A006049 -id:A006049 - cp's OEIS frontend

A062974 Numbers k such that omega(k+1) < 2*omega(k), where omega(k) is the number of distinct prime divisors of k.

2, 3, 4, 6, 7, 8, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 28, 30, 31, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100

Offset: 1

Jason Earls, Jul 26 2001

nonn

A001221(a(n)+1) < 2*A001221(a(n)).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A006049 (subsequence).

import Data.List (findIndices)
a062974 n = a062974_list !! (n-1)
a062974_list = map (+ 1) $ findIndices (< 0) $
   zipWith (-) (tail a001221_list) $ map (* 2) a001221_list
-- Reinhard Zumkeller, Jan 22 2013

j=[]; for(n=1,200, if(omega(n+1)<2*omega(n),j=concat(j,n))); j

A063464 Numbers k such that omega(k) = omega(k+2), where omega(k) is the number of distinct prime divisors of k.

Original entry on oeis.org

2, 3, 5, 7, 9, 10, 11, 12, 17, 18, 20, 22, 23, 24, 25, 26, 27, 29, 33, 34, 36, 38, 41, 44, 46, 47, 48, 50, 52, 54, 55, 56, 59, 63, 71, 72, 74, 75, 79, 80, 81, 85, 86, 91, 92, 93, 94, 96, 98, 101, 104, 106, 107, 115, 116, 117, 122, 125, 130, 133, 134, 137, 138, 141, 142

Offset: 1

Jason Earls, Jul 26 2001

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harry J. Smith)

Cf. A001221, A006049.

Select[Range[200],PrimeNu[#]==PrimeNu[#+2]&] (* Harvey P. Dale, Jun 20 2011 *)

j=[]; for(n=1,350, if(omega(n)==omega(n+2),j=concat(j,n))); j

{ n=0; for (m=1, 10^9, if (omega(m) == omega(m + 2), write("b063464.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 22 2009

A075044 a(0) = 1; a(n) = the smallest number k such that n numbers from k to k+n-1 have n distinct prime divisors, or 0 if no such number exists.

Original entry on oeis.org

1, 2, 14, 644, 134043, 129963314, 626804494291

Offset: 0

Amarnath Murthy, Sep 03 2002

a(7) <= 45164156742722455667280. - Giorgos Kalogeropoulos, Apr 10 2025

Cf. A006049, A045983, A006073.

Do[k = 1; While[Length /@ FactorInteger /@ Range[k, k+n-1] != Table[n, {n}], k++ ]; Print[k], {n, 0, 5}] (* Ryan Propper, Oct 01 2005 *)

Corrected and extended by Ryan Propper, Oct 01 2005

Offset corrected and a(6) from Donovan Johnson, Aug 03 2009

A076253 a(n) = the least positive integer solution of the "n-th omega recurrence" omega(k) = omega(k-1) + ... + omega(k-n), if such k exists; = 0 otherwise. (omega(n) denotes the number of distinct prime factors of n.)

Original entry on oeis.org

3, 3, 2310, 746130, 601380780, 89419589469210, 489423552293946270

Offset: 1

Joseph L. Pe, Nov 04 2002

Question: Is a(n) > 0 for all n, i.e. can the n-th omega recurrence be solved for all n?

Note that 601380780 is not squarefree. Using primorials, I easily found candidates up to a(8). - Lambert Klasen (lambert.klasen(AT)gmx.net), Nov 05 2005

			k=3 is the smallest solution of omega(k)=omega(k-1), so a(1)=3.
k=3 is the smallest solution of omega(k)=omega(k-1)+omega(k-2), so a(2)=3.
k=2310 is the smallest solution of omega(k)=omega(k-1)+omega(k-2)+omega(k-3), so a(3)=2310.

Cf. A001221, A006049, A076251, A076252.

(*code to find a(4)*) omega[n_] := Length[FactorInteger[n]]; ub = 2*10^6; For[i = 2, i <= ub, i++, a[i] = omega[i]]; start = 5; For[j = start, j <= ub, j++, If[a[j] == a[j - 1] + a[j - 2] + a[j - 3] + a[j - 4], Print[j]]]

/* find a(5) */ v=[0,0,0,0,0]; s=0;for(i=1,5,v[i]=omega(i);s+=v[I])
for(i=6,10^10,o=omega(i);if(o==s,print(i);break);s-=v[i%5+1];s+=o;v[i%5+1]=o) \\ Lambert Klasen (lambert.klasen(AT)gmx.net), Nov 05 2005

a(5) from Lambert Klasen (lambert.klasen(AT)gmx.net), Nov 05 2005

a(6)-a(7) from Donovan Johnson, Feb 07 2009

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.

A355709 Numbers k such that k and k+1 have the same number of 3-smooth divisors.

Original entry on oeis.org

2, 14, 21, 33, 38, 44, 50, 57, 69, 74, 80, 86, 93, 99, 105, 110, 116, 122, 129, 135, 141, 146, 158, 165, 171, 177, 182, 194, 201, 213, 218, 230, 237, 249, 254, 260, 266, 273, 285, 290, 296, 302, 309, 315, 321, 326, 332, 338, 345, 357, 362, 374, 381, 387, 393, 398

Offset: 1

Amiram Eldar, Jul 15 2022

nonn

Numbers k such that A072078(k) = A072078(k+1).

This sequence is infinite since it includes all the numbers of the form 3*(2^(2*k+1)-1), with k>=1.

			2 is a term since A072078(2) = A072078(3) = 2.

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

Cf. A072078, A355710 (5-smooth analog).

Similar sequences: A005237, A006049, A332386, A343819, A344312, A344313, A344314, A348345.

s[n_] := Times @@ (1 + IntegerExponent[n, {2, 3}]); Select[Range[400], s[#] == s[#+1] &]

s(n) = (valuation(n, 2) + 1) * (valuation(n, 3) + 1);
s1 = s(1); for(k = 2, 400, s2 = s(k); if(s1 == s2, print1(k-1,", ")); s1 = s2);

A293460 a(n) = Sum_{k=1..n} sign(omega(n+1) - omega(n)) (where omega(m) = A001221(m), the number of distinct primes dividing m).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 3, 2, 3, 3, 4, 3, 4, 3, 4, 4, 4, 4, 5, 4, 5, 4, 5, 4, 5, 4, 4, 4

Offset: 0

Rémy Sigrist, Oct 26 2017

sign

The sign function is defined by:
- sign(0) = 0,
- sign(n) = +1 for any n > 0,
- sign(n) = -1 for any n < 0.
a(n) corresponds to the number of integers up to n in A294277 minus the number of integers up to n in A294278.
The first negative value occurs at a(178) = -1.
Will this sequence change sign indefinitely?

			The following table shows the first terms of the sequence, alongside sign(omega(n+1)-omega(n)), omega(n+1) and omega(n):
n       a(n)    sign    w(n+1)  w(n)
-       ----    ----    ------  ----
0       0
1       1       1       1       0
2       1       0       1       1
3       1       0       1       1
4       1       0       1       1
5       2       1       2       1
6       1       -1      1       2
7       1       0       1       1
8       1       0       1       1
9       2       1       2       1
10      1       -1      1       2
11      2       1       2       1
12      1       -1      1       2
13      2       1       2       1
14      2       0       2       2
15      1       -1      1       2
16      1       0       1       1
17      2       1       2       1
18      1       -1      1       2
19      2       1       2       1
20      2       0       2       2

Georg Fischer, Table of n, a(n) for n = 0..1000
Rémy Sigrist, Line graph of the first 10000 terms
Rémy Sigrist, Line graph of the first 100000000 terms
Rémy Sigrist, Line graph of the first 1000000000 terms
Rémy Sigrist, Line graph of the first 10000000000 terms

Cf. A001221, A006049, A294277, A294278.

Accumulate[Join[{0},Sign[Differences[PrimeNu[Range[90]]]]]] (* Harvey P. Dale, Dec 25 2024 *)

s = 0; for (n=1, 87, print1 (s ", "); s += sign(omega(n+1)-omega(n)))

a(0) = 0, and for any n > 0:
- a(A294277(n)) = a(A294277(n)-1) + 1,
- a(A006049(n)) = a(A006049(n)-1),
- a(A294278(n)) = a(A294278(n)-1) - 1.
Also: a(n) = #{ k / A294277(k) <= n } - #{ k / A294278(k) <= n }.

A375287 Square array T(n, k), n > 1 and k >= 1, read by upward antidiagonals, give the smallest number that starts a sequence of exactly k consecutive numbers, each having exactly n distinct prime factors (counted without multiplicity), or -1 if no such number exists.

Original entry on oeis.org

6, 30, 14, 210, 230, 20, 2310, 7314, 644, 33, 30030, 254540, 37960, 1308, 54, 510510, 11243154, 1042404, 134043, 2664, 91, 9699690, 965009045, 323567034, 21871365, 357642, 6850, 323, 223092870, 65893166030, 30989984674, 7933641735, 129963314, 2713332, 10280, 141

Offset: 2

Jean-Marc Rebert, Aug 10 2024

All positive terms are composite.

			T(2,3) = 20 = 2^2 * 5, because both 21 and 22 have the same omega. Thus, 20 is the starting number of a run of 3 numbers that each have same omega, i.e. 2. No lesser number has this property, so T(2,3) = 20.
Table begins (upper left corner = T(2,1)):
      6       14        20         33 ...
     30      230       644       1308 ...
    210     7314     37960     134043 ...
   2310   254540   1042404   21871365 ...
  30030 11243154 323567034 7933641735 ...
    ...      ...       ...        ...

Cf. A001221, A002110 (col 1), A006049, A006073, A045932-A045938, A064709 (row 2), A185032 (row 3), A185042 (row 4), A384507 (row 5).

T(n,1) = A002110(n) for n > 1.

a(29) corrected by and a(30)-a(37) from Jinyuan Wang, Sep 05 2025

cp's OEIS Frontend

A062974 Numbers k such that omega(k+1) < 2*omega(k), where omega(k) is the number of distinct prime divisors of k.

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A063464 Numbers k such that omega(k) = omega(k+2), where omega(k) is the number of distinct prime divisors of k.

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A075044 a(0) = 1; a(n) = the smallest number k such that n numbers from k to k+n-1 have n distinct prime divisors, or 0 if no such number exists.

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A076253 a(n) = the least positive integer solution of the "n-th omega recurrence" omega(k) = omega(k-1) + ... + omega(k-n), if such k exists; = 0 otherwise. (omega(n) denotes the number of distinct prime factors of n.)

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

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A355709 Numbers k such that k and k+1 have the same number of 3-smooth divisors.

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A293460 a(n) = Sum_{k=1..n} sign(omega(n+1) - omega(n)) (where omega(m) = A001221(m), the number of distinct primes dividing m).

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A375287 Square array T(n, k), n > 1 and k >= 1, read by upward antidiagonals, give the smallest number that starts a sequence of exactly k consecutive numbers, each having exactly n distinct prime factors (counted without multiplicity), or -1 if no such number exists.

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