The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A080765 Integers m such that m+1 divides lcm(1 through m). 10
5, 9, 11, 13, 14, 17, 19, 20, 21, 23, 25, 27, 29, 32, 33, 34, 35, 37, 38, 39, 41, 43, 44, 45, 47, 49, 50, 51, 53, 54, 55, 56, 57, 59, 61, 62, 64, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 81, 83, 84, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 97, 98, 99, 101, 103, 104, 105, 107 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Integers m for which A003418(m) = A003418(m+1).
a(n) = A024619(n) - 1. Proof:
If N+1 is a power of a prime (N+1=P^K), then only smaller powers of that prime divide numbers up to N and so lcm(1..N) doesn't have K powers of P; that is, N+1=P^K doesn't divide lcm(1..N).
From Don Reble, Mar 12 2003: (Start)
If N+1 is not a power of a prime, then it has at least two prime factors. Call one of them P, let K be such that P^K divides N+1, but P^(K+1) doesn't, and let N+1=P^K*R. Then
- R is greater than 1 because it is divisible by another prime factor of N+1;
- P^K and R are each less than N+1 because the other is greater than one;
- lcm(P^K,R) divides lcm(1..N) because 1..N includes both numbers;
- lcm(P^K,R)=N+1 because P doesn't divide R;
- N+1 divides lcm(1..N). (End)
LINKS
Andrei Asinowski, Cyril Banderier, Benjamin Hackl, Flip-sort and combinatorial aspects of pop-stack sorting, arXiv:2003.04912 [math.CO], 2020.
FORMULA
a(n) ~ n. - David A. Corneth, Aug 30 2019
EXAMPLE
17 is the sequence because lcm(1,2,...,17)=12252240 and 17+1=18 divides 12252240.
MATHEMATICA
Select[Range[120], Divisible[LCM @@ Range[#], #+1]&] (* Jean-François Alcover, Jun 21 2018 *)
PROG
(PARI) a=1; for(n=1, 108, a=lcm(a, n); if(a%(n+1)==0, print1(n, ", "))) \\ Klaus Brockhaus, Jun 11 2004
(PARI) first(n) = {my(u = max(2*n, 50), charact = vector(u, i, 1), res = List()); forprime(p = 2, 2*n, for(t = 1, logint(u, p), charact[p^t - 1] = 0)); for(i = 1, u, if(charact[i] == 1, listput(res, i); if(#res >= n, return(res)))); res } \\ David A. Corneth, Aug 30 2019
(Sage)
[x - 1 for x in (1..108) if not is_prime_power(n)] # Peter Luschny, May 23 2013
CROSSREFS
Cf. A003418.
Sequence in context: A049049 A118358 A101731 * A226039 A257292 A234285
KEYWORD
nonn,easy
AUTHOR
Lekraj Beedassy, Mar 10 2003
EXTENSIONS
More terms from Klaus Brockhaus, Jun 11 2004
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified May 17 12:10 EDT 2024. Contains 372600 sequences. (Running on oeis4.)