A113150 -id:A113150 - cp's OEIS frontend

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

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

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

			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.

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

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.

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).

(* 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 *)

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

A248594 Semiprimes whose next four consecutive integers have exactly three, four, five, and six prime factors, respectively (allowing multiplicity of factors).

Original entry on oeis.org

153221, 196621, 222422, 230261, 288437, 307373, 340421, 400082, 657302, 660713, 706073, 723461, 777773, 838562, 843521, 954581, 961621, 988601, 1009985, 1031846, 1034933, 1190822, 1215821, 1246802, 1384621, 1409873, 1612321, 1723082, 1737122, 1886441

Offset: 1

Gil Broussard, Oct 09 2014

This sequence is related to A113150; for instance, a(14) = 838562 = A113150(1) + 1, since 838561 is prime. - Michel Marcus, Oct 23 2014

			a(1)=153221 because 153221 is a product of 2 primes (17*9013) and
153222 is a product of 3 primes (2 * 3 * 25537) and
153223 is a product of 4 primes (7 * 7 * 53 * 59) and
153224 is a product of 5 primes (2 * 2 * 2 * 107 * 179) and
153225 is a product of 6 primes (3 * 3 * 3 * 5 * 5 * 227).

Cf. A001358, A113150.

isok(n) = bigomega(n)==2 && bigomega(n+1)==3 && bigomega(n+2)==4 && bigomega(n+3)==5 && bigomega(n+4)==6; \\ Michel Marcus, Oct 23 2014

cp's OEIS Frontend

A072875 Smallest start for a run of n consecutive numbers of which the i-th has exactly i prime factors.

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