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A007684
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Prime(n)*...*prime(a(n)) is the least product of consecutive primes that is non-deficient.
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8
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2, 6, 11, 21, 35, 51, 73, 98, 130, 167, 204, 249, 296, 347, 406, 471, 538, 608, 686, 768, 855, 950, 1050, 1156, 1266, 1377, 1495, 1621, 1755, 1898, 2049, 2194, 2347, 2504, 2670, 2837, 3013, 3194, 3380, 3573, 3771, 3974, 4187, 4401, 4625, 4856
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OFFSET
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1,1
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COMMENTS
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Subscript of the smallest primorial number that when divided by the (n-1)-th primorial number gives an abundant number.
Products of consecutive primes started with prime(a) up to prime(b) result in abundant squarefree numbers if b is large enough and provides perhaps the least squarefree solutions to Rivera Puzzle 329 and its generalization.
Adding a new prime p to the product increases the relative abundancy sigma(N)/N by a factor 1+1/p. This leads to a simple and fast algorithm, see the PARI code. - M. F. Hasler, Jul 30 2016
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LINKS
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FORMULA
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a(n) is the minimal x such that floor(sigma(p#(x)/p#(n-1)) / (p#(x)/p#(n-1))) = 2, where p#(w) is the w-th primorial number, the product of first w prime numbers. For a>b, the p#(a)/p#(b)=A002110(a)/A002110(b) quotients are prime(b+1)*prime(b+2)*...*prime(a).
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EXAMPLE
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n=1: a(1)=2 means that primorial(2)=6 divided by primorial(1-1)=1 gives the quotient 6/1=6 which is just non-deficient (being a perfect number);
n=3: a(n)=11 because prime(3)=5, primorial(11) = 2*3*5*...*29*31, primorial(3-1) = 2*3 = 6.
p#(11)/p#(2) = 3*5*7*11*13*17*19*23*29*31 = 33426748355 = q and sigma(q)/q = 2.00097 > 2 so q is an abundant number. Also p#(10)/p#(3-1) is not yet abundant.
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MATHEMATICA
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spr[x_, y_] :=Apply[Times, Table[(Prime[w]+1)/(Prime[w]), {w, x, y}]];
Table[Min[Flatten[Position[Table[Floor[spr[n, w]], {w, 1, 1000}], 2]]], {n, 1, 20}] (* Labos Elemer, Sep 19 2005 *)
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PROG
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(PARI) a=1; i=0; for(n=1, 99, while(2>a*=1+1/prime(i++), ); print1(i", "); a/=1+1/prime(n)) \\ M. F. Hasler, Jul 30 2016
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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