A352475 -id:A352475 - cp's OEIS frontend

A354178 Numbers whose number of divisors is coprime to 30.

1, 64, 729, 1024, 4096, 15625, 46656, 59049, 65536, 117649, 262144, 531441, 746496, 1000000, 1771561, 2985984, 3779136, 4194304, 4826809, 7529536, 9765625, 11390625, 16000000, 24137569, 34012224, 43046721, 47045881, 47775744, 60466176, 64000000, 85766121, 113379904

Offset: 1

Amiram Eldar, May 18 2022

nonn

Numbers k such that gcd(d(k), 30) = 1, where d(k) is the number of divisors of k (A000005).

All the terms are squares since their number of divisors is odd.

			64 is a term since A000005(64) = 7 and gcd(7, 30) = 1.

Michael De Vlieger, Table of n, a(n) for n = 1..10000 (first 1709 terms from Amiram Eldar)
Titus Hilberdink, How often is d(n) a power of a given integer?, Journal of Number Theory, Vol. 236 (2022), pp. 261-279.
Index entries for sequences computed from exponents in factorization of n.

Subsequence of other sequences of numbers k such that gcd(d(k), m) = 1: A000290 (m=2), A336590 (m=3), A352475 (m=6).

Cf. A000005, A007775, A354179.

Select[Range[10^4]^2, CoprimeQ[DivisorSigma[0, #], 30] &]

isok(k) = gcd(numdiv(k), 30) == 1;
for(k=1, 10650, if(isok(k^2), print1(k^2,", ")))

a(n) = A354179(n)^2.
The number of terms <= x is (zeta(5)*zeta(5/3))/(zeta(4)*zeta(10/3))*x^(1/6) + (zeta(3)*zeta(3/5))/(zeta(2)*zeta(12/5))*x^(1/10) + O(x^(1/20 + eps)) for all eps > 0 (Hilberdink, 2022).
Sum_{n>=1} 1/a(n) = Product_{p prime} (p^2 + p^8 + p^12 + p^14 + p^18 + p^20 + p^24 + p^30)/(p^30 - 1) = 1.0183538548...

A355269 Lexicographically earliest infinite sequence of distinct positive integers such that a(n+1) is prime to the number of divisors of a(n).

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 4, 8, 11, 13, 15, 17, 19, 21, 23, 25, 10, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 14, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 6, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119

Offset: 1

David James Sycamore and Michael De Vlieger, Jun 26 2022

nonn

1,2,3 are the earliest consecutive numbers satisfying the definition, therefore these are the initial terms. The sequence is infinite since there is always a number prime to d(a(n)), and we take the least such as a(n+1).
Tau(k) is odd iff k is a square, therefore if a(n) is nonsquare, a(n+1) is odd. Consequently the sequence displays runs of consecutive odd nonsquare numbers, extending between successive odd squares, whereupon the parity of tau(a(n)) changes from even to odd. While odd numbers occur early, even numbers are very delayed because (in general) an even number can only be admitted following an odd square, or (on rarer occasions) following an even square subsequent to an odd square (the only time we observe two consecutive even terms).
Conjectures: When a(n) is square a(n+1) is even; no two squares of equal parity can appear as adjacent terms; sequence is a permutation of the positive integers with primes in natural order.
From Michael De Vlieger, Jun 28 2022: (Start)
Let j = a(n-1), k = a(n), tau(m) = A000005(m), and P(m) = A002110(m).
Theorem: prime p | tau(j) implies gcd(p, k) = 1. Consequence of sequence definition.
Even tau(j) implies nonsquare j (in A000037), which in turn implies odd k. Prime j implies tau(j) = 2, which in turn implies odd k. Therefore there is a significant tendency for k to be odd.
Square j (in A000290) implies odd tau(j) but do not necessarily lead to even k. For n > 3, the smallest missing numbers are even, therefore even k is likely to enter the sequence following square j.
Terms j such that tau(j) is coprime to 6, i.e., j in A352475, allow successor k such that 6 | k. For n > 7, the smallest missing number is divisible by 6 because we usually have gcd(tau(j), 6) > 1. Therefore k such that 6 | k likely follows j such that tau(j) is coprime to 6. A similar argument can be made regarding j such that tau(j) is coprime to 30, i.e., j in A354178.
Primorials appear latest in this sequence because they are divisible by the smallest primes. For example, a(1132) = 30 and a(4884401) = 210.
Generally, we have k : k | P(m) iff j : (tau(j), P(m)) = 1. Since gcd(tau(j), P(m)) > 1 is more common than gcd(tau(j), P(m)) = 1 for m > 1, we are apt to see k | P(m). (End)

			a(4)=5 because a(3)=3 has 2 divisors and 5 is the least unused term prime to 2. Likewise a(5) = 7, and a(6) = 9, the first odd square. Since 9 has 3 divisors a(7) is 4, the smallest unused number prime to 3, and the first occurrence of an even square, also having 3 divisors. Consequently a(8)=8, the smallest unused number prime to 3. Thus we see 4,8 the first occasion of a pair of adjacent even terms, consequence of odd square (9) followed by even square (4). The next occasion is a(67,68,69)=121,16,12.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Annotated log log scatterplot of a(n), n = 1..2^16, showing records in red and local minima in blue. Three prominent trajectories appear; that of odd and early terms mostly in red, that of terms divisible by 6 mostly in blue, and even numbers that are not divisible by 6 in green. (It may be that numbers divisible by 6 are not always minima; instead, numbers divisible by a larger primorial may be delayed to a greater degree than those merely divisible by 6 for large n.)

Cf. A000005, A000037, A000290, A002110, A352475, A354178, A354903, A016742, A016754.

Block[{a, c, j, k, u, v, nn}, nn = 120; a[1] = c[1] = j = 1; u = 2; v = 3; Do[k = u; If[EvenQ[j], k = v; While[Nand[c[k] == 0, CoprimeQ[j, k]], k += 2], While[Nand[c[k] == 0, CoprimeQ[j, k]], k++]]; Set[{a[i], c[k]}, {k, i}]; j = DivisorSigma[0, k]; If[k == u, While[c[u] != 0, u++]]; If[k == v, While[c[v] != 0, v += 2]], {i, 2, nn}]; Array[a, nn] ] (* Michael De Vlieger, Jun 28 2022 *)

from math import gcd
from sympy import divisor_count
from itertools import count, islice
def agen(): # generator of terms
    aset, k, mink = {1, 2}, 1, 3; yield from [1, 2]
    for n in count(3):
        an, k = k, mink
        while k in aset or not gcd(divisor_count(an), k) == 1: k += 1
        aset.add(k); yield k
        while mink in aset: mink += 1
print(list(islice(agen(), 66))) # Michael S. Branicky, Jun 28 2022

A355058 Numbers m such that d(m) mod 6 = 3, where d(m) is the number of divisors of m.

Original entry on oeis.org

4, 9, 25, 36, 49, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 676, 784, 841, 900, 961, 1089, 1156, 1225, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364

Offset: 1

Michael De Vlieger, Jul 04 2022

All terms are square; contains squares of primes.

			Let p be a prime; p^2 has 3 divisors {1, p, p^2}, therefore all squares of primes {4, 9, 25, 49, ...} are in the sequence.
36 is in the sequence because d(36) = 9, and 9 mod 6 = 3.
16 is not in the sequence because it has 5 divisors, and 5 mod 6 = 5.

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

Cf. A000005, A000290, A001248, A086463, A352475.

Select[Range[2^12], Mod[DivisorSigma[0, #], 6] == 3 &]

isok(m) = (numdiv(m) % 6) == 3; \\ Michel Marcus, Jul 05 2022

from itertools import count, islice
from sympy import factorint, prod
def A355058_gen(): # generator of terms
    return map(lambda n:n**2,filter(lambda n:prod((2*e+1)%6 for e in factorint(n).values())%6==3,count(1)))
A355058_list = list(islice(A355058_gen(),30)) # Chai Wah Wu, Jul 06 2022

{ A000290 \ A352475 }.

Sum_{n>=1} 1/a(n) = Pi^2/18 (A086463). - Amiram Eldar, Jul 06 2022

A358001 Numbers whose number of divisors is coprime to 210.

Original entry on oeis.org

1, 1024, 4096, 59049, 65536, 262144, 531441, 4194304, 9765625, 43046721, 60466176, 241864704, 244140625, 268435456, 282475249, 387420489, 544195584, 1073741824, 2176782336, 3869835264, 10000000000, 13841287201, 15479341056, 25937424601, 31381059609, 34828517376

Offset: 1

Michael De Vlieger, Dec 03 2022

nonn

210 is the product of the smallest 4 primes.
Numbers k such that gcd(d(k), 210) = 1, where d(k) is the number of divisors of k (A000005).
The square roots of terms are in A001694.

Michael De Vlieger, Table of n, a(n) for n = 1..5000

Subsequence of other sequences of numbers k such that gcd(d(k), m) = 1: A000290 (m=2), A336590 (m=3), A352475 (m=6), A354178 (m=30).

Cf. A000005, A001694, A008364.

With[{nn = 200000}, Select[Union@ Flatten@ Table[a^2*b^3, {b, nn^(1/3)}, {a, Sqrt[nn/b^3]}], CoprimeQ[DivisorSigma[0, #^2], 210] &]^2]

a(n) = A358250(n)^2.

Sum_{n>=1} 1/a(n) = Product_{p prime} (Sum_{k=2..210, gcd(k-1,210)=1} p^k)/(p^210-1) = 1.001258995976... . - Amiram Eldar, Dec 06 2022

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A354178 Numbers whose number of divisors is coprime to 30.

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A355269 Lexicographically earliest infinite sequence of distinct positive integers such that a(n+1) is prime to the number of divisors of a(n).

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A355058 Numbers m such that d(m) mod 6 = 3, where d(m) is the number of divisors of m.

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A358001 Numbers whose number of divisors is coprime to 210.

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