A064708 -id:A064708 - cp's OEIS frontend

A087977 a(n) is the first term in the first chain of at least n consecutive numbers each having exactly four distinct prime factors.

210, 7314, 37960, 134043, 357642, 1217250, 1217250, 14273478, 44939642, 76067298, 163459742, 547163235, 2081479430, 2771263512, 11715712410, 17911205580, 56608713884, 118968284928, 118968284928, 585927201062, 585927201062, 585927201062, 585927201062

Offset: 1

Labos Elemer, Sep 26 2003

Eggleton and MacDougall show that there are no more than 419 terms in this sequence. - T. D. Noe, Oct 13 2008

a(28) > 2 * 10^15. - Toshitaka Suzuki, Jun 22 2025

			a(6) = a(7) = 1217250 because the relevant 7 successive numbers have 4 distinct prime factors:
  1217250 = 2   *  3^2 *   5^3 * 541;
  1217251 = 7   * 17   *  53   * 193;
  1217252 = 2^2 * 23   * 101   * 131;
  1217253 = 3   * 47   *  89   *  97;
  1217254 = 2   * 19   * 103   * 311;
  1217255 = 5   * 13   *  61   * 307;
  1217256 = 2^3 *  3   *  67   * 757.

Toshitaka Suzuki, Table of n, a(n) for n = 1..27 (terms 1..23 from Donovan Johnson).
Roger B. Eggleton and James A. MacDougall, Consecutive integers with equally many principal divisors, Math. Mag. 81 (2008), 235-248.

Cf. A080569 (m=3), A064708 (m=2).

Cf. A087978, A138206, A138207, A154573.

k=1; Do[While[Union[Table[Length[FactorInteger[i]], {i, k, k+n-1}]]!={4}, k++ ]; Print[k], {n, 1, 8}]
Module[{d4=Table[If[PrimeNu[n]==4,1,0],{n,143*10^5}]},Flatten[Table[ SequencePosition[d4,PadRight[{},n,1],1],{n,8}],1][[All,1]]] (* Requires Mathematica version 10 or later *) (* This generates the first 8 terms of the sequence *) (* Harvey P. Dale, Aug 25 2017 *)

More terms from Don Reble, Sep 29 2003
a(13)-a(19) from Donovan Johnson, Mar 06 2008
a(20)-a(23) from Donovan Johnson, Jan 15 2009

A064709 Initial term of run of (exactly) n consecutive numbers with just 2 distinct prime factors.

Original entry on oeis.org

6, 14, 20, 33, 54, 91, 323, 141

Offset: 1

Robert G. Wilson v, Oct 13 2001

The given terms up to a(8) = 141 are the only terms less than 10^18. To speed the search, note that any string of 6 or more consecutive numbers contains a multiple of 6 and hence must contain a number of the form 2^a * 3^b. Conjecture: 141 is the last term, because numbers with only two different prime factors get pretty rare, so having several in a row near a number of the form 2^a * 3^b is pretty unlikely. - Joshua Zucker, May 05 2006
Sequence cannot have any terms for n > 29, since a run of 30 or more consecutive numbers must contain a multiple of 30, divisible by at least 3 primes. - Franklin T. Adams-Watters, Oct 23 2006
I searched numbers of the form n=2^a * 3^b through 10^700 and could not find any solution where even 4 numbers (n+2, n-2, n+3, n-3) had omega=2. The last such number through 10^700 is only 169075682574336=2^33 * 3^9. So a full set of 9 numbers seems quite unlikely. - Fred Schneider, Jan 05 2008
From Vim Wenders, Apr 02 2008: (Start)
The sequence is complete. The argument of Franklin T. Adams-Watters is easily extended: if 2^a*3^b, a,b, >= 1 is a term then omega(2^a*3^b+-6) > 2 (because the exponents of 2 and 3 follow a ruler like sequence). So the last possible term would be a(11).
Also, if 2*p, p prime, is in the run of an initial value to check, then p+2, p+4, ... has to be prime too (for the values 2p+4 = 2(p+2), 2p+8 = 2(p+4) ...), which is impossible for obvious reason.
The two arguments limit the maximum length of a run to 8. (End)
Wenders's argument is incomplete because the consecutive even numbers can have the form 2^a p^b. As stated in the paper by Eggleton and MacDougall, it is still a conjecture that 9 consecutive omega-2 numbers do not exist. - T. D. Noe, Oct 13 2008
For a(10) to exist, one of the consecutive terms must be in A033846. Also, the sequence cannot have any terms for n > 14. If a(15) exists, one term has to be of the form A = 2^n*5^m. Then, there also must be two other terms divisible by 5, excluding A. Call them B and C.
       Case 1: If A is the smallest of these terms, then B = 5*(2^n*5^(m-1)+1) and C = 10*(2^(n-1)*5^(m-1)+1). However, for large enough values of m and n, either B/5 or C/10 is divisible by 3 and another prime > 5.
       Case 2: If A is the middle term, then B = 5*(2^n*5^(m-1)-1) and C = 5*(2^n*5^(m-1)+1). For large enough values of m and n, either B/5 or C/5 is a multiple of 3 and another prime > 5.
       Case 3: If A is the highest term, B = 5*(2^n*5^(m-1)-1) and C = 10*(2^(n-1)*5^(m-1)-1). Again, for large enough values of m and n, either B/5 or C/10 is a multiple of 3 and another prime > 5.
       Thus, if there are more terms in the sequence, they can only be a(9), a(10), a(11), a(12), a(13) or a(14).
     Other comments: The only numbers that satisfy a(8) are 141 and 212. The reason a(9) wasn't satisfied in either of these is because one end of the run of numbers was a prime and the other end stopped at a multiple of 10. I believe it is possible to show that a(10) can never exist because there cannot be a multiple of 10 in the run of consecutive numbers, perhaps because there cannot be two multiples of 5. - Derek Orr, May 24 2014
Eggleton and MacDougall prove that no terms exist beyond a(9) and conjecture that a(9) does not exist. - Jason Kimberley, Jul 08 2017

			6 = 2*3;
14 = 2*7 and 15 = 3*5;
20 = 2^2*5, 21 = 3*7 and 22 = 2*11;
33 = 3*11, 34 = 2*17, 35 = 5*7 and 36 = (2*3)^2; etc.

Roger B. Eggleton and James A. MacDougall, Consecutive integers with equally many principal divisors, Math. Mag. 81 (2008), 235-248. [_T. D. Noe_, Oct 13 2008]
Carlos Rivera, Prime Puzzle 427

Cf. A064708.

Function[s, Function[t, Map[t[[First@ FirstPosition[t[[All, -1]], k_ /; k == #] ]] &, Range[0, Max@ t[[All, -1]] ] ][[All, 1]] ]@ Join[{{First@ s, 0}, {#[[1, 1, 1]], 1}}, Rest@ Map[{#[[1, 1]], Length@ # + 1} &, #, {1}]] &@ SplitBy[Partition[Select[#, Last@ # == 1 &][[All, 1]], 2, 1], Differences] &@ Map[{First@ #, First@ Differences@ #} &, Partition[s, 2, 1]]]@ Select[Range[10^5], PrimeNu[#] == 2 &] (* Michael De Vlieger, Jul 17 2017 *)
With[{t=Table[If[PrimeNu[n]==2,1,0],{n,350}]},Flatten[Table[ SequencePosition[ t,Join[{0},PadRight[{},k,1],{0}],1],{k,8}],1]][[All,1]]+1 (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 16 2020 *)

A080569 a(n) is the first number in the first run of at least n successive numbers, all having exactly 3 distinct prime factors.

Original entry on oeis.org

30, 230, 644, 1308, 2664, 6850, 10280, 39693, 44360, 48919, 218972, 526095, 526095, 526095, 17233173, 127890362, 29138958036, 146216247221, 23671413563491, 36966736685739

Offset: 1

Randy L. Ekl, Feb 21 2003

The 19th term, if it exists, is at least 1.1 * 10^12. - Fred Schneider, Jan 05 2008
There can be at most 209 terms in this sequence. Any list of 210 consecutive numbers must contain a number n which is multiple of 2*3*5*7 = 210. So omega(n) would be >3. - Fred Schneider, Jan 05 2008
Eggleton and MacDougall show that there are no more than 59 terms in this sequence. [From T. D. Noe, Oct 13 2008]
a(19) > 10^13. - Donovan Johnson, Jun 11 2013
a(19) <= 7523987244435061. - Donovan Johnson, Jul 08 2013
a(21) > 2 * 10^15, if it exists. - Toshitaka Suzuki, Jun 23 2025

			a(3) = 644 because 644 = 2^2 * 7 * 23, so omega(644) = 3, 645 = 3*5*43, so omega(645) = 3 and 646 = 2*17*19, so omega(646) = 3 and no other number n < 644 has omega(n)=omega(n+1)=omega(n+2)=3.

Roger B. Eggleton and James A. MacDougall, Consecutive integers with equally many principal divisors, Math. Mag. 81 (2008), 235-248.
Carlos Rivera, Prime Puzzle 427. Runs of consecutive numbers such that... (I), The Prime Puzzles & Problems Connection.

Cf. A064708, A064709, A185032, A048932.

k = 1; Do[ While[ Union[ Table[ Length[ FactorInteger[i]], {i, k, k + n - 1}]] != {3}, k++ ]; Print[k], {n, 1, 16}]

k=1; for(i=1,600000,s=1; for(j=1,k,if(omega(i+j-1)!=3,s=0,)); if(s==1,print1(i,", "); k++; i--,) )

Edited and extended by Robert G. Wilson v, Feb 22 2003
More terms from Don Reble, Mar 02 2003
a(19)-a(20) from Toshitaka Suzuki, Apr 01 2025

A087978 a(n) is the first term in a chain of at least n consecutive numbers, each having exactly m = 5 distinct prime factors.

Original entry on oeis.org

2310, 254540, 1042404, 21871365, 129963314, 830692265, 4617927894, 18297409143, 41268813542, 287980277114, 1182325618032, 6455097761454, 14207465691240, 54049709480208, 90987640183352, 546525829796442, 546525829796442

Offset: 1

Labos Elemer, Sep 26 2003

Every chain of 30030 consecutive numbers has exactly one number divisible by 30030 = 2 * 3 * 5 * 7 * 11 * 13 hence is divisible by more than five distinct primes. Therefore the sequence is finite. - David A. Corneth, Jul 19 2023

a(18) > 2 * 10^15. - Toshitaka Suzuki, Jun 23 2025

Roger B. Eggleton and James A. MacDougall, Consecutive Integers with Equally Many Principal Divisors, Math. Mag. 81 (2008), 235-248. [_T. D. Noe_, Oct 13 2008]

Cf. A064708 (m=2), A080569 (m=3), A087977 (m=4).
Cf. A138206, A138207, A154573. - Donovan Johnson, Jan 15 2009
Cf. A046387.

k=1; Do[While[Union[Table[Length[FactorInteger[i]], {i, k, k+n-1}]]!={5}, k++ ]; Print[k], {n, 1, 8}]

More terms from Don Reble, Sep 29 2003
a(7)-a(10) from Donovan Johnson, Mar 06 2008
a(11)-a(12) from Donovan Johnson, Jan 15 2009
a(13)-a(15) from Toshitaka Suzuki, Apr 06 2025
a(16)-a(17) from Toshitaka Suzuki, Jun 23 2025

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A087977 a(n) is the first term in the first chain of at least n consecutive numbers each having exactly four distinct prime factors.

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A064709 Initial term of run of (exactly) n consecutive numbers with just 2 distinct prime factors.

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A080569 a(n) is the first number in the first run of at least n successive numbers, all having exactly 3 distinct prime factors.

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A087978 a(n) is the first term in a chain of at least n consecutive numbers, each having exactly m = 5 distinct prime factors.

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