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A046021
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Least inverse of the Kempner function A002034.
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9
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1, 2, 3, 4, 5, 9, 7, 32, 27, 25, 11, 243, 13, 49, 125, 4096, 17, 2187, 19, 625, 343, 121, 23, 59049, 3125, 169, 177147, 2401, 29, 78125, 31, 134217728, 1331, 289, 16807, 43046721, 37, 361, 2197, 1953125, 41, 117649, 43, 14641, 9765625, 529, 47
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OFFSET
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1,2
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COMMENTS
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To compute the n-th term for n > 1: For each prime p that divides n, find the highest power p^E(p) that divides (n-1)!. Then a(n) is the smallest of the numbers p^(E(p)+1). - Jonathan Sondow, Mar 03 2004
p^(E(p)+1) is smallest when p = P(n), the largest prime dividing n (since E(p) is approximately p^((n-1)/(p-1)), which decreases as p increases). So a(n) = P(n)^(E(P(n))+1) = A006530(n)^A102048(n) for n>1. - Jonathan Sondow, Dec 26 2004
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REFERENCES
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R. L. Graham, D. E. Knuth and O. Patashnik, "Factorial Factors" Sect. 4.4 in Concrete Mathematics: A Foundation for Computer Science, 2nd ed. Reading, MA: Addison-Wesley, pp. 111-115, 1994.
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LINKS
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Eric Weisstein's World of Mathematics, Factorial
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FORMULA
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a(n) = P^(1+Sum(k=1 to [log(n-1)/log(P)], [(n-1)/P^k])) for n>1, where P = A006530(n) is the largest prime dividing n. E.g. a(6) = 9 because 9 is the least integer m with A002034(m) = 6, that is, m divides 6!, but m does not divide k! for k < 6. - Jonathan Sondow, Dec 26 2004
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MATHEMATICA
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With[{p=First[Last[FactorInteger[n, FactorComplete->True]]]}, p^(1+Sum[Floor[(n-1)/p^k], {k, Floor[Log[n-1]/Log[p]]}])] (* Jonathan Sondow, Dec 26 2004 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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