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A258409
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Greatest common divisor of all (d-1)'s, where the d's are the positive divisors of n.
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17
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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
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OFFSET
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2,2
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COMMENTS
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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 is true, because the set of numbers == 1 (mod g) is closed under multiplication. - Robert Israel, Sep 14 2016
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LINKS
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EXAMPLE
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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.
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MAPLE
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f:= n -> igcd(op(map(`-`, numtheory:-factorset(n), -1))):
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MATHEMATICA
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Table[GCD @@ (Divisors[n] - 1), {n, 2, 100}]
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PROG
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(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
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CROSSREFS
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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.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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