A033846 -id:A033846 - cp's OEIS frontend

A284311 Array T(n,k) read by antidiagonals (downward): T(1,k) = A005117(k+1) (squarefree numbers > 1); for n > 1, columns are nonsquarefree numbers (in ascending order) with exactly the same prime factors as T(1,k).

2, 3, 4, 5, 9, 8, 6, 25, 27, 16, 7, 12, 125, 81, 32, 10, 49, 18, 625, 243, 64, 11, 20, 343, 24, 3125, 729, 128, 13, 121, 40, 2401, 36, 15625, 2187, 256, 14, 169, 1331, 50, 16807, 48, 78125, 6561, 512, 15, 28, 2197, 14641, 80, 117649, 54, 390625, 19683, 1024

Offset: 1

Bob Selcoe, Mar 24 2017

A permutation of the natural numbers > 1.

T(1,k)= A005117(m) with m > 1; terms in column k = T(1,k) * A162306(T(1,k)) only not bounded by T(1,k). Let T(1,k) = b. Terms in column k are multiples of b and numbers c such that c | b^e with e >= 0. Alternatively, terms in column k are multiples bc with c those numbers whose prime divisors p also divide b. - Michael De Vlieger, Mar 25 2017

			Array starts:
    2    3     5  6      7  10       11        13  14  15
    4    9    25 12     49  20      121       169  28  45
    8   27   125 18    343  40     1331      2197  56  75
   16   81   625 24   2401  50    14641    371293  98 135
   32  243  3125 36  16807  80   161051   4826809 112 225
   64  729 15625 48 117649 100  1771561  62748517 196 375
  128 2187 78125 54 823543 160 19487171 815730721 224 405
Column 6 is: T(1,6) = 2*5; T(2,6) = 2^2*5; T(3,6) = 2^3*5; T(4,6) = 2*5^2; T(5,6) = 2^4*5, etc.

Alois P. Heinz, Antidiagonals n = 1..141, flattened

Cf. A005117 (squarefree numbers), A033845 (column 4), columns 1,2,3,5 are powers of primes, A033846 (column 6), A033847 (column 9), A033849 (column 10).
The columns that are powers of primes have indices A071403(n) - 1. - Michel Marcus, Mar 24 2017
See also A007947; the k-th column of the array corresponds to the numbers with radical A005117(k+1). - Rémy Sigrist, Mar 24 2017
Cf. A284457 (this sequence read by antidiagonals upwards), A285321 (a similar array, but columns come in different order).
Cf. A065642.
Cf. A008479 (index of the row where n is located), A285329 (of the column).

f[n_, k_: 1] := Block[{c = 0, sgn = Sign[k], sf}, sf = n + sgn; While[c < Abs[k], While[! SquareFreeQ@ sf, If[sgn < 0, sf--, sf++]]; If[sgn < 0, sf--, sf++]; c++]; sf + If[sgn < 0, 1, -1]] (* after Robert G. Wilson v at A005117 *); T[n_, k_] := T[n, k] = Which[And[n == 1, k == 1], 2, k == 1, f@ T[n - 1, k], PrimeQ@ T[n, 1], T[n, 1]^k, True, Module[{j = T[n, k - 1]/T[n, 1] + 1}, While[PowerMod[T[n, 1], j, j] != 0, j++]; j T[n, 1]]]; Table[T[n - k + 1, k], {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Mar 25 2017 *)

(define (A284311 n) (A284311bi  (A002260 n) (A004736 n)))
(define (A284311bi row col) (if (= 1 row) (A005117 (+ 1 col)) (A065642 (A284311bi (- row 1) col))))
;; Antti Karttunen, Apr 17 2017

From Antti Karttunen, Apr 17 2017: (Start)
A(1,k) = A005117(1+k), A(n,k) = A065642(A(n-1,k)).
A(A008479(n), A285329(n)) = n for all n >= 2.
(End)

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 *)

A288162 Numbers whose prime factors are 2 and 13.

Original entry on oeis.org

26, 52, 104, 208, 338, 416, 676, 832, 1352, 1664, 2704, 3328, 4394, 5408, 6656, 8788, 10816, 13312, 17576, 21632, 26624, 35152, 43264, 53248, 57122, 70304, 86528, 106496, 114244, 140608, 173056, 212992, 228488, 281216, 346112, 425984, 456976, 562432, 692224, 742586, 851968, 913952

Offset: 1

Bernard Schott, Jun 06 2017

nonn

Numbers k such that phi(k)/k = 6/13.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A033845, A033846, A033847, A033848, A033849, A033850, A033851, A086780, A107326, A143207.

[n:n in [1..100000] | Set(PrimeDivisors(n)) eq {2,13}];  // Marius A. Burtea, May 10 2019

Select[Range[920000],FactorInteger[#][[All,1]]=={2,13}&] (* Harvey P. Dale, Jun 18 2021 *)

is(n) = factor(n)[, 1]~==[2, 13] \\ Felix Fröhlich, Jun 06 2017

list(lim)=my(v=List(),t); for(n=1,logint(lim\2,13), t=13^n; while((t<<=1)<=lim, listput(v,t))); Set(v) \\ Charles R Greathouse IV, Jun 11 2017

a(n) = 26 * A107326(n). - David A. Corneth, Jun 06 2017

Sum_{n>=1} 1/a(n) = 1/12. - Amiram Eldar, Dec 22 2020

A382248 Smallest number k that is neither squarefree nor a prime power such that k is coprime to n.

Original entry on oeis.org

12, 45, 20, 45, 12, 175, 12, 45, 20, 63, 12, 175, 12, 45, 28, 45, 12, 175, 12, 63, 20, 45, 12, 175, 12, 45, 20, 45, 12, 539, 12, 45, 20, 45, 12, 175, 12, 45, 20, 63, 12, 275, 12, 45, 28, 45, 12, 175, 12, 63, 20, 45, 12, 175, 12, 45, 20, 45, 12, 539, 12, 45, 20

Offset: 1

Michael De Vlieger, Mar 31 2025

Let p be the smallest prime that is coprime to n and let q be the second smallest prime that is coprime to n. Then a(n) = p^2 * q.

Records in this sequence are set by n in A002110.

			a(1) = 12 = 2^2*3, since p = 2, q = 3.
a(2) = 45 = 3^2*5, since p = 3, q = 5.
a(3) = 20 = 2^2*5, since p = 2, q = 5.
a(4) = 45 = 3^2*5, since p = 3, q = 5, a(2^i) = 45 for i > 0.
a(6) = 175 = 5^2*7, since p = 5, q = 7.
a(9) = 20 = 2^2*5, since p = 2, q = 5, a(3^i) = 20 for i > 0.
a(10) = 63 = 3^2*7, since p = 3, q = 7.
a(12) = 175 = 5^2*7, since p = 5, q = 7, a(k) = 175 for n in A033845 (i.e., n such that rad(n) = 6).
a(20) = 63 = 3^2*7, since p = 3, q = 7, a(k) = 63 for n in A033846 (i.e., n such that rad(n) = 10).
a(30) = 539 = 7^2*11, since p = 7, q = 11, etc.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A002110, A007947, A053669, A126706, A380539.

Table[c = 0; q = 2; Times @@ Reap[While[c < 2, While[Divisible[n, q], q = NextPrime[q]]; Sow[q^(2 - c)]; q = NextPrime[q]; c++] ][[-1, 1]], {n, 120}]

a(n) = my(k=2); while (isprimepower(k) || issquarefree(k) || (gcd(k, n) != 1) , k++); k; \\ Michel Marcus, Apr 01 2025

a(n) = A053669(n)^2 * A380539(n).

For k and m such that rad(k) = rad(m), a(k) = a(m), where rad = A007947.

A343300 a(n) is p1^1 + p2^2 + ... + pk^k where {p1,p2,...,pk} are the distinct prime factors in ascending order in the prime factorization of n.

Original entry on oeis.org

0, 2, 3, 2, 5, 11, 7, 2, 3, 27, 11, 11, 13, 51, 28, 2, 17, 11, 19, 27, 52, 123, 23, 11, 5, 171, 3, 51, 29, 136, 31, 2, 124, 291, 54, 11, 37, 363, 172, 27, 41, 354, 43, 123, 28, 531, 47, 11, 7, 27, 292, 171, 53, 11, 126, 51, 364, 843, 59, 136, 61, 963, 52, 2, 174, 1342, 67, 291, 532, 370, 71, 11, 73

Offset: 1

Giorgos Kalogeropoulos, Apr 11 2021

nonn

From Bernard Schott, May 07 2021: (Start)
a(n) depends only on prime factors of n (see formulas).
Primes are fixed points of this sequence.
Terms are in increasing order in A344023. (End)

			a(60) = 136 because the distinct prime factors of 60 are {2, 3, 5} and 2^1 + 3^2 + 5^3 = 136.

Michel Marcus, Table of n, a(n) for n = 1..10000

Cf. A027748, A344023 (terms ordered).

A339794 a(n) is the least integer k satisfying rad(k)^2 < sigma(k) and whose prime factors set is the same as the prime factors set of A005117(n+1).

Original entry on oeis.org

4, 9, 25, 18, 49, 80, 121, 169, 112, 135, 289, 361, 441, 352, 529, 416, 841, 360, 961, 891, 1088, 875, 1369, 1216, 1053, 1681, 672, 1849, 1472, 2209, 2601, 2809, 3025, 3249, 1856, 3481, 3721, 1984, 4225, 1584, 4489, 4761, 1960, 5041, 5329, 4736, 5929, 2496, 6241

Offset: 1

Michel Marcus, Dec 17 2020

nonn

Equivalently, subsequence of terms of A339744 excluding terms whose prime factor set has already been encountered.
a(n) = A005117(n + 1)^2 when A005117(n + 1) is prime. Proof: if A005117(n + 1) is a prime p then rad(A005117(n + 1))^2 = rad(p)^2 = p^2 and so integers whose prime factors set is the same as the prime factors set of A005117(n + 1) = p are p^m where m >= 1. p^2 > sigma(p^1) = p + 1 but p^2 < sigma(p^2) = p^2 + p + 1. Q.E.D. - David A. Corneth, Dec 19 2020
From Bernard Schott, Jan 19 2021: (Start)
Indeed, a(n) satisfies the double inequality A005117(n+1) < a(n) <= A005117(n+1)^2.
It is also possible that a(n) = A005117(n+1)^2, even when A005117(n+1) is not prime; the smallest such example is for a(13) = 441 = 21^2 = A005117(14)^2. (End)

			   n  a(n) prime factor set
   1    4  [2]           A000079
   2    9  [3]           A000244
   3   25  [5]           A000351
   4   18  [2, 3]        A033845
   5   49  [7]           A000420
   6   80  [2, 5]        A033846
   7  121  [11]          A001020
   8  169  [13]          A001022
   9  112  [2, 7]        A033847
  10  135  [3, 5]        A033849
  11  289  [17]          A001026
  12  361  [19]          A001029
  13  441  [3, 7]        A033850
  14  352  [2, 11]       A033848
  15  529  [23]          A009967
  16  416  [2, 13]       A288162
  17  841  [29]          A009973
  18  360  [2, 3, 5]     A143207

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from Michel Marcus)

Cf. A000203 (sigma), A007947 (rad).
Cf. A005117 (squarefree numbers), A027748, A265668, A339744.
Subsequence: A001248 (squares of primes).

u(n) = {my(fn=factor(n)[,1]); for (k = n, n^2, my(fk = factor(k)); if (fk[,1] == fn, if (factorback(fk[,1])^2 < sigma(fk), return (k));););}
lista(nn) = {for (n=2, nn, if (issquarefree(n), print1(u(n), ", ");););}

a(n) <= A005117(n+1)^2. - David A. Corneth, Dec 19 2020

A370266 Numbers k that are not prime powers, such that k/rad(k) >= rad(k), where rad(k) = A007947(k).

Original entry on oeis.org

36, 48, 54, 72, 96, 100, 108, 144, 160, 162, 192, 196, 200, 216, 224, 225, 250, 288, 320, 324, 375, 384, 392, 400, 405, 432, 441, 448, 484, 486, 500, 567, 576, 640, 648, 675, 676, 686, 704, 768, 784, 800, 832, 864, 896, 900, 960, 968, 972, 1000, 1029, 1080, 1089

Offset: 1

Michael De Vlieger, Feb 18 2024

Numbers k = m * s, where s is composite and squarefree, rad(m) | s, and m >= s.

A177492 is a proper subset.

			For s = 6, this sequence contains {36, 48, 54, 72, 96, ...}, i.e., A033845(n) for n >= A010846(6).
For s = 10, this sequence contains {100, 160, 200, 250, 320, ...}, i.e., A033846(n) for n >= A010846(10).
For s = 14, this sequence contains {196, 224, 392, 448, 686, ...}, i.e., A033847(n) for n >= A010846(14).
For s = 15, this sequence contains {225, 375, 405, 675, 1125, ...}, i.e., A033849(n) for n >= A010846(15), etc.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A007947, A010846, A024619, A120944, A126706, A177492, A246547, A286708, A341645, A366250.

Select[Select[Range[2, 1100], Not@*PrimePowerQ], #1/#2 >= #2 & @@ {#, Times @@ FactorInteger[#][[All, 1]]} &]

Set difference of A341645 and A246547.
Intersection of A341645 and A126706.
Union of A286708 and A366250.

A372864 Numbers k such that A372720(k) = 0.

Original entry on oeis.org

1, 500, 578, 722, 750, 1058, 1500, 1682, 1922, 2646, 2744, 3430, 3645, 4800, 5202, 5346, 5476, 5488, 5625, 6318, 6400, 6724, 7168, 7396, 8000, 8836, 10092, 10976, 11236, 11532, 11979, 12005, 13068, 13924, 14450, 14884, 15309, 16810, 16875, 16896, 18050, 18225

Offset: 1

Michael De Vlieger, Jun 02 2024

nonn

Let tau = A000005, let omega = A001221, let f = A008479, and let g = A372720.
For squarefree k, A372720(k) >= 0, since f(k) = 1 while tau(k) = 2^omega(k).
For prime power p^m, A372720(p^m) = 1, since f(p^m) = m while tau(k) = m+1.
Therefore, apart from a(1) = 1, this sequence is a proper subset of A126706.
In the sequence R = {k = m*s : rad(m) | s, s > 1 in A120944}, there is a smallest term k such that g(k) <= 0 and a largest term k such that g(k) is positive. For instance, in A033845 where s = 6, only {6, 12, 18, 24, 36, 48, 54, 72, 96, 108, 144, 192, 216, 288, 432, 576, 864} are such that g(k) > 0.
Apart from terms in this sequence, all the rest of the terms k in R are such that g(k) is negative.
There are no 3-smooth numbers k > 1 in this sequence, however there are 3 terms {500, 6400, 8000} in A033846 (with s = rad(k) = 10). For s = 2*3*23, there are 6 terms {19044, 25392, 38088, 70656, 536544, 953856}.
Conjecture: proper subset of A361098, hence of A360765 and A360768. This is to say that k = a(n) is such that A003557(k) >= A119288(k), i.e., k/rad(k) >= second smallest prime factor of k, and A003557(k) > A053669(k), where A053669(k) is the smallest prime q that does not divide k.

			a(1) = 1 since tau(1) - f(1) = 1 - 1 = 0.
a(2) = 500 = 2^2 * 5*3, since tau(500) - f(500)
     = (2+1)*(3+1) - card({10,20,40,50,80,100,160,200,250,320,400,500})
     = 12 - 12 = 0.
a(3) = 578 = 2*17^2, since tau(578) - f(578)
     = (1+1)*(2+1) - card({34,68,136,272,544,578})
     = 6 - 6 = 0, etc.

Michael De Vlieger, Table of n, a(n) for n = 1..2000

Cf. A000005, A001221, A007947, A008479, A010846, A372720, A372972.

rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]]; Position[Table[r = rad[n]; DivisorSigma[0, n] - Count[Range[n/r], ?(Divisible[r, rad[#]] &)], {n, 20000}], ?(# == 0 &)][[All, 1]]

A381805 Smallest composite squarefree number that is coprime to n.

Original entry on oeis.org

6, 15, 10, 15, 6, 35, 6, 15, 10, 21, 6, 35, 6, 15, 14, 15, 6, 35, 6, 21, 10, 15, 6, 35, 6, 15, 10, 15, 6, 77, 6, 15, 10, 15, 6, 35, 6, 15, 10, 21, 6, 55, 6, 15, 14, 15, 6, 35, 6, 21, 10, 15, 6, 35, 6, 15, 10, 15, 6, 77, 6, 15, 10, 15, 6, 35, 6, 15, 10, 33, 6, 35

Offset: 1

Michael De Vlieger, Mar 31 2025

Let p be the smallest prime that is coprime to n and let q be the second smallest prime that is coprime to n. Then a(n) = p*q.

Records in this sequence are set by n in A002110.

			a(1) = 6 = 2*3, since p = 2, q = 3.
a(2) = 15 = 3*5, since p = 3, q = 5.
a(3) = 10 = 2*5, since p = 2, q = 5.
a(4) = 15 = 3*5, since p = 3, q = 5, a(2^i) = 15 for i > 0.
a(6) = 35 = 5*7, since p = 5, q = 7.
a(9) = 20 = 2*5, since p = 2, q = 5, a(3^i) = 10 for i > 0.
a(10) = 21 = 3*7, since p = 3, q = 7.
a(12) = 35 = 5*7, since p = 5, q = 7, a(k) = 35 for n in A033845 (i.e., n such that rad(n) = 6).
a(20) = 21 = 3*7, since p = 3, q = 7, a(k) = 21 for n in A033846 (i.e., n such that rad(n) = 10).
a(30) = 77 = 7*11, since p = 7, q = 11, etc.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A002110, A007947, A051250, A053669, A120944, A380539, A382248.

Table[c = 0; q = 2; Times @@ Reap[While[c < 2, While[Divisible[n, q], q = NextPrime[q]]; Sow[q]; q = NextPrime[q]; c++] ][[-1, 1]], {n, 120}]

a(n) = my(k=2); while (isprime(k) || !issquarefree(k) || (gcd(k, n) != 1) , k++); k; \\ Michel Marcus, Apr 01 2025

a(n) = A053669(n) * A380539(n) = A382248(n)/A020639(n).
For k and m such that rad(k) = rad(m), a(k) = a(m), where rad = A007947.
n < a(n) for n in A051250, a finite sequence whose largest term is 60.

A382659 Numbers k such that k < A053669(k)^2 * A380539(k), i.e., k < A382248(k).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 48, 50, 54, 60, 66, 70, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 210, 240, 252, 270, 300, 330, 360, 390

Offset: 1

Michael De Vlieger, Apr 14 2025

Numbers k whose reduced residue system (RRS) does not intersect A126706 (i.e., the sequence of numbers that are neither squarefree nor prime powers). Alternatively, numbers k whose RRS is a subset of A303554 (i.e., the union of powers of primes and squarefree numbers).
Let p = A053669(k), and let q = A380539(k). Thus, p and q are the smallest and second smallest primes, respectively, that do not divide k. Let m = p^2 * q = A382248(k). Then this sequence is that of k such that k < m.
There are 72 terms in this sequence.
Sequences A048597 and A051250 are proper subsets of this sequence.

			Let omega = A001221.
For omega = 0, we have the subset {1}. 1 is in the sequence since 1 < m, m = 2^2 * 3 = 12.
For omega = 1, we have the subset {2, 3, 4, 5, 7, 8, 9, 11, 16, 32}.
  11 is in the sequence since 11 < m, m = 2^2 * 3 = 12, but 13 is not, since 13 > 12.
  9 is in the sequence since 9 < m, m = 2^2 * 5 = 20.
  25 is not a term since 25 > 12, and 27 is not a term since 27 > 20.
For omega = 2, we have the subset {6, 10, 12, 14, 15, 18, 20, 22, 24, 26, 28, 34, 36, 38, 40, 44, 48, 50, 54, 72, 96, 108, 144, 162}.
  38 = 2*19 is a term since 38 < 45, 45 = 3^2 * 5, but 46 = 2*23 is not, since 46 > 45.
  15 = 3*5 is a term since 15 < 20, but 21 is not, since 21 > 20 and 35 is not, since 35 > 12.
  Intersection with A033845 = {k : rad(k) = 6} is {6, 12, 18, 24, 36, 48, 54, 72, 96, 108, 144, 162}, since m = 5^2 * 7 = 175.
  Intersection with A033846 = {k : rad(k) = 10} is {10, 20, 40, 50}, since m = 3^2 * 7 = 63.
  Intersection with A033847 = {k : rad(k) = 14} is {14, 28}, since m = 3^2 * 5 = 45, etc.
For omega = 3, we have the subset {30, 42, 60, 66, 70, 78, 84, 90, 102, 114, 120, 126, 132, 138, 150, 156, 168, 174, 180, 240, 252, 270, 300, 360, 450, 480}, of which {30, 42, 66, 70, 78, 102, 114, 138, 174} are squarefree.
  Intersection with A143207 = {k : rad(k) = 30} is {30, 60, 90, .., 480} because m = 7^2 * 11 = 539.
  Intersection with 42*A108319 = {k : rad(k) = 42} is {42, 84, 126, 168}, since m = 5^2 * 11 = 275, etc.
For omega = 4, we have the subset {210, 330, 390, 420, 510, 630, 840, 1050, 1260, 1470}, of which {210, 330, 390, 510} are squarefree.
  Intersection with A147571 = {k : rad(k) = 210} is {210, 420, 630, 840, 1050, 1260, 1470} since m = 11^2 * 13 = 1573, etc.
For omega = 5, we have 2310 = 2*3*5*7*11, a term since 2310 < 13*17 = 2873; 2730 = 2*3*5*7*13 is not a term.
There are no terms larger than 2310, since the intersection with A147572 = {2310}, 2730 is not a term, and k = Product_{i=1..j} prime(i), k > prime(j+1)^2 * prime(j+2) for j > 5. Therefore the sequence is finite like A051250.

Michael De Vlieger, Table of n, a(n) for n = 1..72

Cf. A048597 (k such that k < p^2), A051250 (k such that k < p*q), A053669, A126706, A303554, A380539, A382248, A382960.

Select[Range[30030], Function[n, c = 0; q = 2; n < Times @@ Reap[While[c < 2, While[Divisible[n, q], q = NextPrime[q]]; Sow[q^(2 - c)]; q = NextPrime[q]; c++]][[-1, 1]] ] ]

cp's OEIS Frontend

A284311 Array T(n,k) read by antidiagonals (downward): T(1,k) = A005117(k+1) (squarefree numbers > 1); for n > 1, columns are nonsquarefree numbers (in ascending order) with exactly the same prime factors as T(1,k).

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

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A288162 Numbers whose prime factors are 2 and 13.

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Magma

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A382248 Smallest number k that is neither squarefree nor a prime power such that k is coprime to n.

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A343300 a(n) is p1^1 + p2^2 + ... + pk^k where {p1,p2,...,pk} are the distinct prime factors in ascending order in the prime factorization of n.

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Maple

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A339794 a(n) is the least integer k satisfying rad(k)^2 < sigma(k) and whose prime factors set is the same as the prime factors set of A005117(n+1).

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A370266 Numbers k that are not prime powers, such that k/rad(k) >= rad(k), where rad(k) = A007947(k).

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