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 34 results. Next

A056169 Number of unitary prime divisors of n.

Original entry on oeis.org

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

Views

Author

Labos Elemer, Jul 27 2000

Keywords

Comments

The zeros of this sequences are the powerful numbers (A001694). There are no arbitrarily long subsequences with a given upper bound; for example, every sequence of 4 values includes one divisible by 2 but not 4, so there are no more than 3 consecutive zeros. Similarly, there can be no more than 23 consecutive values with none divisible by both 2 and 3 but neither 4 nor 9 (so a(n) >= 2), etc. In general, this gives an upper bound that is a (relatively) small multiple of the k-th primorial number (prime(k)#). One suspects that the actual upper bounds for such subsequences are quite a bit lower; e.g., Erdős conjectured that there are no three consecutive powerful numbers. - Franklin T. Adams-Watters, Aug 08 2006
In particular, for every A048670(k)*A002110(k) consecutive terms, at least one is greater than or equal to k. - Charlie Neder, Jan 03 2019
Following Catalan's conjecture (which became Mihăilescu's theorem in 2002), the first case of two consecutive zeros in this sequence is for a(8) and a(9), because 8 = 2^3 and 9 = 3^2, and there are no other consecutive zeros for consecutive powers. However, there are other pairs of consecutive zeros at powerful numbers (A001694, A060355). The next example is a(288) = a(289) = 0, because 288 = 2^5 * 3^2 and 289 = 17^2, then also a(675) and a(676). - Bernard Schott, Jan 06 2019
a(2k-1) is the number of primes p such that p || x + y and p^2 || x^(2k-1) + y^(2k-1) for some positive integers x and y. For any positive integers x, y and k > 1, there is no prime p such that p || x + y and p^2 || x^(2k) + y^(2k). - Jinyuan Wang, Apr 08 2020

Examples

			9 = 3^2 so a(9) = 0; 10 = 2 * 5 so a(10) = 2; 11 = 11^1 so a(11) = 1.
		

Crossrefs

Programs

  • Haskell
    a056169 = length . filter (== 1) . a124010_row
    -- Reinhard Zumkeller, Sep 10 2013
    
  • Maple
    a:= n-> nops(select(i-> i[2]=1, ifactors(n)[2])):
    seq(a(n), n=1..120);  # Alois P. Heinz, Mar 27 2017
  • Mathematica
    Join[{0},Table[Count[Transpose[FactorInteger[n]][[2]],1],{n,2,110}]] (* Harvey P. Dale, Mar 15 2012 *)
    Table[DivisorSum[n, 1 &, And[PrimeQ@ #, CoprimeQ[#, n/#]] &], {n, 105}] (* Michael De Vlieger, Nov 28 2017 *)
  • PARI
    a(n)=my(f=factor(n)[,2]); sum(i=1,#f,f[i]==1) \\ Charles R Greathouse IV, Apr 29 2015
    
  • Python
    from sympy import factorint
    def a(n):
        f=factorint(n)
        return 0 if n==1 else sum(1 for i in f if f[i]==1)
    print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 19 2017
    
  • Scheme
    ;; With memoization-macro definec.
    (definec (A056169 n) (if (= 1 n) 0 (+ (if (= 1 (A067029 n)) 1 0) (A056169 (A028234 n))))) ;; Antti Karttunen, Nov 28 2017

Formula

A prime factor of n is unitary iff its exponent is 1 in prime factorization of n. In general, gcd(p, n/p) = 1 or = p.
Additive with a(p^e) = 1 if e = 1, 0 otherwise.
a(n) = #{k: A124010(n,k) = 1, k = 1..A001221}. - Reinhard Zumkeller, Sep 10 2013
From Antti Karttunen, Nov 28 2017: (Start)
a(1) = 0; for n > 1, a(n) = A063524(A067029(n)) + a(A028234(n)).
a(n) = A001221(A055231(n)) = A001222(A055231(n)).
a(n) = A001221(n) - A056170(n) = A001221(n) - A001221(A000188(n)).
a(n) = A001222(n) - A275812(n).
a(n) = A162642(n) - A295662(n).
a(n) <= A162642(n) <= a(n) + A295659(n).
a(n) <= A295664(n).
(End)
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B - C), where B is Mertens's constant (A077761) and C = Sum_{p prime} (1/p^2) = 0.452247... (A085548). - Amiram Eldar, Sep 28 2023

A316523 Number of odd multiplicities minus number of even multiplicities in the canonical prime factorization of n.

Original entry on oeis.org

0, 1, 1, -1, 1, 2, 1, 1, -1, 2, 1, 0, 1, 2, 2, -1, 1, 0, 1, 0, 2, 2, 1, 2, -1, 2, 1, 0, 1, 3, 1, 1, 2, 2, 2, -2, 1, 2, 2, 2, 1, 3, 1, 0, 0, 2, 1, 0, -1, 0, 2, 0, 1, 2, 2, 2, 2, 2, 1, 1, 1, 2, 0, -1, 2, 3, 1, 0, 2, 3, 1, 0, 1, 2, 0, 0, 2, 3, 1, 0, -1, 2, 1, 1
Offset: 1

Views

Author

Gus Wiseman, Jul 05 2018

Keywords

Crossrefs

Cf. A187039 (where a(n)=0). - Michel Marcus, Jul 08 2018

Programs

  • Maple
    f:= proc(n) local F;
      F:= map(t -> t[2],ifactors(n)[2]);
      2*nops(select(type,F,odd))-nops(F);
    end proc:
    map(f, [$1..100]); # Robert Israel, Aug 27 2018
  • Mathematica
    Table[Total[-(-1)^If[n==1,{},FactorInteger[n][[All,2]]]],{n,100}]
  • PARI
    a(n) = my(f=factor(n)); -sum(k=1, #f~, (-1)^(f[k,2])); \\ Michel Marcus, Jul 08 2018; corrected Jun 13 2022

Formula

If i and j are coprime, a(i*j) = a(i)+a(j). - Robert Israel, Aug 27 2018
From Amiram Eldar, Oct 05 2023: (Start)
Additive with a(p^e) = (-1)^(e+1).
a(n) = A162642(n) - A162641(n).
Sum_{k=1..n} a(k) = n * log(log(n)) + c * n + O(n/log(n)), where c = A077761 - 2*A179119 = -0.398962... . (End)

A366528 Sum of odd prime indices of n.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Oct 22 2023

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, sum A056239(n).

Examples

			The prime indices of 198 are {1,2,2,5}, so a(198) = 1+5 = 6.
		

Crossrefs

Zeros are A066207, counted by A035363.
The triangle for this rank statistic is A113685, without zeros A365067.
For count instead of sum we have A257991, even A257992.
Nonzeros are A366322, counted by A086543.
The even version is A366531, halved A366533, triangle A113686.
A000009 counts partitions into odd parts, ranks A066208.
A053253 = partitions with all odd parts and conjugate parts, ranks A352143.
A066967 adds up sums of odd parts over all partitions.
A112798 lists prime indices, reverse A296150, length A001222, sum A056239.
A162641 counts even prime exponents, odd A162642.
A352142 = odd indices with odd exponents, counted by A117958.

Programs

  • Mathematica
    Table[Total[Cases[FactorInteger[n], {p_?(OddQ@*PrimePi),k_}:>PrimePi[p]*k]],{n,100}]

Formula

a(n) = A056239(n) - A366531(n).

A162641 Number of even exponents in canonical prime factorization of n.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, Jul 08 2009

Keywords

Crossrefs

Cf. A268335 (positions of zeros), A295316.

Programs

Formula

a(n) = A001221(n) - A162642(n).
a(A002035(n)) = 0.
a(A072587(n)) > 0.
Additive with a(p^e) = A059841(e). - Antti Karttunen, Jul 23 2017
From Antti Karttunen, Nov 28 2017: (Start)
a(n) = A162642(A003557(n)).
a(n) <= A056170(n).
(End)
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} 1/(p*(p+1)) = 0.3302299262... (A179119). - Amiram Eldar, Dec 25 2021

A293442 Multiplicative with a(p^e) = A019565(e).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 6, 3, 4, 2, 6, 2, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 12, 3, 4, 6, 6, 2, 8, 2, 10, 4, 4, 4, 9, 2, 4, 4, 12, 2, 8, 2, 6, 6, 4, 2, 10, 3, 6, 4, 6, 2, 12, 4, 12, 4, 4, 2, 12, 2, 4, 6, 15, 4, 8, 2, 6, 4, 8, 2, 18, 2, 4, 6, 6, 4, 8, 2, 10, 5, 4, 2, 12, 4, 4, 4, 12, 2, 12, 4, 6, 4, 4, 4, 20, 2, 6, 6, 9, 2, 8, 2, 12, 8
Offset: 1

Views

Author

Antti Karttunen, Oct 31 2017

Keywords

Comments

From Peter Munn, Apr 06 2021: (Start)
a(n) is determined by the prime signature of n.
Compare with the multiplicative, self-inverse A225546, which also maps 2^e to the squarefree number A019565(e). However, this sequence maps p^e to the same squarefree number for every prime p, whereas A225546 maps the e-th power of progressively larger primes to progressively greater powers of A019565(e).
Both sequences map powers of squarefree numbers to powers of squarefree numbers.
(End)

Crossrefs

Sequences used in a definition of this sequence: A000188, A003961, A019565, A028234, A059895, A067029, A162642.
Sequences with related definitions: A225546, A293443, A293444.
Cf. also A293214.
Sequences used to express relationship between terms of this sequence: A006519, A007913, A008833, A064989, A334747.
Sequences related via this sequence: (A001222, A048675, A064547), (A007814, A162642), (A087207, A267116), (A248663, A268387).

Programs

  • Mathematica
    f[n_] := If[n == 1, 1, Apply[Times, Prime@ Flatten@ Position[Reverse@ IntegerDigits[Last@ #, 2], 1]] * f[n/Apply[Power, #]] &@ FactorInteger[n][[1]]]; Array[f, 105] (* Michael De Vlieger, Oct 31 2017 *)

Formula

a(1) = 1; for n > 1, a(n) = A019565(A067029(n)) * a(A028234(n)).
Other identities. For all n >= 1:
a(a(n)) = A293444(n).
A048675(a(n)) = A001222(n).
A001222(a(n)) = A064547(n) = A048675(A293444(n)).
A007814(a(n)) = A162642(n).
A087207(a(n)) = A267116(n).
A248663(a(n)) = A268387(n).
From Peter Munn, Mar 14 2021: (Start)
Alternative definition: a(1) = 1; a(2) = 2; a(n^2) = A003961(a(n)); a(A003961(n)) = a(n); if A059895(n, k) = 1, a(n*k) = a(n) * a(k).
For n >= 3, a(n) < n.
a(2n) = A334747(a(A006519(n))) * a(n/A006519(n)), where A006519(n) is the largest power of 2 dividing n.
a(2n+1) = a(A064989(2n+1)).
a(n) = a(A007913(n)) * a(A008833(n)) = 2^A162642(n) * A003961(a(A000188(n))).
(End)

A229125 Numbers of the form p * m^2, where p is prime and m > 0: union of A228056 and A000040.

Original entry on oeis.org

2, 3, 5, 7, 8, 11, 12, 13, 17, 18, 19, 20, 23, 27, 28, 29, 31, 32, 37, 41, 43, 44, 45, 47, 48, 50, 52, 53, 59, 61, 63, 67, 68, 71, 72, 73, 75, 76, 79, 80, 83, 89, 92, 97, 98, 99, 101, 103, 107, 108, 109, 112, 113, 116, 117, 124, 125, 127, 128, 131, 137, 139, 147, 148, 149
Offset: 1

Views

Author

Chris Boyd, Sep 14 2013

Keywords

Comments

No term is the product of two other terms.
Squares of terms and pairwise products of distinct terms form a subsequence of A028260.
Numbers n such that A162642(n) = 1. - Jason Kimberley, Oct 10 2016
Numbers k such that A007913(k) is a prime number. - Amiram Eldar, Jul 27 2020

Crossrefs

Programs

  • Mathematica
    With[{nn=70},Take[Union[Flatten[Table[p*m^2,{p,Prime[Range[nn]]},{m,nn}]]], nn]] (* Harvey P. Dale, Dec 02 2014 *)
  • PARI
    test(n)=isprime(core(n))
    for(n=1,200,if(test(n), print1(n",")))
    
  • Python
    from math import isqrt
    from sympy import primepi
    def A229125(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//y**2) for y in range(1,isqrt(x)+1))
        return bisection(f,n,n) # Chai Wah Wu, Jan 30 2025

Formula

The number of terms not exceeding x is (Pi^2/6) * x/log(x) + O(x/(log(x))^2) (Cohen, 1962). - Amiram Eldar, Jul 27 2020

A055076 Multiplicity of Max{gcd(d, n/d)} when d runs over divisors of n.

Original entry on oeis.org

1, 2, 2, 1, 2, 4, 2, 2, 1, 4, 2, 2, 2, 4, 4, 1, 2, 2, 2, 2, 4, 4, 2, 4, 1, 4, 2, 2, 2, 8, 2, 2, 4, 4, 4, 1, 2, 4, 4, 4, 2, 8, 2, 2, 2, 4, 2, 2, 1, 2, 4, 2, 2, 4, 4, 4, 4, 4, 2, 4, 2, 4, 2, 1, 4, 8, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 2, 2, 1, 4, 2, 4, 4, 4, 4, 4, 2, 4, 4, 2, 4, 4, 4, 4, 2, 2, 2, 1, 2, 8, 2, 4, 8
Offset: 1

Views

Author

Labos Elemer, Jun 13 2000

Keywords

Comments

Number of distinct values of gcd(d, n!/d) if d runs over divisors of n! seems to be A046951(n).
a(n) = 1 iff n is a square. - Bernard Schott, Oct 22 2019
a(n) is the number of the unitary divisors (cf. A077610) of n that are exponentially odd (A268335). - Amiram Eldar, Nov 11 2022
The number of infinitary divisors of n that are squarefree (A005117). - Amiram Eldar, Jan 09 2024

Examples

			n=120, the set of gcd(d, 120/d) values for the 16 divisors of 120 is {1,2,1,2,1,2,1,2,2,1,2,1,2,1,2,1}. The max is 2 and it occurs 8 times, so a(120)=8. This sequence seems to consist of powers of 2.
		

Crossrefs

Programs

  • Maple
    with(numtheory):
    a:= n->(p->coeff(p, x, degree(p)))(add(x^igcd(d, n/d), d=divisors(n))):
    seq(a(n), n=1..105);  # Alois P. Heinz, Jul 21 2015
  • Mathematica
    a[n_] := With[{g = GCD[#, n/#]& /@ Divisors[n]}, Count[g, Max[g]]];
    Array[a, 105] (* Jean-François Alcover, Mar 28 2017 *)
    f[p_, e_] := 2^Mod[e, 2]; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Nov 11 2022 *)
  • PARI
    A055076(n) = if(1==n,n,my(es=factor(n)[,2]~); prod(i=1,#es,2^(es[i]%2))); \\ Antti Karttunen, Apr 05 2021
    
  • Scheme
    ;; With memoization-macro definec.
    (definec (A055076 n) (if (= 1 n) n (* (+ 1 (A000035 (A067029 n))) (A055076 (A028234 n))))) ;; Antti Karttunen, Dec 02 2017

Formula

Multiplicative with a(p^e) = 2^(e mod 2). - Vladeta Jovovic, Dec 13 2002
a(n) = 2^A162642(n). - Antti Karttunen, Dec 02 2017
a(n) = A034444(A007913(n)). [Found by LODA miner, see C. Krause link. Essentially the same formula as the above ones] - Antti Karttunen, Apr 05 2021
From Amiram Eldar, Sep 09 2023: (Start)
a(n) = A034444(A350389(n)).
Dirichlet g.f.: zeta(2*s) * Product_{p prime} (1 + 2/p^s). (End)
From Vaclav Kotesovec, Sep 09 2023: (Start)
Let f(s) = Product_{p prime} (1 - 3/p^(2*s) + 2/p^(3*s)).
Dirichlet g.f.: zeta(s)^2 * zeta(2*s) * f(s).
Sum_{k=1..n} a(k) ~ (Pi^2 * f(1) * n / 6) * (log(n) + 2*gamma - 1 + 12*zeta'(2)/Pi^2 + f'(1)/f(1)), where
f(1) = A065473 = Product_{primes p} (1 - 3/p^2 + 2/p^3) = 0.286747428434478734107892712789838446434331844097056995641477859336652243...,
f'(1) = f(1) * Sum_{primes p} 6*log(p) / (p^2 + p - 2) = f(1) * 2.798014228561519243358371276385174449737670294137200281334256087932048625...
and gamma is the Euler-Mascheroni constant A001620. (End)

A187039 Numbers that have equal counts of even and odd exponents of primes in their factorization.

Original entry on oeis.org

1, 12, 18, 20, 28, 44, 45, 48, 50, 52, 63, 68, 72, 75, 76, 80, 92, 98, 99, 108, 112, 116, 117, 124, 147, 148, 153, 162, 164, 171, 172, 175, 176, 188, 192, 200, 207, 208, 212, 236, 242, 244, 245, 261, 268, 272, 275, 279, 284, 288, 292, 304, 316, 320, 325, 332
Offset: 1

Views

Author

Vladimir Shevelev, Mar 02 2011

Keywords

Comments

Numbers k such that A162641(k) = A162642(k). - Amiram Eldar, Sep 27 2021

Examples

			108 = 2^2*3^3 has one even and one odd exponent in its factorization and therefore qualifies.
		

Crossrefs

Programs

A295664 Exponent of the highest power of 2 dividing number of divisors of n: a(n) = A007814(A000005(n)); 2-adic valuation of tau(n).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Nov 28 2017

Keywords

Comments

In the prime factorization of n = p1^e1 * ... pk^ek, add together the number of trailing 1-bits in each exponent e when they are written in binary.

Crossrefs

Cf. A000290 (positions of zeros).

Programs

  • Mathematica
    Table[IntegerExponent[DivisorSigma[0, n], 2], {n, 120}] (* Michael De Vlieger, Nov 28 2017 *)
  • PARI
    a(n) = valuation(numdiv(n), 2); \\ Michel Marcus, Nov 30 2017
    
  • Python
    from sympy import divisor_count
    def A295664(n): return (~(m:=int(divisor_count(n))) & m-1).bit_length() # Chai Wah Wu, Jul 05 2022

Formula

Additive with a(p^e) = A007814(1+e).
a(1) = 0; for n > 1, a(n) = A007814(1+A067029(n)) + a(A028234(n)).
a(n) = A007814(A000005(n)).
a(n) >= A162642(n) >= A056169(n).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) =-0.223720656976344505701..., where f(x) = -x + (1-x) * Sum_{k>=1} x^(2^k-1)/(1-x^(2^k)). - Amiram Eldar, Sep 28 2023

A295659 Number of exponents larger than 2 in the prime factorization of n.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Nov 28 2017

Keywords

Examples

			For n = 120 = 2^3 * 3^1 * 5^1 there is only one exponent larger than 2, thus a(120) = 1.
For n = 216 = 2^3 * 3^3 there are two exponents larger than 2, thus a(216) = 2.
		

Crossrefs

Cf. A004709 (positions of zeros), A046099 (of nonzeros), A212793.

Programs

Formula

Additive with a(p^e) = 1 if e>2, 0 otherwise.
a(n) = 0 iff A212793(n) = 1.
a(n) = A001221(A053150(n)).
a(n) = A056170(A003557(n)).
a(n) >= A295662(n) = A162642(n) - A056169(n).
a(n) = A295883(n) + A295884(n).
Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = Sum_{p prime} 1/p^3 = 0.174762... (A085541). - Amiram Eldar, Nov 01 2020
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