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A060681 - OEIS
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A060681
Largest difference between consecutive divisors of n (ordered by size).
70
0, 1, 2, 2, 4, 3, 6, 4, 6, 5, 10, 6, 12, 7, 10, 8, 16, 9, 18, 10, 14, 11, 22, 12, 20, 13, 18, 14, 28, 15, 30, 16, 22, 17, 28, 18, 36, 19, 26, 20, 40, 21, 42, 22, 30, 23, 46, 24, 42, 25, 34, 26, 52, 27, 44, 28, 38, 29, 58, 30, 60, 31, 42, 32, 52, 33, 66, 34, 46, 35, 70, 36, 72, 37
OFFSET
1,3
COMMENTS
Is a(n) the least m > 0 such that n - m divides n! + m? - Clark Kimberling, Jul 28 2012
Is a(n) the least m > 0 such that L(n-m) divides L(n+m), where L = A000032 (Lucas numbers)? - Clark Kimberling, Jul 30 2012
Records give A006093. - Omar E. Pol, Oct 26 2013
Divide n by its smallest prime factor p, then multiply with (p-1), with a(1) = 0 by convention. Compare also to A366387. - Antti Karttunen, Oct 23 2023
a(n) is also the smallest LCM of positive integers x and y where x + y = n. - Felix Huber, Aug 28 2024
LINKS
A. Balog, P. Erdős and G. Tenenbaum, On Arithmetic Functions Involving Consecutive Divisors, In: Analytical Number Theory, pp. 77-90, Birkhäuser, Basel, 1990.
FORMULA
a(n) = n - n/A020639(n).
a(n) = n - A032742(n). - Omar E. Pol, Aug 31 2011
a(2n) = n, a(3*(2n+1)) = 2*(2n+1) = 4n + 2. - Antti Karttunen, Oct 23 2023
EXAMPLE
For n = 35, divisors are {1, 5, 7, 35}; differences are {4, 2, 28}; a(35) = largest difference = 28 = 35 - 35/5.
MAPLE
read("transforms") :
A060681 := proc(n)
if n = 1 then
0 ;
else
sort(convert(numtheory[divisors](n), list)) ;
DIFF(%) ;
max(op(%)) ;
end if;
end proc:
seq(A060681(n), n=1..60) ; # R. J. Mathar, May 23 2018
# second Maple program:
A060681:=n->if(n=1, 0, min(map(x->ilcm(x, n-x), [$1..1/2*n]))); seq(A060681(n), n=1..74); # Felix Huber, Aug 28 2024
MATHEMATICA
a[n_ ] := n - n/FactorInteger[n][[1, 1]]
Array[Max[Differences[Divisors[#]]] &, 80, 2] (* Harvey P. Dale, Oct 26 2013 *)
PROG
(Haskell)
a060681 n = div n p * (p - 1) where p = a020639 n
-- Reinhard Zumkeller, Apr 06 2015
(PARI) diff(v)=vector(#v-1, i, v[i+1]-v[i])
a(n)=vecmax(diff(divisors(n))) \\ Charles R Greathouse IV, Sep 02 2015
(PARI) a(n) = if (n==1, 0, n - n/factor(n)[1, 1]); \\ Michel Marcus, Oct 24 2015
(PARI) first(n) = n = max(n, 1); my(res = vector(n)); res[1] = 0; forprime(p = 2, n, for(i = 1, n \ p, if(res[p * i] == 0, res[p * i] = i*(p-1)))); res \\ David A. Corneth, Jan 08 2019
(Python)
from sympy import primefactors
def A060681(n): return n-n//min(primefactors(n), default=1) # Chai Wah Wu, Jun 21 2023
CROSSREFS
Cf. A020639, A060680, A060682, A060683, A060685, A064097 (number of iterations needed to reach 1).
Cf. also A171462, A366387.
Sequence in context: A067240 A126080 A302043 * A202479 A161660 A343327
KEYWORD
nonn,easy
AUTHOR
Labos Elemer, Apr 19 2001
EXTENSIONS
Edited by Dean Hickerson, Jan 22 2002
a(1)=0 added by N. J. A. Sloane, Oct 01 2015 at the suggestion of Antti Karttunen
STATUS
approved