A030515
Numbers with exactly 6 divisors.
Original entry on oeis.org
12, 18, 20, 28, 32, 44, 45, 50, 52, 63, 68, 75, 76, 92, 98, 99, 116, 117, 124, 147, 148, 153, 164, 171, 172, 175, 188, 207, 212, 236, 242, 243, 244, 245, 261, 268, 275, 279, 284, 292, 316, 325, 332, 333, 338, 356, 363, 369, 387, 388, 404, 412, 423, 425, 428
Offset: 1
- Amarnath Murthy, A note on the Smarandache Divisor sequences, Smarandache Notions Journal, Vol. 11, 1-2-3, Spring 2000.
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N:= 1000: # to get all terms <= N
Primes:= select(isprime, {2,seq(i,i=3..floor(N/4))}):
S:= select(`<=`,{seq(p^5, p = Primes),seq(seq(p*q^2, p=Primes minus {q}),q=Primes)},N):
sort(convert(S,list)); # Robert Israel, Feb 10 2016
-
f[n_]:=Length[Divisors[n]]==6; lst={};Do[If[f[n],AppendTo[lst,n]],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 14 2009 *)
Select[Range[500],DivisorSigma[0,#]==6&] (* Harvey P. Dale, Oct 02 2014 *)
-
is(n)=numdiv(n)==6 \\ Charles R Greathouse IV, Jan 23 2014
-
from sympy import divisor_count
def ok(n): return divisor_count(n) == 6
print([k for k in range(429) if ok(k)]) # Michael S. Branicky, Dec 18 2021
-
from math import isqrt
from sympy import primepi, primerange, integer_nthroot
def A030515(n):
def bisection(f,kmin=0,kmax=1):
while f(kmax) > kmax: kmax <<= 1
kmin = kmax >> 1
while kmax-kmin > 1:
kmid = kmax+kmin>>1
if f(kmid) <= kmid:
kmax = kmid
else:
kmin = kmid
return kmax
def f(x): return n+x-sum(primepi(x//p**2) for p in primerange(isqrt(x)+1))+primepi(integer_nthroot(x,3)[0])-primepi(integer_nthroot(x,5)[0])
return bisection(f,n,n) # Chai Wah Wu, Feb 21 2025
A074172
Smaller of two consecutive numbers of the form p^2*q where p and q are distinct primes.
Original entry on oeis.org
44, 75, 98, 116, 147, 171, 244, 332, 387, 507, 548, 603, 604, 724, 844, 908, 931, 963, 1075, 1083, 1251, 1324, 1412, 1467, 1556, 1587, 1675, 1772, 2523, 2524, 2636, 2644, 2763, 3283, 3356, 3411, 3508, 3788, 3987, 4075, 4203, 4204, 4418, 4491, 4804, 4868
Offset: 1
44 is a member as 44 = 2^2*11 and 45 = 3^2*5.
-
filter:= proc(n) local F;
F:= map(t -> t[2],ifactors(n)[2]);
F = [2,1] or F = [1,2]
end proc:
A054753:= select(filter, {$1..10000}):
sort(convert(A054753 intersect map(`-`,A054753,1),list)); # Robert Israel, Dec 06 2018
-
lst={}; Do[f1=FactorInteger[n]; If[Sort[Transpose[f1][[2]]]=={1, 2}, f2=FactorInteger[n+1]; If[Sort[Transpose[f2][[2]]]=={1, 2}, AppendTo[lst, n]]], {n, 3, 10000}]; lst
-
isok1(n) = vecsort(factor(n)[,2]) == [1,2]~;
isok(n) = isok1(n) && isok1(n+1); \\ Michel Marcus, Sep 20 2017
A178032
Numbers n such that n, n+1, n+2 are all of the form p*q^2 for distinct primes p,q.
Original entry on oeis.org
603, 2523, 4203, 4923, 7442, 10467, 18027, 20402, 54475, 58923, 79011, 97675, 104211, 118323, 120787, 122571, 124891, 132723, 134307, 148075, 200491, 229707, 243602, 246571, 249307, 258507, 303651, 324331, 331387, 370827, 385675
Offset: 1
603=3*3*67, 604=2*2*151, 605=5*11*11
2523=3*29*29, 2524=2*2*631, 2525=5*5*101
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SequencePosition[Table[If[Sort[FactorInteger[n][[All,2]]]=={1,2},1,0],{n,400000}],{1,1,1}][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Mar 30 2019 *)
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is_A178032 = lambda n: all(sorted(m for p,m in factor(k)) == [1, 2] for k in (n, n+1, n+2)) # D. S. McNeil, Dec 17 2010
A308683
Numbers k such that k, k+1, k+2, k+3 are all of the form p*q^2 for distinct primes p,q.
Original entry on oeis.org
17042641441, 180383003522, 231242161921, 313187053922, 349881182161, 717767333521, 740230547281, 907336510801, 1080505860722, 1646310367681, 1682581387922, 1896109379522, 2638936079522, 2919248012401, 3121595927522
Offset: 1
17042641441 = 7 * 7 * 347809009,
17042641442 = 2 * 92311 * 92311,
17042641443 = 3 * 3 * 1893626827,
17042641444 = 2 * 2 * 4260660361.
A178810
Largest possible number of consecutive integers with the same prime signature as A025487(n).
Original entry on oeis.org
1, 2, 1, 3, 1, 5, 1, 3, 3, 1, 1, 3, 7, 1
Offset: 1
A025487(2) = 2, prime signature {1}. There are a maximum of 2 consecutive integers with that prime signature: 2 and 3.
A025487(4) = 6, prime signature {1,1}. There are a maximum of 3 consecutive integers with that prime signature (e.g., 33, 34 and 35).
A025487(6) = 12, prime signature {1,2}. There are a maximum of 5 consecutive integers with that prime signature (e.g., 10093613546512321, 10093613546512322, 10093613546512323, 10093613546512324, and 10093613546512325). Compare A141621.
A025487(13) = 60, prime signature {1,1,2}. There are a maximum of 7 possible consecutive integers, between two multiples of 8, with that prime signature; the smallest such run starts at 932537185321. - _Bernard Schott_, Feb 16 2021
A178811
The smallest integer that begins the longest run of consecutive integers with the prime signature of A025487(n).
Original entry on oeis.org
1, 2, 4, 33, 8, 10093613546512321, 16, 28375, 1309, 32, 36, 7939375, 932537185321, 64
Offset: 1
For n = 3, A025487(3) = 4, corresponding to a prime signature of {2}. Since the maximum number of consecutive integers with that prime signature is 1, a(3) is 4, the smallest integer that starts a "run" of 1.
A025487(4) = 6 whose prime signature is {1,1}; a(4) = 33 because 33 is the smallest integer where starts a run of A178810(4) = 3 consecutive integers with prime signature {1,1}: (33=3*11, 34=2*17, 35=5*7). - _Bernard Schott_, Feb 16 2021
A323743
Table read by rows: row n lists the numbers k for which there exist only finitely many runs of n consecutive integers whose number-of-divisors function sums to k.
Original entry on oeis.org
1, 3, 4, 5, 5, 7, 8, 9, 8, 9, 11, 12, 13, 14, 15, 10, 13, 15, 17, 18, 19, 14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 16, 19, 20, 21, 22, 23, 25, 26, 27, 29, 30, 31, 20, 22, 24, 27, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39
Offset: 1
There is only one number with exactly 1 divisor (namely, k=1), but there are infinitely many numbers with j divisors for every j >= 2, so row 1 consists only of the single term 1.
The sequence of values tau(k) for k >= 1 is A000005, which begins 1, 2, 2, 3, 2, 4, 2, 4, 3, 4, ..., from which the sums of two consecutive terms are 1+2=3, 2+2=4, 2+3=5, 3+2=5, 2+4=6, 4+2=6, 2+4=6, 4+3=7, 3+4=7, ...; no number j < 3 appears as such a sum, every j >= 6 appears infinitely many times as such a sum, and each j in {3,4,5} appears as such a sum only finitely many times, so row 2 is {3, 4, 5}.
Row 3 does not contain 6 as a term because there exists no run of 3 consecutive numbers whose sum of tau values is exactly 6.
The first six rows of the table are as follows:
row 1: {1};
row 2: {3, 4, 5};
row 3: {5, 7, 8, 9};
row 4: {8, 9, 11, 12, 13, 14, 15};
row 5: {10, 13, 15, 17, 18, 19};
row 6: {14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27}.
Cf.
A000005,
A005237,
A006558,
A048892,
A072507,
A100366,
A119479,
A141621,
A284596,
A284597,
A292580,
A319037,
A319045,
A319046.
A373560
a(n) is the smallest multiple of prime(n)^2 that starts a run of 5 consecutive integers with 6 divisors, or -1 if no such multiple exists.
Original entry on oeis.org
-1, -1, -1, 10093613546512321, -1, -1, 7700031346933907521, -1, 5344962129269790721, -1, 20453982425165652721, -1, 8163195338222675521, -1, 2467958104789157112721, -1, -1, -1, -1, 14666767069023896053921, 212170739123852995921, 287954235303137500060321, -1, 84769922583214545304321
Offset: 1
a(1) = a(2) = a(3) = -1 because the first of five consecutive integers having six divisors is never a multiple of 2^2, 3^2, or 5^2.
a(4) = 10093613546512321 because it is the smallest term in A141621 that is a multiple of prime(4)^2 = 49.
a(9) = 5344962129269790721 because it is the smallest term in A141621 that is a multiple of prime(9)^2 = 23^2.
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