cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 15 results. Next

A284017 Square root of the smallest square referenced in A038109 (Divisible exactly by the square of a prime).

Original entry on oeis.org

2, 3, 2, 3, 2, 5, 2, 2, 2, 3, 7, 5, 2, 2, 3, 2, 3, 5, 2, 2, 3, 2, 7, 3, 2, 2, 2, 3, 11, 2, 3, 2, 2, 3, 7, 2, 5, 3, 2, 2, 13, 3, 2, 5, 2, 2, 2, 3, 5, 2, 3, 2, 2, 3, 2, 3, 2, 11, 2, 7, 2, 2, 3, 2, 5, 2, 3, 2, 3, 17, 2, 7, 2, 3, 2, 3, 2, 2, 5, 2, 3, 13, 2, 3, 2
Offset: 1

Views

Author

Robert Price, Mar 18 2017

Keywords

Comments

a(n) is the least prime p whose exponent in the prime factorization of A038109(n) is exactly 2. - Robert Israel, Mar 28 2017

Examples

			A038109(3)=12, 12 = 2*2*3, so 12 is divisible by the square of 2.
		

Crossrefs

Programs

  • Maple
    N:= 1000: # to use the members of A038109 <= N
    P:= select(isprime, [$1..floor(sqrt(N))]):
    S:= {}:
    for p in P do
      Ks:= select(t -> t mod p <> 0, {$1..floor(N/p^2)});
      R:= map(`*`,Ks,p^2) minus S;
      for r in R do B[r]:= p od:
      S:= S union R;
    od:
    A038109:= sort(convert(S,list)):
    seq(B[A038109[i]], i=1..nops(A038109)); # Robert Israel, Mar 28 2017
  • Mathematica
    s[n_] := If[(pos = Position[(f = FactorInteger[n])[[;; , 2]], 2]) == {}, 1, f[[pos[[1, 1]], 1]]]; Select[Array[s, 300], # > 1 &] (* Amiram Eldar, Nov 14 2020 *)

Formula

a(n) = sqrt(A284018(n)). - Amiram Eldar, Nov 14 2020

Extensions

Corrected by Robert Israel, Mar 28 2017

A284018 The smallest square referenced in A038109 (Divisible exactly by the square of a prime).

Original entry on oeis.org

4, 9, 4, 9, 4, 25, 4, 4, 4, 9, 49, 25, 4, 4, 9, 4, 9, 25, 4, 4, 9, 4, 49, 9, 4, 4, 4, 9, 121, 4, 9, 4, 4, 9, 49, 4, 25, 9, 4, 4, 169, 9, 4, 25, 4, 4, 4, 9, 25, 4, 9, 4, 4, 9, 4, 9, 4, 121, 4, 49, 4, 4, 9, 4, 25, 4, 9, 4, 9, 289, 4, 49, 4, 9, 4, 9, 4, 4, 25, 4
Offset: 1

Views

Author

Robert Price, Mar 18 2017

Keywords

Comments

a(n) = p^2 where p is the least prime whose exponent in the prime factorization of A038109(n) is exactly 2. - Robert Israel, Mar 28 2017

Examples

			A038109(3)=12, 12 = 2*2*3, so 12 is divisible by the square of 2 which is 4.
		

Crossrefs

Programs

  • Maple
    N:= 1000: # to use the members of A038109 <= N
    P:= select(isprime, [$1..floor(sqrt(N))]):
    S:= {}:
    for p in P do
      Ks:= select(t -> t mod p <> 0, {$1..floor(N/p^2)});
      R:= map(`*`, Ks, p^2) minus S;
      for r in R do B[r]:= p^2 od:
      S:= S union R;
    od:
    A038109:= sort(convert(S, list)): seq(B[A038109[i]], i=1..nops(A038109));# Robert Israel, Mar 28 2017
  • Mathematica
    s[n_] := If[(pos = Position[(f = FactorInteger[n])[[;; , 2]], 2]) == {}, 1, f[[pos[[1, 1]], 1]]]; Select[Array[s, 300], # > 1 &]^2 (* Amiram Eldar, Nov 14 2020 *)

Formula

a(n) = A284017(n)^2. - Amiram Eldar, Nov 14 2020

Extensions

Corrected by Robert Israel, Mar 28 2017

A013929 Numbers that are not squarefree. Numbers that are divisible by a square greater than 1. The complement of A005117.

Original entry on oeis.org

4, 8, 9, 12, 16, 18, 20, 24, 25, 27, 28, 32, 36, 40, 44, 45, 48, 49, 50, 52, 54, 56, 60, 63, 64, 68, 72, 75, 76, 80, 81, 84, 88, 90, 92, 96, 98, 99, 100, 104, 108, 112, 116, 117, 120, 121, 124, 125, 126, 128, 132, 135, 136, 140, 144, 147, 148, 150, 152, 153, 156, 160
Offset: 1

Views

Author

Keywords

Comments

Sometimes misnamed squareful numbers, but officially those are given by A001694.
This is different from the sequence of numbers k such that A007913(k) < phi(k). The two sequences differ at the values: 420, 660, 780, 840, 1320, 1560, 4620, 5460, 7140, ..., which is essentially A070237. - Ant King, Dec 16 2005
Numbers k such that Sum_{d|k} (d/phi(d))*mu(k/d) = 0. - Benoit Cloitre, Apr 28 2002
Also, k with at least one x < k such that A007913(x) = A007913(k). - Benoit Cloitre, Apr 28 2002
Numbers k for which there exists a partition into two parts p and q such that p + q = k and p*q is a multiple of k. - Amarnath Murthy, May 30 2003
Numbers k such that there is a solution 0 < x < k to x^2 == 0 (mod k). - Franz Vrabec, Aug 13 2005
Numbers k such that moebius(k) = 0.
a(n) = k such that phi(k)/k = phi(m)/m for some m < k. - Artur Jasinski, Nov 05 2008
Appears to be numbers such that when a column with index equal to a(n) in A051731 is deleted, there is no impact on the result in the first column of A054525. - Mats Granvik, Feb 06 2009
Numbers k such that the number of prime divisors of (k+1) is less than the number of nonprime divisors of (k+1). - Juri-Stepan Gerasimov, Nov 10 2009
Orders for which at least one non-cyclic finite abelian group exists: A000688(a(n)) > 1. This follows from the fact that not all exponents in the prime factorization of a(n) are 1 (moebius(a(n)) = 0). The number of such groups of order a(n) is A192005(n) = A000688(a(n)) - 1. - Wolfdieter Lang, Jul 29 2011
Subsequence of A193166; A192280(a(n)) = 0. - Reinhard Zumkeller, Aug 26 2011
It appears that terms are the numbers m such that Product_{k=1..m} (prime(k) mod m) <> 0. See Maple code. - Gary Detlefs, Dec 07 2011
A008477(a(n)) > 1. - Reinhard Zumkeller, Feb 17 2012
A057918(a(n)) > 0. - Reinhard Zumkeller, Mar 27 2012
A056170(a(n)) > 0. - Reinhard Zumkeller, Dec 29 2012
Numbers k such that A001221(k) != A001222(k). - Felix Fröhlich, Aug 13 2014
Numbers k such that A001222(k) > A001221(k), since in this case at least one prime factor of k occurs more than once, which implies that k is divisible by at least one perfect square > 1. - Carlos Eduardo Olivieri, Aug 02 2015
Lexicographically least sequence such that each term has a positive even number of proper divisors not occurring in the sequence, cf. the sieve characterization of A005117. - Glen Whitney, Aug 30 2015
There are arbitrarily long runs of consecutive terms. Record runs start at 4, 8, 48, 242, ... (A045882). - Ivan Neretin, Nov 07 2015
A number k is a term if 0 < min(A000010(k) + A023900(k), A000010(k) - A023900(k)). - Torlach Rush, Feb 22 2018
Every squareful number > 1 is nonsquarefree, but the converse is false and the nonsquarefree numbers that are not squareful (see first comment) are in A332785. - Bernard Schott, Apr 11 2021
Integers m where at least one k < m exists such that m divides k^m. - Richard R. Forberg, Jul 31 2021
Consider the Diophantine equation S(x,y) = (x+y) + (x-y) + (x*y) + (x/y) = z, when x and y are both positive integers with y | x. Then, there is a solution (x,y) iff z is a term of this sequence; in this case, if x = K*y, then z = S(K*y,y) = K*(y+1)^2 (see A351381, link and references Perelman); example: S(12,4) = 75 = a(28). The number of solutions for S(x,y) = a(n) is A353282(n). - Bernard Schott, Mar 29 2022
For each positive integer m, the number of unitary divisors of m = the number of squarefree divisors of m (see A034444); but only for the terms of this sequence does the set of unitary divisors differ from the set of squarefree divisors. Example: the set of unitary divisors of 20 is {1, 4, 5, 20}, while the set of squarefree divisors of 20 is {1, 2, 5, 10}. - Bernard Schott, Oct 15 2022

Examples

			For the terms up to 20, we compute the squares of primes up to floor(sqrt(20)) = 4. Those squares are 4 and 9. For every such square s, put the terms s*k^2 for k = 1 to floor(20 / s). This gives after sorting and removing duplicates the list 4, 8, 9, 12, 16, 18, 20. - _David A. Corneth_, Oct 25 2017
		

References

  • I. Perelman, L'Algèbre récréative, Deux nombres et quatre opérations, Editions en langues étrangères, Moscou, 1959, pp. 101-102.
  • Ya. I. Perelman, Algebra can be fun, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131-132.

Crossrefs

Complement of A005117. Subsequences: A130897, A190641, A332785.
Partitions into: A114374, A256012.

Programs

  • Haskell
    a013929 n = a013929_list !! (n-1)
    a013929_list = filter ((== 0) . a008966) [1..]
    -- Reinhard Zumkeller, Apr 22 2012
    
  • Magma
    [ n : n in [1..1000] | not IsSquarefree(n) ];
    
  • Maple
    a := n -> `if`(numtheory[mobius](n)=0,n,NULL); seq(a(i),i=1..160); # Peter Luschny, May 04 2009
    t:= n-> product(ithprime(k),k=1..n): for n from 1 to 160 do (if t(n) mod n <>0) then print(n) fi od; # Gary Detlefs, Dec 07 2011
    with(NumberTheory): isQuadrateful := n -> irem(Radical(n), n) <> 0:
    select(isQuadrateful, [`$`(1..160)]);  # Peter Luschny, Jul 12 2022
  • Mathematica
    Union[ Flatten[ Table[ n i^2, {i, 2, 20}, {n, 1, 400/i^2} ] ] ]
    Select[ Range[2, 160], (Union[Last /@ FactorInteger[ # ]][[ -1]] > 1) == True &] (* Robert G. Wilson v, Oct 11 2005 *)
    Cases[Range[160], n_ /; !SquareFreeQ[n]] (* Jean-François Alcover, Mar 21 2011 *)
    Select[Range@160, ! SquareFreeQ[#] &] (* Robert G. Wilson v, Jul 21 2012 *)
    Select[Range@160, PrimeOmega[#] > PrimeNu[#] &] (* Carlos Eduardo Olivieri, Aug 02 2015 *)
    Select[Range[200], MoebiusMu[#] == 0 &] (* Alonso del Arte, Nov 07 2015 *)
  • PARI
    {a(n)= local(m,c); if(n<=1,4*(n==1), c=1; m=4; while( cMichael Somos, Apr 29 2005 */
    
  • PARI
    for(n=1, 1e3, if(omega(n)!=bigomega(n), print1(n, ", "))) \\ Felix Fröhlich, Aug 13 2014
    
  • PARI
    upto(n)=my(res = List()); forprime(p = 2, sqrtint(n), for(k = 1, n \ p^2, listput(res, k * p^2))); listsort(res, 1); res \\ David A. Corneth, Oct 25 2017
    
  • Python
    from sympy.ntheory.factor_ import core
    def ok(n): return core(n, 2) != n
    print(list(filter(ok, range(1, 161)))) # Michael S. Branicky, Apr 08 2021
    
  • Python
    from math import isqrt
    from sympy import mobius
    def A013929(n):
        def f(x): return n+sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
        m, k = n, f(n)
        while m != k:
            m, k = k, f(k)
        return m # Chai Wah Wu, Jul 20 2024

Formula

A008966(a(n)) = 0. - Reinhard Zumkeller, Apr 22 2012
Sum_{n>=1} 1/a(n)^s = (zeta(s)*(zeta(2*s)-1))/zeta(2*s). - Enrique Pérez Herrero, Jul 07 2012
a(n) ~ n/k, where k = 1 - 1/zeta(2) = 1 - 6/Pi^2 = A229099. - Charles R Greathouse IV, Sep 13 2013
A001222(a(n)) > A001221(a(n)). - Carlos Eduardo Olivieri, Aug 02 2015
phi(a(n)) > A003958(a(n)). - Juri-Stepan Gerasimov, Apr 09 2019

Extensions

More terms from Erich Friedman
More terms from Franz Vrabec, Aug 13 2005

A337050 Numbers without an exponent 2 in their prime factorization.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 24, 26, 27, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 48, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 64, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87
Offset: 1

Views

Author

Amiram Eldar, Aug 12 2020

Keywords

Comments

Numbers k such that the powerful part (A057521) of k is a cubefull number (A036966).
Numbers k such that A003557(k) = k/A007947(k) is a powerful number (A001694).
The asymptotic density of this sequence is Product_{primes p} (1 - 1/p^2 + 1/p^3) = 0.748535... (A330596).
A304364 is apparently a subsequence.
These numbers were named semi-2-free integers by Suryanarayana (1971). - Amiram Eldar, Dec 29 2020

Examples

			6 = 2^1 * 3^1 is a term since none of the exponents in its prime factorization is equal to 2.
9 = 3^2 is not a term since it has an exponent 2 in its prime factorization.
		

Crossrefs

Complement of A038109.
A005117, A036537, A036966, A048109, A175496, A268335 and A336590 are subsequences.
Numbers without an exponent k in their prime factorization: A001694 (k=1), this sequence (k=2), A386799 (k=3), A386803 (k=4), A386807 (k=5).
Numbers that have exactly m exponents in their prime factorization that are equal to 2: this sequence (m=0), A386796 (m=1), A386797 (m=2), A386798 (m=3).

Programs

  • Maple
    q:= n-> andmap(i-> i[2]<>2, ifactors(n)[2]):
    select(q, [$1..100])[];  # Alois P. Heinz, Aug 12 2020
  • Mathematica
    Select[Range[100], !MemberQ[FactorInteger[#][[;;, 2]], 2] &]
  • PARI
    is(n) = {my(f = factor(n)); for(i = 1, #f~, if(f[i, 2] == 2, return(0))); 1; } \\ Amiram Eldar, Oct 21 2023

Formula

Sum_{n>=1} 1/a(n)^s = zeta(s) * Product_{p prime} (1 - 1/p^(2*s) + 1/p^(3*s)), for s > 1. - Amiram Eldar, Oct 21 2023

A059269 Numbers m for which the number of divisors, tau(m), is divisible by 3.

Original entry on oeis.org

4, 9, 12, 18, 20, 25, 28, 32, 36, 44, 45, 49, 50, 52, 60, 63, 68, 72, 75, 76, 84, 90, 92, 96, 98, 99, 100, 108, 116, 117, 121, 124, 126, 132, 140, 144, 147, 148, 150, 153, 156, 160, 164, 169, 171, 172, 175, 180, 188, 196, 198, 200, 204, 207, 212, 220, 224, 225, 228
Offset: 1

Views

Author

Avi Peretz (njk(AT)netvision.net.il), Jan 24 2001

Keywords

Comments

tau(n) is divisible by 3 iff at least one prime in the prime factorization of n has exponent of the form 3*m + 2. This sequence is an extension of the sequence A038109 in which the numbers has at least one prime with exponent 2 (the case of m = 0 here ) in their prime factorization.
The union of A211337 and A211338 is the complementary sequence to this one. - Douglas Latimer, Apr 12 2012
Numbers whose cubefree part (A050985) is not squarefree (A005117). - Amiram Eldar, Mar 09 2021

Examples

			a(7) = 28 is a term because the number of divisors of 28, d(28) = 6, is divisible by 3.
		

Crossrefs

Characteristic function: A353470.

Programs

  • Maple
    with(numtheory): for n from 1 to 1000 do if tau(n) mod 3 = 0 then printf(`%d,`,n) fi: od:
  • Mathematica
    Select[Range[230], Divisible[DivisorSigma[0, #], 3] &] (* Amiram Eldar, Jul 26 2020 *)
  • PARI
    is(n)=vecmax(factor(n)[,2]%3)==2 \\ Charles R Greathouse IV, Apr 10 2012
    
  • PARI
    is(n)=numdiv(n)%3==0 \\ Charles R Greathouse IV, Sep 18 2015

Formula

Conjecture: a(n) ~ k*n where k = 1/(1 - Product(1 - (p-1)/(p^(3*i)))) = 3.743455... where p ranges over the primes and i ranges over the positive integers. - Charles R Greathouse IV, Apr 13 2012
The asymptotic density of this sequence is 1 - zeta(3)/zeta(2) = 1 - 6*zeta(3)/Pi^2 = 0.2692370305... (Sathe, 1945). Therefore, the above conjecture, a(n) ~ k*n, is true, but k = 1/(1-6*zeta(3)/Pi^2) = 3.7141993349... - Amiram Eldar, Jul 26 2020
A001248 UNION A030515 UNION A030627 UNION A030630 UNION A030633 UNION A030636 UNION ... - R. J. Mathar, May 05 2023

Extensions

More terms from James Sellers, Jan 24 2001

A369427 The number of unitary divisors of n that are squares of primes.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0
Offset: 1

Views

Author

Amiram Eldar, Jan 23 2024

Keywords

Comments

The number of exponents in the prime factorization of n that are equal to 2.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e == 2, 1, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecsum(apply(x -> if(x == 2, 1, 0), factor(n)[, 2]));

Formula

Additive with a(p^e) = 1 if e = 2, and 0 otherwise.
a(n) > 0 if and only if n is in A038109.
a(A061742(n)) = n, and a(k) < n for all k < A061742(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} (1/p^2 - 1/p^3) = A085548 - A085541 = 0.27748478074162196208... .

A252849 Numbers with an even number of square divisors.

Original entry on oeis.org

4, 8, 9, 12, 18, 20, 24, 25, 27, 28, 36, 40, 44, 45, 49, 50, 52, 54, 56, 60, 63, 64, 68, 72, 75, 76, 84, 88, 90, 92, 98, 99, 100, 104, 108, 116, 117, 120, 121, 124, 125, 126, 128, 132, 135, 136, 140, 144, 147, 148
Offset: 1

Views

Author

Walker Dewey Anderson, Mar 22 2015

Keywords

Comments

Closed lockers in the locker problem where the student numbers are the set of perfect squares.
The locker problem is a classic mathematical problem. Imagine a row containing an infinite number of lockers numbered from one to infinity. Also imagine an infinite number of students numbered from one to infinity. All of the lockers begin closed. The first student opens every locker that is a multiple of one, which is every locker. The second student closes every locker that is a multiple of two, so all of the even-numbered lockers are closed. The third student opens or closes every locker that is a multiple of three. This process continues for all of the students.
A variant on the locker problem is when not all student numbers are considered; in the case of this sequence, only the square-numbered students open and close lockers. The sequence here is a list of the closed lockers after all of the students have gone.
From Amiram Eldar, Jul 07 2020: (Start)
Numbers k such that the largest square dividing k (A008833) is not a fourth power.
The asymptotic density of this sequence is 1 - Pi^2/15 = 1 - A182448 = 0.342026... (Cesàro, 1885). (End)
Closed under application of A331590: for n, k >= 1, A331590(a(n), k) is in the sequence. - Peter Munn, Sep 18 2020

Crossrefs

Complement of A252895.
A046951, A335324 are used in a formula defining this sequence.
Disjoint union of A336593 and A336594.
A030140, A038109, A082293, A217319 are subsequences.
Ordered 3rd trisection of A225546.

Programs

Formula

From Peter Munn, Sep 18 2020: (Start)
Numbers k such that A046951(k) mod 2 = 0.
Numbers k such that A335324(k) > 1.
(End)

A162872 Primes p such that p-1 and p+1 each contain at least one squared prime in their prime factorization.

Original entry on oeis.org

19, 149, 197, 199, 293, 307, 349, 491, 523, 557, 577, 739, 773, 883, 1013, 1051, 1061, 1151, 1171, 1277, 1451, 1493, 1531, 1549, 1601, 1637, 1667, 1693, 1709, 1733, 1747, 1861, 1949, 2069, 2141, 2179, 2251, 2351, 2357, 2467, 2549, 2683, 2789, 2843, 2851
Offset: 1

Views

Author

Keywords

Comments

The selection criterion is that p-1 and p+1 are in the subsequence 4=2^2, 9=3^2, 12=2^2*3, 18=2*3^2, ... of nonsquarefree numbers (A013929) that actually display at least one square in their standard prime factorization.
So at least one of the e_i in p-1=product p_i^e_i, and at least one of the e_j in p+1=product p_j^e_j must equal 2. This is more stringent than being nonsquarefree, and the sequence becomes a subsequence of A075432.

Examples

			19 is in the sequence because 19 - 1 = 2*3^2 contains 3^2 and because 19 + 1 = 2^2*5 contains 2^2 in the factorization.
		

Crossrefs

Programs

  • Maple
    isA162872 := proc(n)
        if isprime(n) then
            isA038109(n-1) and isA038109(n+1) ;
        else
            false;
        end if;
    end proc:
    n := 1:
    for c from 1 to 50000 do
        if isA162872(c) then
            printf("%d %d\n",n,c) ;
            n := n+1 ;
    end if; # R. J. Mathar, Dec 08 2015
    N:= 10^5: # to get all terms < N, where N is even
    V:= Vector(N/2):
    for i from 1 do
      p:= ithprime(i);
      if p^2 > N+1 then break fi;
      if p = 2 then inds:= 2*[seq(i, i=1..floor(N/4), 2)]
      else inds:= p^2*select(t -> t mod p <> 0, [$1..floor(N/2/p^2)])
      fi;
      V[inds]:= 1;
    od:
    select(t -> V[(t-1)/2] = 1 and V[(t+1)/2] = 1 and isprime(t), [seq(t, t=3..N, 2)]); # Robert Israel, Dec 08 2015
  • Mathematica
    f[n_]:=Module[{a=m=0},Do[If[FactorInteger[n][[m,2]]==2,a=1],{m,Length[FactorInteger[n]]}]; a]; lst={};Do[p=Prime[n];If[f[p-1]==1&&f[p+1]==1,AppendTo[lst,p]], {n,7!}];lst
    ospQ[n_]:=AnyTrue[FactorInteger[n+1][[;;,2]],#==2&]&&AnyTrue[FactorInteger[n-1][[;;,2]],#==2&]; Select[Prime[Range[500]],ospQ] (* Harvey P. Dale, May 11 2025 *)

Formula

{p in A000040: p+1 in A038109 and p-1 in A038109}. - R. J. Mathar, Dec 08 2015

Extensions

Role of squarefree numbers clarified by R. J. Mathar, Jul 31 2007

A304365 Numbers k such that Sum_{d|k, d = 1 or not a perfect power} mu(k/d) is nonzero.

Original entry on oeis.org

1, 4, 9, 12, 18, 20, 25, 28, 36, 44, 45, 49, 50, 52, 60, 63, 68, 72, 75, 76, 84, 90, 92, 98, 99, 100, 108, 116, 117, 121, 124, 126, 132, 140, 144, 147, 148, 150, 153, 156, 164, 169, 171, 172, 175, 180, 188, 196, 198, 200, 204, 207, 212, 216, 220, 225, 228, 234
Offset: 1

Views

Author

Gus Wiseman, May 11 2018

Keywords

Comments

First differs from A038109 at a(53) = 216, A038109(53) = 220.
Contains all prime powers (A000961).
The numbers of terms not exceeding 10^k, for k = 1, 2, ..., are 3, 26, 254, 2557, 25663, 256765, 2567839, 25679023, 256791104, 2567912451, ... . Apparently, the asymptotic density of this sequence exists and equals 0.256791... . - Amiram Eldar, May 20 2023

Crossrefs

Programs

  • Mathematica
    Select[Range[100],Sum[If[GCD@@FactorInteger[d][[All,2]]===1,MoebiusMu[#/d],0],{d,Divisors[#]}]=!=0&]
  • PARI
    ok(n)={sumdiv(n, d, if(ispower(d), 0, moebius(n/d))) <> 0} \\ Andrew Howroyd, Aug 26 2018

A297404 A binary representation of the positive exponents that appear in the prime factorization of a number, shown in decimal.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 4, 2, 1, 1, 3, 1, 1, 1, 8, 1, 3, 1, 3, 1, 1, 1, 5, 2, 1, 4, 3, 1, 1, 1, 16, 1, 1, 1, 2, 1, 1, 1, 5, 1, 1, 1, 3, 3, 1, 1, 9, 2, 3, 1, 3, 1, 5, 1, 5, 1, 1, 1, 3, 1, 1, 3, 32, 1, 1, 1, 3, 1, 1, 1, 6, 1, 1, 3, 3, 1, 1, 1, 9, 8, 1, 1, 3, 1, 1
Offset: 1

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Author

Rémy Sigrist, Dec 29 2017

Keywords

Comments

This sequence is similar to A087207; here we encode the exponents, there the prime numbers appearing in the prime factorization of a number.
The binary representation of a(n) shows which exponents appear in the prime factorization of n, but without multiplicities:
- for any prime number p and k > 0, if p^k divides n but p^(k+1) does not divide n, then a(n) AND 2^(k-1) = 2^(k-1) (where AND denotes the bitwise AND operator),
- conversely, if a(n) AND 2^(k-1) = 2^(k-1) for some k > 0, then there is prime number p such that p^k divides n but p^(k+1) does not divide n.

Examples

			For n = 90:
- 90 = 5^1 * 3^2 * 2^1,
- the exponents appearing in the prime factorization of 90 are 1 and 2,
- hence a(90) = 2^(1-1) + 2^(2-1) = 3.
		

Crossrefs

Programs

  • Mathematica
    Array[Total@ Map[2^(# - 1) &, Union[FactorInteger[#][[All, -1]] ]] - Boole[# == 1] &, 86] (* Michael De Vlieger, Dec 29 2017 *)
  • PARI
    a(n) = my (x=Set(factor(n)[,2]~)); sum(i=1, #x, 2^(x[i]))/2

Formula

a(p^k) = 2^(k-1) for any prime number p and k > 0.
a(n^2) = A000695(2 * a(n)) / 2 for any n > 0.
a(n) <= 1 iff n is squarefree (A005117).
a(n) <= 3 iff n is cubefree (A004709).
a(n) is odd iff n belongs to A052485 (weak numbers).
a(n) is even iff n belongs to A001694 (powerful numbers).
a(n) AND 2 = 2 iff n belongs to A038109 (where AND denotes the bitwise AND operator).
A000120(a(n)) <= 1 iff n belongs to A072774 (powers of squarefree numbers).
A000120(a(n)) > 1 iff n belongs to A059404.
If gcd(m, n) = 1, then a(m * n) = a(m) OR a(n) (where OR denotes the bitwise OR operator).
a(n) = a(A328400(n)). - Peter Munn, Oct 02 2023
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