A060681 Largest difference between consecutive divisors of n (ordered by size).

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

Author

Labos Elemer, Apr 19 2001

Keywords

• nonn
• easy

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

Examples

			For n = 35, divisors are {1, 5, 7, 35}; differences are {4, 2, 28}; a(35) = largest difference = 28 = 35 - 35/5.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Antal Balog, Paul Erdős, and Gérald Tenenbaum,, On Arithmetic Functions Involving Consecutive Divisors, In: Analytical Number Theory, pp. 77-90, Birkhäuser, Basel, 1990.

Crossrefs

Cf. A013661, A020639, A060680, A060682, A060683, A060685, A064097 (number of iterations needed to reach 1).

Cf. also A171462, A366387.

Programs

Haskell

a060681 n = div n p * (p - 1) where p = a020639 n
-- Reinhard Zumkeller, Apr 06 2015

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 *)

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

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

Sum_{k=1..n} a(k) ~ (1/2 - c) * n^2, where c is defined in the corresponding formula in A032742. - Amiram Eldar, Dec 21 2024

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