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.

Previous Showing 51-60 of 104 results. Next

A378087 First-differences of A067535 (least positive integer >= n that is squarefree).

Original entry on oeis.org

1, 1, 2, 0, 1, 1, 3, 0, 0, 1, 2, 0, 1, 1, 2, 0, 2, 0, 2, 0, 1, 1, 3, 0, 0, 3, 0, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 3, 0, 0, 1, 4, 0, 0, 0, 2, 0, 2, 0, 2, 0, 1, 1, 2, 0, 1, 3, 0, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 3, 0, 0, 1, 1, 3, 0, 0, 1, 2, 0, 1, 1, 2
Offset: 1

Views

Author

Gus Wiseman, Nov 20 2024

Keywords

Comments

Does this contain all nonnegative integers? The positions of first appearances begin: 4, 1, 3, 7, 47, 241, 843, 22019, 217069, ...

Crossrefs

Ones are A007674.
Zeros are A013929, complement A005117.
Positions of first appearances are A020754 (except first term) = A045882 - 1.
First-differences of A067535.
Twos are A280892.
For prime-powers we have A377780, differences of A000015.
The nonsquarefree opposite is A378036, differences of A378033.
The restriction to primes + 1 is A378037 (opposite A378038), differences of A112926.
For nonsquarefree numbers we have A378039, see A377783, A377784, A378040.
The opposite is A378085, differences of A070321.
A000040 lists the primes, differences A001223, seconds A036263.
A005117 lists the squarefree numbers.
A013929 lists the nonsquarefree numbers, differences A078147, seconds A376593.
A061398 counts squarefree numbers between primes, zeros A068360.
A061399 counts nonsquarefree numbers between primes, zeros A068361.

Programs

  • Mathematica
    Differences[Table[NestWhile[#+1&,n,#>1&&!SquareFreeQ[#]&],{n,100}]]

A379306 Number of squarefree prime indices of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 0, 3, 2, 2, 1, 3, 1, 1, 2, 4, 1, 3, 0, 3, 1, 2, 0, 4, 2, 2, 3, 2, 1, 3, 1, 5, 2, 2, 1, 4, 0, 1, 2, 4, 1, 2, 1, 3, 3, 1, 1, 5, 0, 3, 2, 3, 0, 4, 2, 3, 1, 2, 1, 4, 0, 2, 2, 6, 2, 3, 1, 3, 1, 2, 0, 5, 1, 1, 3, 2, 1, 3, 1, 5, 4, 2, 1, 3, 2, 2, 2
Offset: 1

Views

Author

Gus Wiseman, Dec 25 2024

Keywords

Comments

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

Examples

			The prime indices of 39 are {2,6}, so a(39) = 2.
The prime indices of 70 are {1,3,4}, so a(70) = 2.
The prime indices of 98 are {1,4,4}, so a(98) = 1.
The prime indices of 294 are {1,2,4,4}, a(294) = 2.
The prime indices of 1911 are {2,4,4,6}, so a(1911) = 2.
The prime indices of 2548 are {1,1,4,4,6}, so a(2548) = 3.
		

Crossrefs

Positions of first appearances are A000079.
Positions of zero are A379307, counted by A114374 (strict A256012).
Positions of one are A379316, counted by A379308 (strict A379309).
A000040 lists the primes, differences A001223.
A005117 lists the squarefree numbers, differences A076259.
A008966 is the characteristic function for the squarefree numbers.
A013929 lists the nonsquarefree numbers, differences A078147.
A055396 gives least prime index, greatest A061395.
A056239 adds up prime indices, row sums of A112798, counted by A001222.
A061398 counts squarefree numbers between primes, zeros A068360.
A377038 gives k-th differences of squarefree numbers.
Other counts of prime indices:
- A087436 postpositive, see A038550.
- A330944 nonprime, see A000586, A000607, A076610, A330945.
- A379310 nonsquarefree, see A302478.
- A379311 old prime, see A204389, A320629, A379312-A379315.

Programs

  • Mathematica
    prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Table[Length[Select[prix[n],SquareFreeQ]],{n,100}]

Formula

Totally additive with a(prime(k)) = A008966(k).

A243285 Number of integers 1 <= k <= n which are not divisible by the square of their largest noncomposite divisor.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 5, 5, 5, 6, 7, 8, 9, 10, 11, 11, 12, 12, 13, 14, 15, 16, 17, 18, 18, 19, 19, 20, 21, 22, 23, 23, 24, 25, 26, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 38, 38, 39, 40, 41, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 50, 51, 52, 53, 54, 55, 56, 57
Offset: 1

Views

Author

Antti Karttunen, Jun 02 2014

Keywords

Comments

a(n) tells how many natural numbers <= n there are which are not divisible by the square of their largest noncomposite divisor.
The largest noncomposite divisor of 1 is 1 itself, and 1 is divisible by 1^2, thus 1 is not included in the count, and a(1)=0.
The "largest noncomposite divisor" for any integer > 1 means the same thing as the largest prime divisor, and thus we are counting the terms of A102750 (Numbers n such that square of largest prime dividing n does not divide n).
Thus this is the partial sums of the characteric function for A102750.

Examples

			For n = 9, there are numbers 2, 3, 5, 6 and 7 which are not divisible by the square of their largest prime factor, while 1 is excluded (no prime factors) and 4 and 8 are divisible both by 2^2 and 9 is divisible by 3^2. Thus a(9) = 5.
		

Crossrefs

Programs

  • Mathematica
    ndsQ[n_]:=Mod[n,Max[Select[Divisors[n],!CompositeQ[#]&]]^2]!=0; Accumulate[Table[If[ ndsQ[n],1,0],{n,80}]] (* Harvey P. Dale, Oct 14 2023 *)
  • Python
    from sympy import primefactors
    def a243285(n): return 0 if n==1 else sum([1 for k in range(2, n + 1) if k%(primefactors(k)[-1]**2)!=0]) # Indranil Ghosh, Jun 15 2017
  • Scheme
    (define (A243285 n) (- n (A243283 n)))
    

Formula

a(n) = n - A243283(n).
For all n, a(A102750(n)) = n, thus this sequence works also as an inverse function for the injection A102750.

A365680 The number of exponentially squarefree divisors of n.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 4, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 5, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 2, 8, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 6, 4, 8, 2, 6, 4, 8, 2, 12, 2, 4, 6, 6, 4, 8, 2, 8, 4, 4, 2, 12, 4, 4
Offset: 1

Views

Author

Amiram Eldar, Sep 15 2023

Keywords

Comments

First differs from A252505 at n = 32.
The number of divisors of n that are exponentially squarefree numbers (A209061), i.e., numbers having only squarefree exponents in their canonical prime factorization.
The sum of these divisors is A365682(n) and the largest of them is A365683(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := Count[Range[e], ?SquareFreeQ] + 1; a[1] = 1; a[n] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    s(n) = sum(k=1, n, issquarefree(k)) + 1;
    a(n) = vecprod(apply(x -> s(x), factor(n)[, 2]));

Formula

Multiplicative with a(p^e) = A013928(e+1) + 1.
a(n) <= A000005(n), with equality if and only if n is a biquadratefree number (A046100).

A262868 Number of squarefree numbers appearing among the larger parts of the partitions of n into two parts.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 2, 3, 3, 3, 3, 4, 3, 4, 4, 5, 5, 6, 6, 7, 6, 7, 7, 8, 8, 8, 8, 8, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15, 14, 14, 14, 15, 15, 15, 15, 16, 15, 16, 16, 17, 17, 18, 18, 19, 18, 19, 19, 19, 19, 20, 20, 21, 20, 21, 21, 22, 22
Offset: 1

Views

Author

Wesley Ivan Hurt, Oct 03 2015

Keywords

Comments

Number of distinct rectangles with squarefree length and integer width such that L + W = n, W <= L. For example, a(14) = 4; the rectangles are 1 X 13, 3 X 11, 4 X 10 and 7 X 7. - Wesley Ivan Hurt, Nov 02 2017
a(10) = 3, a(100) = 30, a(10^3) = 302, a(10^4) = 3041, a(10^5) = 30393, a(10^6) = 303968, a(10^7) = 3039658, a(10^8) = 30396350, a(10^9) = 303963598, a(10^10) = 3039635373, a(10^11) = 30396355273, a(10^12) = 303963551068, a(10^13) = 3039635509338, a(10^14) = 30396355094469, a(10^15) = 303963550926043, a(10^16) = 3039635509271763, a(10^17) = 30396355092700721, and a(10^18) = 303963550927014110. The limit of a(n)/n is 3/Pi^2. - Charles R Greathouse IV, Nov 04 2017

Examples

			a(4)=2; there are two partitions of 4 into two parts: (3,1) and (2,2). Both of the larger parts are squarefree, thus a(4)=2.
a(5)=1; there are two partitions of 5 into two parts: (4,1) and (3,2). Among the larger parts, only 3 is squarefree, thus a(5)=1.
		

Crossrefs

Programs

  • Maple
    with(numtheory): A262868:=n->add(mobius(n-i)^2, i=1..floor(n/2)): seq(A262868(n), n=1..100);
  • Mathematica
    Table[Sum[MoebiusMu[n - i]^2, {i, Floor[n/2]}], {n, 100}]
    Table[Count[IntegerPartitions[n,{2}][[All,1]],?SquareFreeQ],{n,80}] (* _Harvey P. Dale, Jan 03 2022 *)
  • PARI
    a(n) = sum(i=1, n\2, moebius(n-i)^2); \\ Michel Marcus, Oct 04 2015
    
  • PARI
    f(n)=my(s); forfactored(k=1,sqrtint(n),s+=n\k[1]^2*moebius(k)); s
    a(n)=n--; f(n) - f(n\2) \\ Charles R Greathouse IV, Nov 04 2017

Formula

a(n) = Sum_{i=1..floor(n/2)} mu(n-i)^2, where mu is the Möbius function A008683.
a(n) = A262991(n) - A262869(n).
a(n) ~ 3*n/Pi^2. - Charles R Greathouse IV, Nov 04 2017

A262869 Number of squarefree numbers appearing among the smaller parts of the partitions of n into two parts.

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 15, 15, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 18, 18, 19, 19, 20, 20, 20, 20, 21, 21, 22, 22, 23, 23
Offset: 1

Views

Author

Wesley Ivan Hurt, Oct 03 2015

Keywords

Comments

Number of distinct rectangles with integer length and squarefree width such that L + W = n, W <= L. For example, a(14) = 6; the rectangles are 13 X 1, 12 X 2, 11 X 3, 9 X 5, 8 X 6, 7 X 7. - Wesley Ivan Hurt, Nov 04 2017

Examples

			a(5)=2; there are two partitions of 5 into two parts: (4,1) and (3,2). Both of the smaller parts are squarefree, thus a(5)=2.
a(6)=3; there are three partitions of 6 into two parts: (5,1), (4,2) and (3,3). Among the three smaller parts, all are squarefree, thus a(6)=3.
		

Crossrefs

Programs

  • Maple
    with(numtheory): A262869:=n->add(mobius(i)^2, i=1..floor(n/2)): seq(A262869(n), n=1..100);
  • Mathematica
    Table[Sum[MoebiusMu[i]^2, {i, Floor[n/2]}], {n, 100}]
    Table[Count[IntegerPartitions[n,{2}][[All,2]],?SquareFreeQ],{n,80}] (* _Harvey P. Dale, Oct 17 2021 *)
  • PARI
    a(n) = sum(i=1, n\2, moebius(i)^2); \\ Michel Marcus, Oct 04 2015
    
  • PARI
    a(n)=my(s); n\=2; forsquarefree(k=1, sqrtint(n), s += n\k[1]^2*moebius(k)); s \\ Charles R Greathouse IV, Jan 08 2018

Formula

a(n) = Sum_{i=1..floor(n/2)} mu(i)^2, where mu is the Möebius function (A008683).
a(n) = A262991(n) - A262868(n).
a(n) = A013928(floor(n/2)+1). - Georg Fischer, Nov 29 2022

A372306 Cardinality of the largest subset of {1,...,n} such that no three distinct elements of this subset multiply to a square.

Original entry on oeis.org

1, 2, 3, 4, 5, 5, 6, 6, 6, 7, 8, 8, 9, 10, 10, 10, 11, 11, 12, 12, 13, 13, 14, 15, 15, 16, 17, 18, 19, 19, 20, 20, 20, 21, 21, 21, 22, 23, 23, 24, 25, 26, 27, 28, 29, 30, 31, 31, 31, 31, 32, 33, 34, 34, 34, 35, 36, 37, 38, 39, 40, 41, 42, 42, 42, 43, 44, 45, 46
Offset: 1

Views

Author

Terence Tao, May 25 2024

Keywords

Comments

a(n) >= A373114(n).
a(n) ~ n (Erdős-Sárközy-Sós).
a(n+1)-a(n) is either 0 or 1 for any n.
If "three" is replaced by "two" one obtains A013928. If "three" is replaced by "one", one obtains A028391. If "three" is replaced by "any odd", one obtains A373114.

Examples

			a(7)=6, because the set {1,2,3,4,5,7} has no three distinct elements multiplying to a square, but {1,2,3,4,5,6,7} has 2*3*6 = 6^2.
		

Crossrefs

Programs

  • PARI
    \\ See PARI link
  • Python
    from math import isqrt
    def is_square(n):
        return isqrt(n) ** 2 == n
    def valid_subset(A):
        length = len(A)
        for i in range(length):
            for j in range(i + 1, length):
                for k in range(j + 1, length):
                    if is_square(A[i] * A[j] * A[k]):
                        return False
        return True
    def largest_subset_size(N):
        from itertools import combinations
        max_size = 0
        for size in range(1, N + 1):
            for subset in combinations(range(1, N + 1), size):
                if valid_subset(subset):
                    max_size = max(max_size, size)
        return max_size
    for N in range(1, 11):
        print(largest_subset_size(N))
    
  • Python
    from math import prod
    from functools import lru_cache
    from itertools import combinations
    from sympy.ntheory.primetest import is_square
    @lru_cache(maxsize=None)
    def A372306(n):
        if n==1: return 1
        i = A372306(n-1)+1
        if sum(1 for p in combinations(range(1,n),2) if is_square(n*prod(p))) > 0:
            a = [set(p) for p in combinations(range(1,n+1),3) if is_square(prod(p))]
            for q in combinations(range(1,n),i-1):
                t = set(q)|{n}
                if not any(s<=t for s in a):
                    return i
            else:
                return i-1
        else:
            return i # Chai Wah Wu, May 30 2024
    

Formula

From David A. Corneth, May 29 2024: (Start)
a(k^2) = a(k^2 - 1) for k >= 3.
a(p) = a(p - 1) + 1 for prime p. (End)

Extensions

a(18)-a(36) from Michael S. Branicky, May 25 2024
a(37)-a(38) from Michael S. Branicky, May 26 2024
a(39)-a(63) from Martin Ehrenstein, May 26 2024
a(64)-a(76) from David A. Corneth, May 29 2024, May 30 2024

A347149 Dirichlet g.f.: Product_{primes p} (1 + 3/p^s).

Original entry on oeis.org

1, 3, 3, 0, 3, 9, 3, 0, 0, 9, 3, 0, 3, 9, 9, 0, 3, 0, 3, 0, 9, 9, 3, 0, 0, 9, 0, 0, 3, 27, 3, 0, 9, 9, 9, 0, 3, 9, 9, 0, 3, 27, 3, 0, 0, 9, 3, 0, 0, 0, 9, 0, 3, 0, 9, 0, 9, 9, 3, 0, 3, 9, 0, 0, 9, 27, 3, 0, 9, 27, 3, 0, 3, 9, 0, 0, 9, 27, 3, 0, 0, 9, 3, 0, 9, 9, 9, 0, 3, 0, 9, 0, 9, 9, 9, 0, 3, 0, 0, 0
Offset: 1

Views

Author

Vaclav Kotesovec, Aug 20 2021

Keywords

Crossrefs

Programs

  • Mathematica
    Table[MoebiusMu[n]^2 * 3^PrimeNu[n], {n, 1, 100}]
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, (1 + 3*X))[n], ", "))
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, (1 - 6*X^2 + 8*X^3 - 3*X^4)/(1 - X)^3)[n], ", "))

Formula

Dirichlet g.f.: zeta(s)^3 * Product_{primes p} (1 - 6/p^(2*s) + 8/p^(3*s) - 3/p^(4*s)).
Let f(s) = Product_{primes p} (1 - 6/p^(2*s) + 8/p^(3*s) - 3/p^(4*s)), then Sum_{k=1..n} a(k) ~ n * (f(1)*log(n)^2/2 + log(n)*((3*gamma - 1)*f(1) + f'(1)) + f(1)*(1 - 3*gamma + 3*gamma^2 - 3*sg1) + (3*gamma - 1)*f'(1) + f''(1)/2), where f(1) = Product_{primes p} (1 - 6/p^2 + 8/p^3 - 3/p^4) = 0.1148840440802287887292512767015990978487135526872830176248484270625666728..., f'(1) = f(1) * Sum_{primes p} 12*log(p) / ((p-1)*(p+3)) = 0.5497153490016133577871571904347511299324572220423331992393596243955677299..., f''(1) = f'(1)^2/f(1) + f(1) * Sum_{primes p} (-24*p*(p-1) * log(p)^2 / ((p-1)^2 * (p+3)^2)) = 0.9028322988288094236586622799305270026576436536391185119652318723470259904... and gamma is the Euler-Mascheroni constant A001620 and sg1 is the first Stieltjes constant (see A082633).
a(n) = A008966(n) * A048691(n). - Enrique Pérez Herrero, Oct 27 2022
Multiplicative with a(p) = 3, and a(p^e) = 0 for e >= 2. - Amiram Eldar, Dec 25 2022

A070548 a(n) = Cardinality{ k in range 1 <= k <= n such that Moebius(k) = 1 }.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5, 5, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 10, 11, 11, 11, 12, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 17, 18, 18, 18, 18, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 23, 23, 23, 23
Offset: 1

Views

Author

Benoit Cloitre, May 02 2002

Keywords

Comments

Moebius(k)=1 iff k is the product of an even number of distinct primes (cf. A008683). See A057627 for Moebius(k)=0.
There was an old comment here that said a(n) was equal to A072613(n) + 1, but this is false (e.g., at n=210). - N. J. A. Sloane, Sep 10 2008

Crossrefs

Programs

  • Maple
    with(numtheory); M:=10000; c:=0; for n from 1 to M do if mobius(n) = 1 then c:=c+1; fi; lprint(n,c); od; # N. J. A. Sloane, Sep 14 2008
  • Mathematica
    a[n_] := If[MoebiusMu[n] == 1, 1, 0]; Accumulate@ Array[a, 100] (* Amiram Eldar, Oct 01 2023 *)
  • PARI
    for(n=1,150,print1(sum(i=1,n,if(moebius(i)-1,0,1)),","))

Formula

Asymptotics: Let N(i) = number of k in the range [1,n] with mu(k) = i, for i = 0, 1, -1. Then we know N(1) + N(-1) ~ 6n/Pi^2 (see A059956). Also, assuming the Riemann hypothesis, | N(1) - N(-1) | < n^(1/2 + epsilon) (see the Mathworld Mertens Conjecture link). Hence a(n) = N(1) ~ 3n/Pi^2 + smaller order terms. - Stefan Steinerberger, Sep 10 2008
a(n) = (1/2)*Sum_{i=1..n} (mu(i)^2 + mu(i)) = (1/2)*(A013928(n+1) + A002321(n)). - Ridouane Oudra, Oct 19 2019
From Amiram Eldar, Oct 01 2023: (Start)
a(n) = A013928(n+1) - A070549(n).
a(n) = A070549(n) + A002321(n). (End)

A070549 a(n) = Cardinality{ k in range 1 <= k <= n such that Moebius(k) = -1 }.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 14, 15, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 19, 19, 20, 20, 20, 20, 20, 21, 22, 22, 22, 23, 24, 24, 25, 25, 25, 25, 25, 26
Offset: 1

Views

Author

Benoit Cloitre, May 02 2002

Keywords

Comments

mu(k)=-1 if k is the product of an odd number of distinct primes. See A057627 for mu(k)=0.

Crossrefs

Partial sums of A252233.

Programs

  • Maple
    ListTools:-PartialSums([seq(-min(numtheory:-mobius(n),0),n=1..100)]); # Robert Israel, Jan 08 2018
  • Mathematica
    a[n_]:=Sum[Boole[MoebiusMu[k]==-1],{k,n}]; Array[a,78] (* Stefano Spezia, Jan 30 2023 *)
  • PARI
    for(n=1,150,print1(sum(i=1,n,if(moebius(i)+1,0,1)),","))

Formula

From Amiram Eldar, Oct 01 2023: (Start)
a(n) = (A013928(n+1) - A002321(n))/2.
a(n) = A013928(n+1) - A070548(n).
a(n) = A070548(n) - A002321(n).
a(n) ~ (3/Pi^2) * n. (End)
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