cp's OEIS Frontend

A080765 Integers m such that m+1 divides lcm(1 through m).

%I A080765 #39 Jun 17 2020 21:20:27
%S A080765 5,9,11,13,14,17,19,20,21,23,25,27,29,32,33,34,35,37,38,39,41,43,44,
%T A080765 45,47,49,50,51,53,54,55,56,57,59,61,62,64,65,67,68,69,71,73,74,75,76,
%U A080765 77,79,81,83,84,85,86,87,89,90,91,92,93,94,95,97,98,99,101,103,104,105,107
%N A080765 Integers m such that m+1 divides lcm(1 through m).
%C A080765 Integers m for which A003418(m) = A003418(m+1).
%C A080765 a(n) = A024619(n) - 1. Proof:
%C A080765 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).
%C A080765 From _Don Reble_, Mar 12 2003: (Start)
%C A080765 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
%C A080765 - R is greater than 1 because it is divisible by another prime factor of N+1;
%C A080765 - P^K and R are each less than N+1 because the other is greater than one;
%C A080765 - lcm(P^K,R) divides lcm(1..N) because 1..N includes both numbers;
%C A080765 - lcm(P^K,R)=N+1 because P doesn't divide R;
%C A080765 - N+1 divides lcm(1..N). (End)
%H A080765 David A. Corneth, <a href="/A080765/b080765.txt">Table of n, a(n) for n = 1..10000</a>
%H A080765 Andrei Asinowski, Cyril Banderier, Benjamin Hackl, <a href="https://arxiv.org/abs/2003.04912">Flip-sort and combinatorial aspects of pop-stack sorting</a>, arXiv:2003.04912 [math.CO], 2020.
%F A080765 a(n) ~ n. - _David A. Corneth_, Aug 30 2019
%e A080765 17 is the sequence because lcm(1,2,...,17)=12252240 and 17+1=18 divides 12252240.
%t A080765 Select[Range[120], Divisible[LCM @@ Range[#], #+1]&] (* _Jean-François Alcover_, Jun 21 2018 *)
%o A080765 (PARI) a=1;for(n=1,108,a=lcm(a,n);if(a%(n+1)==0,print1(n,","))) \\ _Klaus Brockhaus_, Jun 11 2004
%o A080765 (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
%o A080765 (Sage)
%o A080765 [x - 1 for x in (1..108) if not is_prime_power(n)]  # _Peter Luschny_, May 23 2013
%Y A080765 Cf. A003418.
%K A080765 nonn,easy
%O A080765 1,1
%A A080765 _Lekraj Beedassy_, Mar 10 2003
%E A080765 More terms from _Klaus Brockhaus_, Jun 11 2004