A080765
Don Reble
djr at nk.ca
Wed Mar 12 08:37:58 CET 2003
> %I A080765
> %S A080765 5,9,11,13,14,17,19,20,21,23,25,27
> %N A080765 m such that m+1 divides lcm(1 through m).
> %C A080765 Is a(n)=A024619(n) - 1 ?
The answer to that question is yes. (A024619 is the numbers which are
not powers of primes.)
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).
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 one, because it's 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).
--
Don Reble djr at nk.ca
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