A365552 -id:A365552 - cp's OEIS frontend

A375428 The maximum exponent in the unique factorization of n in terms of distinct terms of A115975 using the Zeckendorf representation of the exponents in the prime factorization of n; a(1) = 0.

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 5, 1, 2, 2, 2, 1, 1, 1, 3, 1

Offset: 1

Amiram Eldar, Aug 15 2024

Differs from A095691 and A365552 at n = 1, 32, 36, 64, 72, 96, 100, ... . Differs from A368105 at n = 1, 36, 72, 100, 108, ... .

When the exponents in the prime factorization of n are expanded as sums of distinct Fibonacci numbers using the Zeckendorf representation (A014417), we get a unique factorization of n in terms of distinct terms of A115975, i.e., n is represented as a product of prime powers (A246655) whose exponents are Fibonacci numbers. a(n) is the maximum exponent of these prime powers. Thus all the terms are Fibonacci numbers.

			For n = 16 = 2^4, the Zeckendorf representation of 4 is 101, i.e., 4 = Fibonacci(2) + Fibonacci(4) = 1 + 3. Therefore 16 = 2^(1+3) = 2^1 * 2^3, and a(16) = 3.

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

Cf. A000045, A014417, A051903, A087172, A115975, A246655, A318465, A368781, A375429, A375430.

A087172[n_] := Module[{k = 2}, While[Fibonacci[k] <= n, k++]; Fibonacci[k-1]]; a[n_] := A087172[Max[FactorInteger[n][[;;, 2]]]]; a[1] = 0; Array[a, 100]

A087172(n) = {my(k = 2); while(fibonacci(k) <= n, k++); fibonacci(k-1);}
a(n) = if(n == 1, 0, A087172(vecmax(factor(n)[,2])));

a(n) = A087172(A051903(n)) for n >= 2.
a(n) = A000045(A375429(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2 - 1/zeta(2) + Sum_{k>=4} Fibonacci(k) * (1 - 1/zeta(Fibonacci(k))) = 1.64419054900327345836... .

A368105 The number of bi-unitary divisors of n that are powerful (A001694).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 2, 1, 1, 1

Offset: 1

Amiram Eldar, Dec 12 2023

First differs from A095691 and A365552 at n = 32.

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

Cf. A013661, A001694, A005117, A062503, A286324,
Similar sequences: A005361, A323308, A360721.
Cf. A095691, A365552.

f[p_, e_] := If[e == 2 || OddQ[e], e, e -1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = vecprod(apply(x -> if(x%2 || x == 2, x, x-1), factor(n)[, 2]));

Multiplicative with a(p^e) = e if e = 2 or e is odd, and e-1 otherwise.
a(n) >= 1, with equality if and only if n is squarefree (A005117).
a(n) <= A286324(n), with equality if and only if n equals the square of a squarefree number (A062503).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(2) * Product_{p prime} (1 + 1/p^3 - 1/p^4 + 1/p^5) = 1.87133814920590891161... .

A383292 Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^(2s) + 1/p^(3s)).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 3, 1, 2, 2, 4

Offset: 1

Vaclav Kotesovec, Apr 22 2025

First differs from A095691, A365552 and A368105 at n = 32.

The number of divisors of n that are both biquadratefree (A046100) and powerful (A001694). - Amiram Eldar, Apr 22 2025

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000

Cf. A073184, A365498, A380922.
Cf. A095691, A365552, A368105.
Cf. A330595.
Cf. A046100, A001694.

f[p_, e_] := If[e < 4, e, 3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Apr 22 2025 *)

for(n=1, 100, print1(direuler(p=2, n, 1/(1-X) * (1 + X^2 + X^3))[n], ", "))

Sum_{k=1..n} a(k) ~ c * n, where c = A330595 = Product_{p prime} (1 + 1/p^2 + 1/p^3) = 1.74893299784324530303390699768511480225988349359548...

Multiplicative with a(p^e) = e if e < 4 and 3 otherwise. - Amiram Eldar, Apr 22 2025

cp's OEIS Frontend

A375428 The maximum exponent in the unique factorization of n in terms of distinct terms of A115975 using the Zeckendorf representation of the exponents in the prime factorization of n; a(1) = 0.

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A368105 The number of bi-unitary divisors of n that are powerful (A001694).

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A383292 Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^(2s) + 1/p^(3s)).

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cp's OEIS Frontend

A375428 The maximum exponent in the unique factorization of n in terms of distinct terms of A115975 using the Zeckendorf representation of the exponents in the prime factorization of n; a(1) = 0.

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Author

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Programs

Mathematica

PARI

Formula

A368105 The number of bi-unitary divisors of n that are powerful (A001694).

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Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A383292 Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^(2*s) + 1/p^(3*s)).

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Author

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Mathematica

PARI

Formula

A383292 Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^(2s) + 1/p^(3s)).