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A140761
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Primes p(j)=A000040(j), j>1, such that p(1)*p(2)*...*p(j) is an integral multiple of p(1)+p(2)+...+p(j).
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5
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5, 19, 41, 83, 163, 167, 179, 191, 223, 229, 241, 263, 269, 271, 317, 337, 349, 367, 389, 433, 463, 491, 521, 701, 719, 757, 809, 823, 829, 859, 877, 883, 919, 941, 971, 991, 997, 1021, 1033, 1049, 1091, 1153, 1181, 1193, 1223, 1291, 1301, 1319, 1327, 1361
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OFFSET
| 1,1
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FORMULA
| Find integral quotients of products of consecutive primes divided by their sum.
a(n)=A000040(A051838(n)). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 09 2008
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EXAMPLE
| a(1)=5 because it is the last consecutive prime in the run 2*3*5=30 and 2+3+5=10; since 30/10=3, it is the first integral quotient.
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CROSSREFS
| Cf. A159578, A140763 A116536.
Sequence in context: A125202 A024841 A155737 * A100572 A119534 A033622
Adjacent sequences: A140758 A140759 A140760 * A140762 A140763 A140764
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KEYWORD
| easy,nonn
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AUTHOR
| Enoch Haga (Enokh(AT)comcast.net), May 28 2008
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EXTENSIONS
| Edited by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 09 2008
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