[go: up one dir, main page]

login
A055874
a(n) = largest m such that 1, 2, ..., m divide n.
43
1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2
OFFSET
1,2
COMMENTS
From Antti Karttunen, Nov 20 2013 & Jan 26 2014: (Start)
Differs from A232098 for the first time at n=840, where a(840)=8, while A232098(840)=7. A232099 gives all the differing positions. See also the comments at A055926 and A232099.
The positions where a(n) is an odd prime is given by A017593 up to A017593(34)=414 (so far all 3's), after which comes the first 7 at a(420). (A017593 gives the positions of 3's.)
(Continued on Jan 26 2014):
Only terms of A181062 occur as values.
A235921 gives such n where a(n^2) (= A235918(n)) differs from A071222(n-1) (= A053669(n)-1). (End)
a(n) is the largest m such that A003418(m) divides n. - David W. Wilson, Nov 20 2014
a(n) is the largest number of consecutive integers dividing n. - David W. Wilson, Nov 20 2014
A051451 gives indices where record values occur. - Gionata Neri, Oct 17 2015
Yuri Matiyasevich calls this the maximum inheritable divisor of n. - N. J. A. Sloane, Dec 14 2023
LINKS
FORMULA
a(n) = A007978(n) - 1. - Antti Karttunen, Jan 26 2014
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A064859 (Farhi, 2009). - Amiram Eldar, Jul 25 2022
EXAMPLE
a(12) = 4 because 1, 2, 3, 4 divide 12, but 5 does not.
MAPLE
N:= 1000: # to get a(1) to a(N)
A:= Vector(N, 1);
for m from 2 do
Lm:= ilcm($1..m);
if Lm > N then break fi;
if Lm mod (m+1) = 0 then next fi;
for k from 1 to floor(N/Lm) do
A[k*Lm]:=m
od
od:
convert(A, list); # Robert Israel, Nov 28 2014
MATHEMATICA
a[n_] := Module[{m = 1}, While[Divisible[n, m++]]; m - 2]; Array[a, 100] (* Jean-François Alcover, Mar 07 2016 *)
PROG
(Haskell)
a055874 n = length $ takeWhile ((== 0) . (mod n)) [1..]
-- Reinhard Zumkeller, Feb 21 2012, Dec 09 2010
(Scheme)
(define (A055874 n) (let loop ((m 1)) (if (not (zero? (modulo n m))) (- m 1) (loop (+ 1 m))))) ;; Antti Karttunen, Nov 18 2013
(PARI) a(n) = my(m = 1); while ((n % m) == 0, m++); m - 1; \\ Michel Marcus, Jan 17 2014
(Python)
from itertools import count
def A055874(n):
for m in count(1):
if n % m:
return m-1 # Chai Wah Wu, Jan 02 2022
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Leroy Quet, Jul 16 2000
STATUS
approved