A237201 Smallest integer m such that the n consecutive numbers m, m+1, ..., m+n-1 have n prime factors each, counted with multiplicity; a(n) = 0 if no such number exists.

2, 9, 170, 4023, 632148, 4843161124, 1981162639374

Offset: 1

Author

Derek Orr, Feb 04 2014

Keywords

• nonn
• hard
• more

Crossrefs

Cf. A001222.

Programs

Mathematica

Table[First@Select[Range[10^6],Union[PrimeOmega[(#+Range[n]-1)]]==={n}&,1],{n,5}] (* Wouter Meeussen, Feb 09 2014 *)
With[{po=PrimeOmega[Range[633000]]},Table[SequencePosition[po,PadRight[{},n,n],1][[1,1]],{n,5}]] (* Requires Mathematica version 10 or later *) (* The program generates the first 5 terms of the sequence. *) (* Harvey P. Dale, Jun 15 2021 *)

PARI

for(n=1, 5, for(k=2^n-1, oo, my(found=1); for(j=1, n, if(bigomega(k+j)!=n, found=0; break)); if(found, print1(k+1,", "); break))) \\ Hugo Pfoertner, Oct 21 2020

Python

import sympy
from sympy import isprime
from sympy import factorint
def PrimeFact(x):
  n = 9930000
  lst = []
  while n < 10**10:
    if not isprime(n):
      count = 0
      for i in range(n, n+x):
        if sum(factorint(i).values()) == x:
          count += 1
        else:
          n += 1
          break
      if count == x:
        return n
    else:
      n += 1

Extensions

a(6) from Giovanni Resta, Feb 09 2014

a(7) from Giovanni Resta, Feb 10 2014