A074851 -id:A074851 - cp's OEIS frontend

A255346 Numbers k such that k and k+1 both have at least two distinct prime factors.

14, 20, 21, 33, 34, 35, 38, 39, 44, 45, 50, 51, 54, 55, 56, 57, 62, 65, 68, 69, 74, 75, 76, 77, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 98, 99, 104, 105, 110, 111, 114, 115, 116, 117, 118, 119, 122, 123, 129, 132, 133, 134, 135, 140, 141, 142, 143, 144, 145, 146, 147, 152, 153, 154

Offset: 1

M. F. Hasler, Feb 21 2015

These numbers provide solutions to the problem of finding (x,y) such that x(x+1) | y(y+1) but none of x or x+1 divides any of y or y+1. Namely, these solutions are given for (x,y) being members of the sequence such that x(x+1) divides y(y+1), the smallest of which are (14,20), (14,35), (20,35), ... but, e.g., (14,69) is excluded since 14 | 70.

Contains A074851 as a subsequence.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
T. Korimort, How many (x,y) satisfy x(x+1)|y(y+1),..., Number Theory group on LinkedIn.com, Feb. 2014.

Cf. A074851.

SequencePosition[Table[If[PrimeNu[n]>1,1,0],{n,200}],{1,1}][[;;,1]] (* Harvey P. Dale, Jul 30 2025 *)

for(n=2,199,omega(n)>=2||(n++&&next);omega(n-1)>=2&&print1((n-1)","))

A140078 Numbers k such that k and k+1 have 4 distinct prime factors.

Original entry on oeis.org

7314, 8294, 8645, 9009, 10659, 11570, 11780, 11934, 13299, 13629, 13845, 14420, 15105, 15554, 16554, 16835, 17204, 17390, 17654, 17765, 18095, 18290, 18444, 18920, 19005, 19019, 19095, 19227, 20349, 20405, 20769, 21164, 21489, 21735

Offset: 1

Artur Jasinski, May 07 2008

nonn

Goldston, Graham, Pintz, & Yildirim prove that this sequence is infinite. - Charles R Greathouse IV, Jun 02 2016

The subsequence of terms where k and k+1 are also squarefree is A318896. - R. J. Mathar, Jul 15 2023

David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 161 (entry for 7314).

Seiichi Manyama, Table of n, a(n) for n = 1..10000
D. A. Goldston, S. W. Graham, J. Pintz and C. Y. Yildirim, Small gaps between almost primes, the parity problem and some conjectures of Erdos on consecutive integers, arXiv:0803.2636 [math.NT], 2008.

Similar sequences with k distinct prime factors: A074851 (k=2), A140077 (k=3), this sequence (k=4), A140079 (k=5).
Cf. A093548.
Equals A321504 \ A321494.

a = {}; Do[If[Length[FactorInteger[n]] == 4 && Length[FactorInteger[n + 1]] == 4, AppendTo[a, n]], {n, 1, 100000}]; a (* Artur Jasinski, May 07 2008 *)
Transpose[Position[Partition[PrimeNu[Range[20000]],2,1],?(#[[1]] == #[[2]] == 4&),{1},Heads->False]][[1]] (* _Harvey P. Dale, Jun 21 2013 *)
SequencePosition[PrimeNu[Range[22000]],{4,4}][[;;,1]] (* Harvey P. Dale, Jun 20 2024 *)

isok(n) = (omega(n)==4) && (omega(n+1)==4); \\ Michel Marcus, Sep 04 2015

{k: k in A033993 and k+1 in A033993}. - R. J. Mathar, Jul 19 2023

Link provided by Harvey P. Dale, Jun 21 2013

A140077 Numbers n such that n and n+1 have 3 distinct prime factors.

Original entry on oeis.org

230, 285, 429, 434, 455, 494, 560, 594, 609, 615, 644, 645, 650, 665, 740, 741, 759, 804, 805, 819, 825, 854, 860, 884, 902, 935, 945, 969, 986, 987, 1001, 1014, 1022, 1034, 1035, 1044, 1064, 1065, 1070, 1085, 1104, 1105, 1130, 1196, 1209, 1220, 1221

Offset: 1

Artur Jasinski, May 07 2008

nonn

Goldston, Graham, Pintz, & Yildirim prove that this sequence is infinite. - Charles R Greathouse IV, Sep 14 2015

See A321503 for numbers n such that n & n+1 have at least 3 prime divisors, disjoint union of this and A321493, the terms of A321503 which are not in this sequence. A321493 has A140078 as a subsequence, which in turn is subsequence of A321504, and so on. Since n and n+1 can't share a prime factor, we have a(1) > sqrt(p(3+3)#) > A000196(A002110(3+3)). Note that A000196(A002110(3+4)) = A321493(1) exactly! - M. F. Hasler, Nov 13 2018

Harvey P. Dale, Table of n, a(n) for n = 1..10000
D. A. Goldston, S. W. Graham, J. Pintz, C. Y. Yildirim, Small gaps between almost primes, the parity problem and some conjectures of Erdos on consecutive integers, arXiv:0803.2636 [math.NT], 2008.

Cf. A074851, A140078, A140079.

Equals A321503 \ A321493.

a = {}; Do[If[Length[FactorInteger[n]] == 3 && Length[FactorInteger[n + 1]] == 3, AppendTo[a, n]], {n, 1, 100000}]; a (*Artur Jasinski*)
SequencePosition[PrimeNu[Range[1250]],{3,3}][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Feb 27 2017 *)

is(n)=omega(n)==3&&omega(n+1)==3 \\ Charles R Greathouse IV, Sep 14 2015

{k: k in A033992 and k+1 in A033992}. - R. J. Mathar, Jul 19 2023

A140079 Numbers n such that n and n+1 have 5 distinct prime factors.

Original entry on oeis.org

254540, 310155, 378014, 421134, 432795, 483405, 486590, 486794, 488565, 489345, 507129, 522444, 545258, 549185, 558789, 558830, 567644, 577940, 584154, 591260, 598689, 627095, 634809, 637329, 663585, 666995, 667029, 678755, 687939, 690234

Offset: 1

Artur Jasinski, May 07 2008

nonn

For the smallest number r such that r and r+1 have n distinct prime factors, see A093548.
Goldston, Graham, Pintz, & Yildirim prove that this sequence is infinite. - Charles R Greathouse IV, Jun 02 2016
Subsequence of the variant A321505 defined with "at least 5" instead of "exactly 5" distinct prime factors. See A321495 for the differences. - M. F. Hasler, Nov 12 2018
The subset of numbers where n and n+1 are also squarefree gives A318964. - R. J. Mathar, Jul 15 2023

Seiichi Manyama, Table of n, a(n) for n = 1..10000
D. A. Goldston, S. W. Graham, J. Pintz, C. Y. Yildirim., Small gaps between almost primes, the parity problem and some conjectures of Erdos on consecutive integers, arXiv:0803.2636 [math.NT], 2008.

Cf. A074851, A140077, A140078, A273897.
Cf. A093548.
Equals A321505 \ A321495.

a = {}; Do[If[Length[FactorInteger[n]] == 5 && Length[FactorInteger[n + 1]] == 5, AppendTo[a, n]], {n, 1, 100000}]; a (*Artur Jasinski*)
Transpose[SequencePosition[Table[If[PrimeNu[n]==5,1,0],{n,700000}],{1,1}]][[1]] (* The program uses the SequencePosition function from Mathematica version 10 *) (* Harvey P. Dale, Jul 25 2015 *)

is(n)=omega(n)==5 && omega(n+1)==5 \\ Charles R Greathouse IV, Jun 02 2016

{k: k in A051270 and k+1 in A051270}. - R. J. Mathar, Jul 19 2023

A321503 Numbers m such that m and m+1 both have at least 3 distinct prime factors.

Original entry on oeis.org

230, 285, 429, 434, 455, 494, 560, 594, 609, 615, 644, 645, 650, 665, 714, 740, 741, 759, 804, 805, 819, 825, 854, 860, 884, 902, 935, 945, 969, 986, 987, 1001, 1014, 1022, 1034, 1035, 1044, 1064, 1065, 1070, 1085, 1104, 1105, 1130, 1196, 1209, 1220, 1221, 1235, 1239, 1245, 1265

Offset: 1

M. F. Hasler, Nov 13 2018

nonn

Disjoint union of A140077 (omega({m, m+1}) = {3}) and A321493 (not both have exactly 3 prime divisors). The latter contains terms with indices {15, 60, 82, 98, 99, 104, ...} of this sequence.

Numbers m and m+1 can never have a common prime factor (consider them mod p), therefore the terms are > sqrt(A002110(3+3)), A002110 = primorial.

M. F. Hasler, Table of n, a(n) for n = 1..10000

Subsequence of A000977.
Cf. A255346, A321504 .. A321506, A321489 (analog for k = 2, ..., 7 prime divisors).
Cf. A321493, A321494 .. A321497 (subsequences of the above: m or m+1 has more than k prime divisors).
Cf. A074851, A140077, A140078, A140079 (complementary subsequences: m and m+1 have exactly k = 2, 3, 4, 5 prime divisors).

aQ[n_]:=Module[{v={PrimeNu[n], PrimeNu[n+1]}}, Min[v]>2]; Select[Range[1300], aQ] (* Amiram Eldar, Nov 12 2018 *)

select( is(n)=omega(n)>2&&omega(n+1)>2, [1..1300])

a(n) ~ n. - Charles R Greathouse IV, Jan 25 2025

A273879 Numbers k such that k and k+1 have 6 distinct prime factors.

Original entry on oeis.org

11243154, 13516580, 16473170, 16701684, 17348330, 19286805, 20333495, 21271964, 21849905, 22054515, 22527141, 22754589, 22875489, 24031370, 25348070, 25774329, 28098245, 28618394, 28625960, 30259229, 31846269, 32642805

Offset: 1

Charles R Greathouse IV, Jun 02 2016

nonn

Goldston, Graham, Pintz, & Yildirim prove that this sequence is infinite (Theorem 2).

			13516580 = 2^2 * 5 * 7 * 11 * 67 * 131 and 13516581 = 3 * 13 * 17 * 19 * 29 * 37 so 13516580 is in this sequence.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
D. A. Goldston, S. W. Graham, J. Pintz, C. Y. Yildirim, Small gaps between almost primes, the parity problem and some conjectures of Erdős on consecutive integers, arXiv:0803.2636 [math.NT], 2008.
D. A. Goldston, S. W. Graham, J. Pintz, C. Y. Yildirim, Small gaps between almost primes, the parity problem and some conjectures of Erdős on consecutive integers, International Mathematics Research Notices 7 (2011), pp. 1439-1450.

Numbers k such that k and k+1 have j distinct prime factors: A006549 (j=1, apart from the first term), A074851 (j=2), A140077 (j=3), A140078 (j=4), A140079 (j=5).

Cf. A006049, A093548.

SequencePosition[PrimeNu[Range[3265*10^4]],{6,6}][[All,1]] (* Harvey P. Dale, Nov 20 2021 *)

PARI
```
is(n)=omega(n)==6 && omega(n+1)==6
```

a(1) = A138206(2). - R. J. Mathar, Jul 15 2023

{k: k in A074969 and k+1 in A074969.} - R. J. Mathar, Jul 19 2023

A364307 Numbers k such that k, k+1 and k+2 have exactly 2 distinct prime factors.

Original entry on oeis.org

20, 33, 34, 38, 44, 50, 54, 55, 56, 74, 75, 85, 86, 91, 92, 93, 94, 98, 115, 116, 117, 122, 133, 134, 141, 142, 143, 144, 145, 146, 158, 159, 160, 175, 176, 183, 187, 200, 201, 205, 206, 207, 212, 213, 214, 215, 216, 217, 224, 235, 247, 248, 295, 296

Offset: 1

R. J. Mathar, Jul 18 2023

nonn

			44 = 2^2*11 has 2 distinct prime factors, and so has 45 = 3^2*5 and so has 46 = 2*23, so 44 is in the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Subsequence of A006073 and of A074851.
Cf. A364308 (3 factors), A364309 (4 factors), A364266 (5 factors), A364265 (6 factors), A001221.
A039833 is a subsequence.

q[n_] := q[n] = PrimeNu[n] == 2; Select[Range[300], q[#] && q[#+1] && q[#+2] &] (* Amiram Eldar, Oct 01 2024 *)

{k: A001221(k) = A001221(k+1) = A001221(k+2) = 2}.

A321489 Numbers m such that both m and m+1 have at least 7 distinct prime factors.

Original entry on oeis.org

965009045, 1068044054, 1168008204, 1177173074, 1209907985, 1218115535, 1240268490, 1338753129, 1344185205, 1408520805, 1477640450, 1487720234, 1509981395, 1663654629, 1693460405, 1731986894, 1758259425, 1819458354, 1821278459, 1826445984, 1857332840

Offset: 1

Amiram Eldar and M. F. Hasler, Nov 12 2018

nonn

The first 300 terms of this sequence are such that m and m+1 both have exactly 7 prime divisors. See A321497 for the terms m such that m or m+1 has more than 7 prime factors: the smallest such term is 5163068910.
Numbers m and m+1 can never have a common prime factor (consider them mod p), therefore the terms are > sqrt(p(7+7)#) = A003059(A002110(7+7)). (Here we see that sqrt(p(7+8)#) is a more realistic estimate of a(1), but for smaller values of k we may have sqrt(p(2k+1)#) > m(k) > sqrt(p(2k)#), where m(k) is the smallest of two consecutive integers each having at least k prime divisors. For example, A321503(1) < sqrt(p(3+4)#) ~ A321493(1).)
From M. F. Hasler, Nov 28 2018: (Start)
The first 100 terms and beyond are all congruent to one of {14, 20, 35, 49, 50, 69, 84, 90, 104, 105, 110, 119, 125, 129, 134, 140, 144, 170, 174, 189, 195} mod 210. Here, 35, 195, 189, 14 140, 20 and 174 (in order of decreasing frequency) occur between 6 and 13 times, and {49, 50, 110, 129, 134, 144, 170} occur only once.
However, as observed by Charles R Greathouse IV, one can construct a term of this sequence congruent to any given m > 0, modulo any given n > 0.
The first terms of this sequence which are multiples of 210 are in A321497. An example of a term that is a multiple of 210 but not in A321497 is 29759526510, due to Charles R Greathouse IV. Such examples can be constructed by solving A*210 + 1 = B for A having 3 distinct prime factors not among {2, 3, 5, 7}, B having 7 distinct prime factors and gcd(B, 210*A) = 1. (End)

			a(1) = 5 * 7 * 11 * 13 * 23 * 83 * 101, a(1)+1 = 2 * 3 * 17 * 29 * 41 * 73 * 109.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 102 terms from M. F. Hasler)

Cf. A255346, A321503 .. A321506 (analog for k = 2, ..., 6 prime divisors).
Cf. A321502, A321493 .. A321497 (m and m+1 have at least but not both exactly k = 2, ..., 7 prime divisors).
Cf. A074851, A140077, A140078, A140079 (m and m+1 both have exactly k = 2, 3, 4, 5 prime divisors).
Cf. A006049, A006549, A093548.
Cf. A002110.

Select[Range[36000000], PrimeNu[#] > 6 && PrimeNu[# + 1] > 6 &]

is(n)=omega(n)>6&&omega(n+1)>6
A321489=List();for(n=965*10^6,1.8e9,is(n)&&listput(A321489,n))

a(n) ~ n. - Charles R Greathouse IV, Nov 29 2018

A088983 Numbers n such that each of the 6 consecutive numbers n through n+5 has exactly two distinct prime factors.

Original entry on oeis.org

91, 141, 142, 143, 212, 213, 214, 323, 324, 2302, 2303

Offset: 1

Labos Elemer, Sep 30 2003

Initial segment of A045934 is identical to this sequence but in A045934 the 12th term is divisible by 3 prime factors. Is the present sequence complete?
No more terms < 3*10^8. - David Wasserman, Aug 29 2005
a(12) > 10^40, if it exists. - Giovanni Resta, May 10 2017
From David A. Corneth, May 14 2017: (Start)
We're looking for at least 6 consecutive positive integers that each have exactly two distinct prime divisors. I.e. 6 consecutive positive integers m with omega(m) = 2. Now of exactly 6 consecutive integers, exactly one of them is divisible by 6, i.e. m is of the form 2*3*k. However m has exactly 2 distinct prime divisors, so k can only have prime divisors 2 or 3. Now, suppose m ends in 6 or higher. Then one of the consecutive integers is divisible by 10 = 2*5. I.e. it's of the form 2*5*t. Then t can only have prime divisors 2 and 5. (End)
This sequence has no run of four consecutive integers, since Eggleton and MacDougall prove that there are no more than 9 consecutive integers with A001221(k) = 2. They conjecture that A007774 contains no runs of 9 consecutive integers, and has only two runs of size 8 (at 141 and 212) and two maximal runs of size 7 (at 323 and 2302); they add that the maximal run of size 6 at 91 might be the only such run, so A088983 might be complete. - Roger Eggleton via Jason Kimberley, Jul 12 2017

Roger B. Eggleton and James A. MacDougall, Consecutive integers with equally many principal divisors, Math. Mag. 81 (2008), 235-248.

Cf. A064709, A045932, A045933, A045934, A045935, A074851, A006073, A003586.

Select[Range[3000], AllTrue[# + Range[0, 5], Length@FactorInteger[#] == 2 &] &] (* Giovanni Resta, May 09 2017 *)

Definition simplified by Roger Eggleton via Jason Kimberley, Jul 12 2017

A321502 Numbers m such that m and m+1 have at least 2, but m or m+1 has at least 3 prime divisors.

Original entry on oeis.org

65, 69, 77, 84, 90, 104, 105, 110, 114, 119, 129, 132, 140, 153, 154, 155, 164, 165, 170, 174, 182, 185, 186, 189, 194, 195, 203, 204, 209, 219, 220, 221, 230, 231, 234, 237, 245, 246, 252, 254, 258, 259, 260, 264, 265, 266, 272, 273, 275, 279, 284, 285, 286, 290, 294, 299, 300, 305

Offset: 1

M. F. Hasler, Nov 27 2018

nonn

Since m and m+1 cannot have a common factor, m(m+1) has at least 2+3 prime divisors (= distinct prime factors), whence m+1 > sqrt(primorial(5)) ~ 48. It turns out that a(1)*(a(1)+1) = 2*3*5*11*13, i.e., the prime factor 7 is not present.

Cf. A321493, A321494, A321495, A321496, A321497 (analog for k = 3, ..., 7 prime divisors).
Cf. A074851, A140077, A140078, A140079 (m and m+1 have exactly k = 2, 3, 4, 5 prime divisors).
Cf. A255346, A321503 .. A321506, A321489 (m and m+1 have at least 2, ..., 7 prime divisors).
Cf. A006049, A006549, A093548.

select( is_A321502(n)=vecmax(n=[omega(n), omega(n+1)])>2&&vecmin(n)>1, [1..500])

Equals A255346 \ A074851.

cp's OEIS Frontend

A255346 Numbers k such that k and k+1 both have at least two distinct prime factors.

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A140078 Numbers k such that k and k+1 have 4 distinct prime factors.

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A140077 Numbers n such that n and n+1 have 3 distinct prime factors.

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A140079 Numbers n such that n and n+1 have 5 distinct prime factors.

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A321503 Numbers m such that m and m+1 both have at least 3 distinct prime factors.

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A273879 Numbers k such that k and k+1 have 6 distinct prime factors.

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A364307 Numbers k such that k, k+1 and k+2 have exactly 2 distinct prime factors.

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A321489 Numbers m such that both m and m+1 have at least 7 distinct prime factors.