A003593 -id:A003593 - cp's OEIS frontend

A095678 Minimal sequence such that all triples of consecutive numbers have no common divisor greater than 1, but all three pairs within the triples are not coprime.

6, 10, 15, 12, 20, 45, 18, 40, 75, 24, 50, 135, 36, 80, 225, 48, 100, 375, 54, 160, 405, 72, 200, 675, 96, 250, 1125, 108, 320, 1215, 144, 400, 1875, 162, 500, 2025, 192, 640, 3375, 216, 800, 3645, 288, 1000, 5625, 324, 1250, 6075, 384, 1280, 9375, 432

Offset: 1

Reinhard Zumkeller, Jul 04 2004

nonn

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

Cf. A001221, A003586, A003592, A003593, A006530, A020639, A033845, A033846, A033849.

seq1[p_, q_, lim_] := Sort[Flatten[Table[p^i * q^j, {i, 1, Log[p, lim]}, {j, 1, Log[q, lim/p^i]}]]];
seq[lim_] := Module[{s1 = seq1[2, 3, lim], s2 = seq1[2, 5, lim], s3 = seq1[3, 5, lim], ns}, ns = Length[s3]; Flatten[Transpose[{s1[[1;;ns]], s2[[1;;ns]], s3}]]]; seq[10^4] (* Amiram Eldar, Sep 29 2024 *)

gcd(a(n),a(n+1),a(n+2)) = 1, gcd(a(n),a(n+1)) > 1, gcd(a(n),a(n+2)) > 1 and gcd(a(n+1),a(n+2)) > 1.
A001221(a(n)) = 2; 2 <= A020639(a(n)) <= 3 <= A006530(a(n)) <= 5.
From Jianing Song, Jun 08 2022: (Start)
a(3n-2) = A033845(n) = 6*A003586(n);
a(3n-1) = A033846(n) = 10*A003592(n);
a(3n) = A033849(n) = 15*A003593(n). (End)
Sum_{n>=1} 1/a(n) = 7/8. - Amiram Eldar, Sep 29 2024

A231114 Numbers k dividing u(k), where the Lucas sequence is defined u(i) = u(i-1) - 4*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 125, 135, 171, 225, 243, 375, 405, 435, 465, 513, 625, 675, 729, 855, 1125, 1215, 1305, 1395, 1539, 1875, 2025, 2175, 2187, 2325, 2565, 3125, 3249, 3375, 3645, 3725, 3915, 4005, 4185, 4275, 4617, 5625, 6075, 6327, 6525, 6561

Offset: 1

Thomas M. Bridge, Nov 06 2013

nonn

Every term (except leading term) is divisible by at least one of 3 or 5.

Furthermore, this sequence contains 3^i*5^j for all i, j >= 0, that is, A003593 is a subsequence.

			The sequence u(i) begins 0, 1, 1, -3, -7, 5, 33. Only for k = 1, 3, 5 does k divides u(k).

Amiram Eldar, Table of n, a(n) for n = 1..1000
C. Smyth, The terms in Lucas sequences divisible by their indices, Journal of Integer Sequences, Vol.13 (2010), Article 10.2.4.
Wikipedia, Lucas sequence

Cf. A003593 (subsequence), A106853 (Lucas sequence).

nn = 10000; s = LinearRecurrence[{1, -4}, {1, 1}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 06 2013 *)

A340268 Composite numbers k>1 such that (s-1) | (d-1) for each d | k, where s = lpf(k) = A020639(k).

Original entry on oeis.org

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 96

Offset: 1

Maxim Karimov, Jan 02 2021

nonn

Not a duplicate of A340058 because the complements A335902 and A340269 differ. - R. J. Mathar, Feb 16 2021

Cf. A000010, A000961, A020639, A340058, A335902, A340269 (complement).

Contains all composite terms of at least A003586, A003591, A003592, A003593, A003596.

n=300; % gives all terms of the sequence not exceeding n
A=[];
for i=2:n
    lpf=2;
    while mod(i,lpf)~=0
        lpf=lpf+1;
    end
    for d=1:floor(i/2)
        if mod(i,d)==0 && mod(d-1,lpf-1)~=0
            break
        elseif d==floor(i/2)
            A=[A i];
        end
    end
end

with(numtheory):
q:= n-> (f-> andmap(d-> irem(d-1, f)=0, divisors(n)))(min(factorset(n))-1):
select(not isprime and q, [$2..96])[];  # Alois P. Heinz, Feb 12 2021

Select[Range[2, 96], Function[{n, s}, And[! PrimeQ@ n, AllTrue[Divisors[n] - 1, Mod[#, s] == 0 &]]] @@ {#, FactorInteger[#][[1, 1]] - 1} &] (* Michael De Vlieger, Feb 12 2021 *)

isok(c) = if ((c>1) && !isprime(c), my(f=factor(c)[,1]); for (k=1, #f~, if ((f[k]-1) % (f[1]-1), return(0))); return(1)); \\ Michel Marcus, Jan 03 2021

A366786 a(n) = A073481(n)*A005117(n).

Original entry on oeis.org

1, 4, 9, 25, 12, 49, 20, 121, 169, 28, 45, 289, 361, 63, 44, 529, 52, 841, 60, 961, 99, 68, 175, 1369, 76, 117, 1681, 84, 1849, 92, 2209, 153, 2809, 275, 171, 116, 3481, 3721, 124, 325, 132, 4489, 207, 140, 5041, 5329, 148, 539, 156, 6241, 164, 6889, 425, 172

Offset: 1

Michael De Vlieger, Dec 16 2023

Define f(x) to be lpf(k)*k, where lpf(k) = A020639(k). This sequence contains the mappings of f(x) across squarefree numbers A005117.
a(1) = 1 by definition. 1 is the empty product and has no least prime factor.
Define sequence R_k = { m : rad(m) | k }, where rad(n) = A007947(n) and k > 1 is squarefree. Then the sequence k*R_k contains all numbers divisible by squarefree k that are also not divisible by any prime q coprime to k.
Plainly, k is the first term in the sequence k*R_k, because 1 is the first term in R_k. Hence a(n) is the second term in k*R_k for n > 1, since lpf(k) is the second term in R_k.

			Let b(n) = A005117(n).
a(2) = 4 = b(2)*lpf(b(2)) = 2*lpf(2) = 2*2. In {2*A000079}, 4 is the second term.
a(5) = 12 = b(5)*lpf(b(5)) = 6*lpf(6) = 6*2. In {6*A003586}, 12 is the second term..
a(11) = 45 = b(11)*lpf(b(11)) = 15*lpf(15) = 15*3. In {15*A003593}, 45 is the second term, etc.

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

Cf. A000040, A001248, A005117, A020639, A065642, A073481, A120944, A285109, A364996, A366807, A366825.

nn = 120; s = Select[Range[nn], SquareFreeQ];
Array[#*FactorInteger[#][[1, 1]] &[s[[#]]] &, Length[s]]

apply(x->(if (x==1,1, x*vecmin(factor(x)[,1]))), select(issquarefree, [1..150])) \\ Michel Marcus, Dec 17 2023

from math import isqrt
from sympy import mobius, primefactors
def A366786(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    return (m:=bisection(f))*min(primefactors(m),default=1) # Chai Wah Wu, Aug 31 2024

a(n) = A065642(A005117(n)), n > 1.
a(n) = A285109(A005117(n)).
a(n) = A020639(A005117(n))*A005117(n).
For prime p, a(p) = p^2.
For composite squarefree k, a(k) = (p^2 * m) such that (p^2 * m) is in A364996.
Permutation of the union of {1}, A001248, and A366825.

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A095678 Minimal sequence such that all triples of consecutive numbers have no common divisor greater than 1, but all three pairs within the triples are not coprime.

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A231114 Numbers k dividing u(k), where the Lucas sequence is defined u(i) = u(i-1) - 4*u(i-2) with initial conditions u(0)=0, u(1)=1.

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A340268 Composite numbers k>1 such that (s-1) | (d-1) for each d | k, where s = lpf(k) = A020639(k).

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A366786 a(n) = A073481(n)*A005117(n).

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