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

1,1

Comments

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

References

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 61, p. 22, Ellipses, Paris 2008.

Links

Table of n, a(n) for n=1..10.

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.

Example

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.

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 *)

Crossrefs

Cf. A001222, A093552, A093550, A086560, A124592.

a(1) = A000040(1), a(2) = A005383(1), a(3) = A112998(1), a(4) = A113000(1), a(5) = A113008(1), a(6) = A113150(1).

Sequence in context: A144545 A085326 A062308 * A093551 A173915 A293041

Adjacent sequences: A072872 A072873 A072874 * A072876 A072877 A072878

Keyword

hard,nice,nonn,more

Author

Rick L. Shepherd, Jun 30 2002 and Jens Kruse Andersen, Jul 28 2002

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

Status

approved