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

1,3

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

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

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

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

Adjacent sequences: A060678 A060679 A060680 * A060682 A060683 A060684

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