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A072875
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Smallest start for a run of n consecutive numbers of which the i-th has exactly i prime factors.
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13
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OFFSET
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1,1
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COMMENTS
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By definition, each term of this sequence is prime.
a(11) <= 1452591346605212407096281241 (Frederick Schneider), see primepuzzles link. - sent by amd64(AT)vipmail.hu, Dec 21 2007
Prime factors are counted with multiplicity. - Harvey P. Dale, Mar 09 2021
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REFERENCES
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J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 61, p. 22, Ellipses, Paris 2008.
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LINKS
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EXAMPLE
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a(3)=61 because 61 (prime), 62 (=2*31), 63 (=3*3*7) have exactly 1, 2, 3 prime factors respectively, and this is the smallest solution;
a(6)=807905281: 807905281 is prime; 807905281+1=2*403952641;
807905281+2=3*15733*17117; 807905281+3=2*2*1871*107951;
807905281+4=5*11*43*211*1619; 807905281+5=2*3*3*3*37*404357;
807905281+6=7*7*7*7*29*41*283; 807905281 is the smallest number m such that m+k is product of k+1 primes for k=0,1,2,3,4,5,6.
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MATHEMATICA
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(* This program is not suitable to compute a large number of terms. *) nmax = 6; kmax = 10^6; a[1] = 2; a[n_] := a[n] = For[k = a[n-1]+n-1, k <= kmax, k++, If[AllTrue[Range[0, n-1], PrimeOmega[k+#] == #+1&], Return[k] ] ]; Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 1, nmax}] (* Jean-François Alcover, Sep 06 2017 *)
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CROSSREFS
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KEYWORD
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hard,nice,nonn,more
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AUTHOR
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EXTENSIONS
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a(7) found by Mark W. Lewis
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STATUS
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approved
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