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A080765
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Integers m such that m+1 divides lcm(1 through m).
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10
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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
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OFFSET
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1,1
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COMMENTS
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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 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)
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LINKS
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FORMULA
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EXAMPLE
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17 is the sequence because lcm(1,2,...,17)=12252240 and 17+1=18 divides 12252240.
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MATHEMATICA
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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