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A048247
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Every prime occurs to this power in some factorial.
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3
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0, 1, 4, 8, 10, 18, 22, 26, 32, 34, 46, 49, 50, 57, 66, 70, 74, 81, 82, 86, 94, 102, 130, 134, 138, 142, 152, 162, 165, 166, 174, 176, 183, 184, 201, 205, 206, 222, 231, 232, 236, 237, 244, 246, 256, 270, 273, 274, 286, 290, 296, 304, 312, 318, 326
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OFFSET
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0,3
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COMMENTS
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There are no primes in the sequence, as the prime p fails the base p test. The set of positive integers failing the base p test for membership has density 1/p. Also, when n is a nonmember of the set, any base p whose test n fails has p<=n. Therefore one conjectural estimate for the number of members of the set <=x would be x*product{primes p<=x}(1-1/p) ~ e^(-gamma)*x/log(x). However, a similar heuristic for the primes fails, as pi(x) ~ x/log(x) and not e^(-gamma)*x/log(x). Here gamma denotes the Euler-Mascheroni constant. - David L. Harden, Aug 24 2002
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REFERENCES
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David L. Harden, posting to sci.math newsgroup, Jun 06 1999.
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LINKS
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FORMULA
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Numbers passing the test for membership for the base p are generated by W_p(x) = product_{n=1..inf} (x^(p*(p^n-1)/(p-1))-1)/(x^((p^n-1)/(p-1))-1). - David L. Harden
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EXAMPLE
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Given any prime p, there exists a positive integer n such that p|n! but p^2 does not divide n!.
Given any prime p, there exists a positive integer n such that p^4|n! but p^5 does not divide n!.
But it is not true that given any prime p, there exists a positive integer n such that p^6|n! but p^7 does not divide n! (for if 64|n! then 128|n!).
For every prime p there is an n such that p^4|n! but p^5 doesn't divide n!: for p=2, we may take n=6; for p=3, we may take n=9; for p>4, we may take n=4p.
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MATHEMATICA
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m = 330;
w[p_] := Product[(x^(p(p^n-1)/(p-1))-1)/(x^((p^n-1)/(p-1))-1), {n, 1, 8}];
T = Select[Table[Exponent[#, x]& /@ List @@ (w[p] + O[x]^m // Normal), {p, Prime[Range[PrimePi[m]]]}], #[[1]] == 0&];
okQ[n_] := AllTrue[T, MemberQ[#, n]&];
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr)
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EXTENSIONS
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STATUS
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approved
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