A070198 Smallest nonnegative number m such that m == i (mod i+1) for all 1 <= i <= n.

0, 1, 5, 11, 59, 59, 419, 839, 2519, 2519, 27719, 27719, 360359, 360359, 360359, 720719, 12252239, 12252239, 232792559, 232792559, 232792559, 232792559, 5354228879, 5354228879, 26771144399, 26771144399, 80313433199, 80313433199

Offset: 0

Author

Benoit Cloitre, May 06 2002

Keywords

• easy
• nonn

Comments

Also, smallest k such that, for 0 <= i < n, i+1 divides k-i.

Suggested by Chinese Remainder Theorem. This sequence can generate others: smallest b(n) such that b(n) == i (mod (i+2)), 1 <= i <= n, gives b(1)=1 and b(n) = a(n+1)-1 for n > 1; smallest c(n) such that c(n) == i (mod (i+3)), 1 <= i <= n, gives c(1)=1, c(2)=17 and c(n) = a(n+2) - 2 for n > 2; smallest d(n) such that c(n) == i (mod (i+4)), 1 <= i <= n, gives d(1)=1, d(2)=26, d(3)=206 and d(n) = a(n+3) - 3 for n > 3, etc.

A208768(n) occurs A057820(n) times. - Reinhard Zumkeller, Mar 01 2012

From Kival Ngaokrajang, Oct 10 2013: (Start)

A070198(n-1) is m such that max(Sum_{i=1..n} m (mod i)) = A000217(n-1).

Example for n = 3:

m\i = 1 2 3 sum

1 0 1 1 2

2 0 0 2 2

3 0 1 0 1

4 0 0 1 1

5 0 1 2 3 <--max remainder sum = 3 = A000217(2)

6 0 0 0 0 first occurs at m = 5 = A070198(2)

(End)

Examples

			a(3) = 11 because 11 == 1 (mod 2), 11 == 2 (mod 3) and 11 == 3 (mod 4).
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• Eric Weisstein's World of Mathematics, Chinese Remainder Theorem
• Wikipedia, Chinese remainder theorem
• Index entries for sequences related to lcm's

Crossrefs

Cf. A053664, A072562.

Cf. A057825 (indices of primes). - R. J. Mathar, Jan 14 2009

Cf. A116151. - Zak Seidov, Mar 11 2014

Programs

Haskell

a070198 n = a070198_list !! n
a070198_list = map (subtract 1) $ scanl lcm 1 [2..]
-- Reinhard Zumkeller, Mar 01 2012

Magma

[Exponent(SymmetricGroup(n))-1 : n in [1..30]]; /* Vincenzo Librandi, Oct 31 2014 - after Arkadiusz Wesolowski in A003418 */

Maple

seq(ilcm($1..n) - 1, n=1..100); # Robert Israel, Nov 03 2014

Mathematica

f[n_] := ChineseRemainder[ Range[0, n - 1], Range[n]]; Array[f, 28] (* or *)
f[n_] := LCM @@ Range@ n - 1; Array[f, 28] (* Robert G. Wilson v, Oct 30 2014 *)

Python

from math import lcm
def A070198(n): return lcm(*range(1,n+2))-1 # Chai Wah Wu, May 02 2023

Formula

a(n) = lcm(1, 2, 3, ..., n+1) - 1 = A003418(n+1) - 1.

Extensions

Edited by N. J. A. Sloane, Nov 18 2007, at the suggestion of Max Alekseyev