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A090521
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Least k such that floor[n!/k] is prime.
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2
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1, 2, 7, 7, 19, 22, 17, 74, 29, 76, 67, 49, 31, 43, 95, 59, 31, 41, 173, 79, 94, 73, 233, 107, 73, 29, 43, 201, 89, 274, 191, 349, 346, 199, 173, 249, 89, 373, 662, 197, 453, 166, 257, 865, 487, 254, 149, 852, 758, 389, 181, 151, 699, 634, 577, 542, 199, 61, 278, 482, 467
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OFFSET
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2,2
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COMMENTS
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Conjecture: There exists a number k such that for all n > k, a(n) is prime. Motivation: If p is the least prime >n then all the numbers from n to p-1 divide n!. And most of the numbers from p+1 to q also divide n! where q is the least prime > p, etc. and the dividend is composite in almost all cases.
I conjecture the opposite: there are infinitely many composites in this sequence. Indeed, heuristics suggest the composites may be of density 1. - Charles R Greathouse IV, Apr 07 2013
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LINKS
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EXAMPLE
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a(5)=7 because the first values of floor(5!/k) for k=1,2,... are 120,60,40,30,24,20,17,15,13,12,... and among these the first prime number is 17, corresponding to k=7.
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MAPLE
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a:= proc(n) local k; for k while not isprime(floor(n!/k)) do od; k end:
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MATHEMATICA
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lk[n_]:=Module[{f=n!, k=1}, While[!PrimeQ[Floor[f/k]], k++]; k]; Array[lk, 70, 2] (* Harvey P. Dale, Mar 06 2015 *)
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PROG
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(PARI) a(n)=my(N=n!); for(k=if(n>3, n+1, n-1), N\3+2, if(ispseudoprime(N\k), return(k))) \\ Charles R Greathouse IV, Apr 04 2013
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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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