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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 (list; graph; refs; listen; history; text; internal format)
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
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+i-1], 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.
Sequence in context: A348953 A095345 A342671 * A353235 A243915 A367482
KEYWORD
nonn
AUTHOR
Michael Kleber, Nov 16 2007
EXTENSIONS
Incorrect formula removed by Thomas Anton, Dec 30 2018
STATUS
approved

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Last modified July 22 07:42 EDT 2024. Contains 374481 sequences. (Running on oeis4.)