

A132468


Longest gap between numbers relatively prime to n.


4



0, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 3, 2, 1, 1, 3, 1, 3, 2, 3, 1, 3, 1, 3, 1, 3, 1, 5, 1, 1, 2, 3, 2, 3, 1, 3, 2, 3, 1, 5, 1, 3, 2, 3, 1, 3, 1, 3, 2, 3, 1, 3, 2, 3, 2, 3, 1, 5, 1, 3, 2, 1, 2, 5, 1, 3, 2, 5, 1, 3, 1, 3, 2, 3, 2, 5, 1, 3, 1, 3, 1, 5, 2, 3, 2, 3, 1, 5, 2, 3, 2, 3, 2, 3, 1, 3, 2, 3, 1, 5, 1, 3, 4
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OFFSET

1,6


COMMENTS

Here "gap" does not include the endpoints.
a(n) is given by the maximum length of a run of numbers satisfying one congruence modulo each of n's distinct prime factors. It follows that if m is the number of distinct prime factors of n and each of n's prime factors is greater than m then a(n) = m.  Thomas Anton, Dec 30 2018


LINKS

N. J. A. Sloane, Table of n, a(n) for n = 1..20000
Mario Ziller, John F. Morack, Algorithmic concepts for the computation of Jacobsthal's function, arXiv:1611.03310 [math.NT], 2016.


FORMULA

a(n) = 1 at every prime power.


EXAMPLE

E.g. n=3: the longest gap in 1, 2, 4, 5, 7, ... is 1, between 2 and 4, so a(3) = 1.


MAPLE

a:=[];
for n from 1 to 120 do
s:=[seq(j, j=1..4*n)];
rec:=0;
for st from 1 to n do
len:=0;
for i from 1 to n while gcd(s[st+i1], n)>1 do len:=len+1; od:
if len>rec then rec:=len; fi;
od:
a:=[op(a), rec];
od:
a; # N. J. A. Sloane, Apr 18 2017


MATHEMATICA

a[ n_ ] := (Max[ Drop[ #, 1 ]Drop[ #, 1 ] ]1&)[ Select[ Range[ n+1 ], GCD[ #, n ]==1& ] ]
Do[Print[n, " ", a[n]], {n, 20000}]


CROSSREFS

Equals A048669(n)  1.
See also A048670, A049298, A070791, A070194.
Sequence in context: A086869 A095345 A342671 * A243915 A309307 A325446
Adjacent sequences: A132465 A132466 A132467 * A132469 A132470 A132471


KEYWORD

nonn


AUTHOR

Michael Kleber, Nov 16 2007


EXTENSIONS

Incorrect formula removed by Thomas Anton, Dec 30 2018


STATUS

approved



