWORD
nonn
AUTHOR
Ivan Neretin, May 29 2015
STATUS
approved