A368781 -id:A368781 - cp's OEIS frontend

A372601 The maximal exponent in the prime factorization of the largest exponentially odd divisor of n.

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

Offset: 1

Amiram Eldar, May 07 2024

First differs from A331273 at n = 64.

Differs from A363332 at n = 1, 216, 432, 648, 864, 1000, ... .

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

Cf. A051903, A109613, A350390, A368711.
Cf. A331273, A363332.
Similar sequences: A007424, A368781, A372602, A372603, A372604.

f[n_] := n - If[EvenQ[n], 1, 0]; a[n_] := f[Max[FactorInteger[n][[;; , 2]]]]; a[1] = 0; Array[a, 100]

s(n) = (n+1) \ 2 * 2 - 1;
a(n) = if(n>1, s(vecmax(factor(n)[,2])), 0);

a(n) = A051903(A350390(n)).
a(n) = A109613(A051903(n)-1) for n >= 2.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + 2 * Sum_{i>=1} (1 - (1/zeta(2*i+1))) = 1.42929441950714075659... .

A372602 The maximal exponent in the prime factorization of the largest square dividing n.

Original entry on oeis.org

0, 0, 0, 2, 0, 0, 0, 2, 2, 0, 0, 2, 0, 0, 0, 4, 0, 2, 0, 2, 0, 0, 0, 2, 2, 0, 2, 2, 0, 0, 0, 4, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 2, 2, 0, 0, 4, 2, 2, 0, 2, 0, 2, 0, 2, 0, 0, 0, 2, 0, 0, 2, 6, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 2, 2, 0, 0, 0, 4, 4, 0, 0, 2, 0, 0, 0

Offset: 1

Amiram Eldar, May 07 2024

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

Cf. A008833, A051903, A052928.

Similar sequences: A007424, A368781, A372601, A372603, A372604.

f[n_] := 2 * Floor[n/2]; a[n_] := f[Max[FactorInteger[n][[;; , 2]]]]; a[1] = 0; Array[a, 100]

s(n) = n \ 2 * 2;
a(n) = if(n>1, s(vecmax(factor(n)[,2])), 0);

a(n) = A051903(A008833(n)).
a(n) = A052928(A051903(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2 * Sum_{i>=1} (1 - (1/zeta(2*i))) = 0.98112786070359477197... .

A372603 The maximal exponent in the prime factorization of the powerful part of n.

Original entry on oeis.org

0, 0, 0, 2, 0, 0, 0, 3, 2, 0, 0, 2, 0, 0, 0, 4, 0, 2, 0, 2, 0, 0, 0, 3, 2, 0, 3, 2, 0, 0, 0, 5, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 2, 2, 0, 0, 4, 2, 2, 0, 2, 0, 3, 0, 3, 0, 0, 0, 2, 0, 0, 2, 6, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 2, 2, 0, 0, 0, 4, 4, 0, 0, 2, 0, 0, 0

Offset: 1

Amiram Eldar, May 07 2024

First differs from A275812 at n = 36, and from A212172 at n = 37.

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

Cf. A051903, A057521, A087156, A368710.
Cf. A013661, A033150, A059956.
Cf. A212172, A275812.
Similar sequences: A007424, A368781, A372601, A372602, A372604.

f[n_] := If[n == 1, 0, n]; a[n_] := f[Max[FactorInteger[n][[;; , 2]]]]; a[1] = 0; Array[a, 100]

s(n) = if(n == 1, 0, n);
a(n) = if(n>1, s(vecmax(factor(n)[,2])), 0);

a(n) = A051903(A057521(n)).
a(n) = A087156(A051903(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 - 1/zeta(2) + Sum_{i>=2} (1 - 1/zeta(i)) = A033150 - A059956 = 1.09728403825134113562... .

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.

Original entry on oeis.org

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... .

A372604 The maximal exponent in the prime factorization of the largest divisor of n whose number of divisors is a power of 2.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1

Offset: 1

Amiram Eldar, May 07 2024

First differs from A331273 at n = 32.
Differs from A368247 at n = 1, 128, 216, 256, 384, 432, 512, ... .
All the terms are of the form 2^k-1 (A000225).

			4 has 3 divisors, 1, 2 and 4. The number of divisors of 4 is 3, which is not a power of 2. The number of divisors of 2 is 2, which is a power of 2. Therefore, A372379(4) = 2 and a(4) = A051903(2) = 1.

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

Cf. A000225, A051903, A092323, A372379, A372466.
Cf. A331273, A368247.
Similar sequences: A007424, A368781, A372601, A372602, A372603.

f[n_] := 2^Floor[Log2[n + 1]] - 1; a[n_] := f[Max[FactorInteger[n][[;; , 2]]]]; a[1] = 0; Array[a, 100]

s(n) = 2^exponent(n+1) - 1;
a(n) = if(n>1, s(vecmax(factor(n)[,2])), 0);

a(n) = A051903(A372379(n)).
a(n) = A092323(A051903(n)+1).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + Sum_{i>=1} 2^i * (1 - 1/zeta(2^(i+1)-1)) = 1.36955053734097783559... .

A375430 The maximum exponent in the unique factorization of n in terms of distinct terms of A115975 using the dual Zeckendorf representation of the exponents in the prime factorization of n; a(1) = 0.

Original entry on oeis.org

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

Offset: 1

Amiram Eldar, Aug 15 2024

First differs from A299090 at n = 128. Differs from A046951 and A159631 at n = 1, 36, 64, 72, ... .

When the exponents in the prime factorization of n are expanded as sums of distinct Fibonacci numbers using the dual Zeckendorf representation (A104326), 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 = 8 = 2^3, the dual Zeckendorf representation of 3 is 11, i.e., 3 = Fibonacci(2) + Fibonacci(3) = 1 + 2. Therefore 8 = 2^(1+2) = 2^1 * 2^2, and a(8) = 2.

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

Cf. A000045, A095791, A104326, A115975, A246655, A331109, A368781, A375428, A375431.

Cf. A046951, A159631, A299090.

A130312[n_] := Module[{k = 0}, While[Fibonacci[k] <= n, k++]; Fibonacci[k-2]]; a[n_] := A130312[1 + Max[FactorInteger[n][[;;, 2]]]]; a[1] = 0; Array[a, 100]

A130312(n) = {my(k = 0); while(fibonacci(k) <= n, k++); fibonacci(k-2);}
a(n) = if(n == 1, 0, A130312(1 + vecmax(factor(n)[,2])));

a(n) = A130312(1 + A051903(n)).
a(n) = A000045(A375431(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + Sum_{k>=4} Fibonacci(k) * (1 - 1/zeta(Fibonacci(k)-1)) = 1.48543763231328442311... .

A375229 Numbers k such that A299090(k) is even.

Original entry on oeis.org

1, 4, 8, 9, 12, 18, 20, 24, 25, 27, 28, 36, 40, 44, 45, 49, 50, 52, 54, 56, 60, 63, 68, 72, 75, 76, 84, 88, 90, 92, 98, 99, 100, 104, 108, 116, 117, 120, 121, 124, 125, 126, 132, 135, 136, 140, 147, 148, 150, 152, 153, 156, 164, 168, 169, 171, 172, 175, 180, 184

Offset: 1

Amiram Eldar, Aug 06 2024

Differs from A252849 by having the terms 1, 256, 512, 768, 1280, ..., and not having the terms 64, 128, 144, 192, 288, ... .
Numbers whose maximum exponent in their unique factorization in terms of distinct "Fermi-Dirac primes" (A368781) is not a power of 4.
The asymptotic density of this sequence is Sum_{k>=1} (1/zeta(4^k) - 1/zeta(2^(2*k-1))) = 0.32005681814901480646... .

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

Cf. A252849, A299090, A368781.

q[n_] := EvenQ[IntegerLength[Max[FactorInteger[n][[;; , 2]]], 2]]; q[1] = True; Select[Range[200], q]

is(n) = if(n == 1, 1, !(#binary(vecmax(factor(n)[, 2])) % 2));

cp's OEIS Frontend

A372601 The maximal exponent in the prime factorization of the largest exponentially odd divisor of n.

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A372602 The maximal exponent in the prime factorization of the largest square dividing n.

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A372603 The maximal exponent in the prime factorization of the powerful part of n.

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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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A372604 The maximal exponent in the prime factorization of the largest divisor of n whose number of divisors is a power of 2.

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A375430 The maximum exponent in the unique factorization of n in terms of distinct terms of A115975 using the dual Zeckendorf representation of the exponents in the prime factorization of n; a(1) = 0.

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A375229 Numbers k such that A299090(k) is even.

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