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Search: id:A080765
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| A080765 |
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m such that m+1 divides lcm(1 through m). |
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+0
6
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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
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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a(n)=A024619(n) - 1. Proof: (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 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). - Don Reble, Mar 12, 2003
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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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PROGRAM
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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
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CROSSREFS
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m for which A003418(m)=A003418(m+1).
Cf. A003418.
Sequence in context: A049049 A118358 A101731 this_sequence A043721 A043727 A043731
Adjacent sequences: A080762 A080763 A080764 this_sequence A080766 A080767 A080768
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KEYWORD
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nonn
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AUTHOR
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Lekraj Beedassy (blekraj(AT)yahoo.com), Mar 10 2003
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EXTENSIONS
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More terms from Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Jun 11 2004
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