A072875 Smallest start for a run of n consecutive numbers of which the i-th has exactly i prime factors.
2, 3, 61, 193, 15121, 838561, 807905281, 19896463921, 3059220303001, 3931520917431241
Offset: 1
Examples
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.
References
- J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 61, p. 22, Ellipses, Paris 2008.
Links
- alt.math.recreational thread, Consecutive numbers with counting prime factors.
- Carlos Rivera, Puzzle 425. Consecutive numbers, increasing quantity of prime factors, The Prime Puzzles & Problems Connection.
Crossrefs
Programs
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Mathematica
(* 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 *)
Extensions
a(7) found by Mark W. Lewis
a(8) and a(9) found by Jens Kruse Andersen
a(10) found by Jens Kruse Andersen; probably a(11) > 10^20. - Aug 24 2002
Entry revised by N. J. A. Sloane, Jan 26 2007
Cross-references and editing by Charles R Greathouse IV, Apr 20 2010
Comments