A008835 -id:A008835 - cp's OEIS frontend

A058035 Largest 4th-power-free number dividing n.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 8, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 8, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 24, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 8, 65, 66, 67, 68, 69, 70, 71, 72

Offset: 1

Henry Bottomley, Nov 16 2000

			a(96) = 24 since the factors of 96 are {1,2,3,4,6,8,12,16,24,32,48,96} but 32, 48 and 96 all contain a 4th power factor (16).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A007947, A007948, A008835, A046100, A053165, A065466.

Cf. A027748, A124010.

a058035 n = product $
   zipWith (^) (a027748_row n) (map (min 3) $ a124010_row n)
-- Reinhard Zumkeller, Jan 06 2012

f[p_, e_] := p^Min[e, 3]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jul 09 2022 *)

a(n) = my(f=factor(n)); for(k=1,#f~,f[k,2]=min(3,f[k,2])); factorback(f); \\ Michel Marcus, Sep 13 2017

Multiplicative with a(p^e) = p ^ min(e,3), p prime, e > 0. - Reinhard Zumkeller, Jan 06 2012

Sum_{k=1..n} a(k) ~ (1/2) * c * n^2, where c = Product_{p prime} (1 - 1/(p^3*(p+1))) = 0.947733... (A065466). - Amiram Eldar, Oct 13 2022

A295884 Number of exponents larger than 3 in the prime factorization of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Antti Karttunen, Nov 29 2017

nonn

a(1296) is the first term greater than 1, and a(810000) is the first term greater than 2. - Harvey P. Dale, Dec 22 2017

			For n = 16 = 2^4, there is one exponent and it is larger than 3, thus a(16) = 1.
For n = 96 = 2^5 * 3^1, there are two exponents, and the other one is larger than 3, thus a(96) = 1.
For n = 1296 = 2^4 * 3^4, there are two exponents larger than 3, thus a(1296) = 2.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences computed from exponents in factorization of n

Cf. A000188, A001221, A003557, A008835, A053164, A056170, A062378, A085964, A295659, A295883.

Array[Total@ Map[Boole[# > 3] &, FactorInteger[#][[All, -1]]] &, 120] (* Michael De Vlieger, Nov 29 2017 *)
Table[Count[FactorInteger[n][[All,2]],?(#>3&)],{n,130}] (* _Harvey P. Dale, Dec 22 2017 *)

Additive with a(p^e) = 1 when e > 3, 0 otherwise.
a(n) = A295659(n) - A295883(n).
a(n) = A056170(A062378(n)) = A056170(A003557(A003557(n))) = A001221(A003557^3(n)).
a(n) = A001221(A053164(n)) = A001221(A008835(n)).
Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = Sum_{p prime} 1/p^4 = 0.076993... (A085964). - Amiram Eldar, Nov 01 2020

A335324 Square part of 4th-power-free part of n.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 4, 1, 1, 1, 1, 1, 9, 1, 4, 1, 1, 1, 4, 25, 1, 9, 4, 1, 1, 1, 1, 1, 1, 1, 36, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 1, 49, 25, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 1, 9, 4, 1, 1, 1, 4, 1, 1, 1, 36, 1, 1, 25, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1

Offset: 1

Peter Munn, May 31 2020

Equivalently, biquadratefree (4th-power-free) part of square part of n.
Multiplicative. The terms are squares of squarefree numbers (A062503).
Every positive integer n is the product of a unique subset S_n of the terms of A050376 (sometimes called Fermi-Dirac primes). a(n) is the product of the members of S_n that are squares of prime numbers (A001248).

			Removing the 4th powers from 192 = 2^6 * 3^1 gives 2^(6 - 4) * 3^1 = 2^2 * 3 = 12. So the 4th-power-free part of 192 is 12. The square part of 12 (largest square dividing 12) is 4. So a(192) = 4.

Michel Marcus, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Biquadratefree.
Eric Weisstein's World of Mathematics, Square part.

A007913, A008833, A008835, A053165 are used in formulas defining the sequence.
Column 1 of A352780.
Range of values is A062503.
Positions of 1's: A252895.
Related to A038500 by A225546.
The formula section details how the sequence maps the terms of A003961, A331590.
Cf. A001248, A050376, A083730.
Cf. A002117, A013662, A078434.

f[p_, e_] := p^(2*Floor[e/2] - 4*Floor[e/4]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Jun 01 2020 *)

A053165(n)=my(f=factor(n)); f[, 2]=f[, 2]%4; factorback(f);
a(n) = my(m=A053165(n)); m/core(m); \\ Michel Marcus, Jun 01 2020

from math import prod
from sympy import factorint
def A335324(n): return prod(p**(e&2) for p, e in factorint(n).items()) # Chai Wah Wu, Aug 07 2024

a(n) = A053165(A008833(n)) = A008833(A053165(n)).
a(n) = A053165(n) / A007913(n).
a(n) = A008833(n) / A008835(n).
n = A007913(n) * a(n) * A008835(n).
a(n) = A225546(A038500(A225546(n))).
a(n^2) = A007913(n)^2.
a(A003961(n)) = A003961(a(n)).
a(A331590(n, k)) = A331590(a(n), a(k)).
a(p^e) = p^(2*floor(e/2) - 4*floor(e/4)). - Amiram Eldar, Jun 01 2020
From Amiram Eldar, Sep 21 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s-2) * zeta(4*s)/(zeta(2*s) * zeta(4*s-4)).
Sum_{k=1..n} a(k) ~ (4*zeta(3/2)*zeta(4))/(21*zeta(3)) * n^(3/2). (End)

A056553 Smallest 4th-power divisible by n divided by largest 4th-power which divides n.

Original entry on oeis.org

1, 16, 81, 16, 625, 1296, 2401, 16, 81, 10000, 14641, 1296, 28561, 38416, 50625, 1, 83521, 1296, 130321, 10000, 194481, 234256, 279841, 1296, 625, 456976, 81, 38416, 707281, 810000, 923521, 16, 1185921, 1336336, 1500625, 1296, 1874161, 2085136

Offset: 1

Henry Bottomley, Jun 25 2000

			a(64) = 16 because smallest 4th power divisible by 64 is 256 and largest 4th power which divides 64 is 16 and 256/16 = 16.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences

Cf. A000190, A008835, A053164, A053165, A053166, A053167, A056551, A056554, A056555.

Cf. A013662, A013678.

f[p_, e_] := p^If[Divisible[e, 4], 0, 1]; a[n_] := (Times @@ (f @@@ FactorInteger[ n]))^4; Array[a, 100] (* Amiram Eldar, Aug 29 2019*)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%4, f[i,1], 1))^4; } \\ Amiram Eldar, Oct 27 2022

a(n) = A053167(n)/A008835(n) = A056556(n)*A053165(n) = A056554(n)^4.
From Amiram Eldar, Oct 27 2022: (Start)
Multiplicative with a(p^e) = 1 if e is divisible by 4, and a(p^e) = p^4 otherwise.
Sum_{k=1..n} a(k) ~ c * n^5, where c = (zeta(20)/(5*zeta(4))) * Product_{p prime} (1 - 1/p^2 + 1/p^4 - 1/p^7 + 1/p^8) = 0.123026157003... . (End)

A056554 Powerfree kernel of 4th-powerfree part of n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 1, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 2, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 3, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 2, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38

Offset: 1

Henry Bottomley, Jun 25 2000

			a(64) = 2 because 4th-power-free part of 64 is 4 and power-free kernel of 4 is 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A000190, A007947, A008835, A013666, A053164, A053165, A053166, A053167, A056552, A056553, A056555.

f[p_, e_] :=  p^If[Divisible[e, 4], 0, 1]; a[n_] := Times @@ (f @@@ FactorInteger[ n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)

a(n) = my(f=factor(n)); for (k=1, #f~, if (frac(f[k,2]/4), f[k,2] = 1, f[k,2] = 0)); factorback(f); \\ Michel Marcus, Feb 28 2019

a(n) = A007947(A053165(n)) = A053166(A053165(n)) = n/(A053164(n)*A000190(n)) = A053166(n)/A053164(n) = A056553(n)^(1/4).
If n = Product_{j} Pj^Ej then a(n) = Product_{j} Pj^Fj, where Fj = 0 if Ej is 0 or a multiple of 4 and Fj = 1 otherwise.
Multiplicative with a(p^e) = p^(if 4|e, then 0, else 1). - Mitch Harris, Apr 19 2005
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(8)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6) = 0.3513111135... . - Amiram Eldar, Oct 27 2022

A056555 Smallest number k (k>0) such that n*k is a perfect 4th power.

Original entry on oeis.org

1, 8, 27, 4, 125, 216, 343, 2, 9, 1000, 1331, 108, 2197, 2744, 3375, 1, 4913, 72, 6859, 500, 9261, 10648, 12167, 54, 25, 17576, 3, 1372, 24389, 27000, 29791, 8, 35937, 39304, 42875, 36, 50653, 54872, 59319, 250, 68921, 74088, 79507, 5324, 1125

Offset: 1

Henry Bottomley, Jun 25 2000

			a(64) = 4 because the smallest 4th power divisible by 64 is 256 and 64*4 = 256.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A000190, A008835, A048798, A053164, A053165, A053166, A053167, A056554.

Cf. A013662, A013674.

f[p_, e_] := p^Mod[4 - e, 4]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 08 2020 *)

a(n,f=factor(n))=f[,2]=-f[,2]%4; factorback(f) \\ Charles R Greathouse IV, Apr 24 2020

a(n) = A053167(n)/n = n^3/A000190(n)^4 = A056553(n)/A053165(n).
Multiplicative with a(p^e) = p^((4 - e) mod 4). - Amiram Eldar, Sep 08 2020
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(16)/(4*zeta(4))) * Product_{p prime} (1 - 1/p^2 + 1/p^4 - 1/p^7 + 1/p^8) = 0.1537848996... . - Amiram Eldar, Oct 27 2022

A062379 n divided by largest 4th-power-free factor of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Henry Bottomley, Jun 18 2001

Antti Karttunen, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.

Cf. A000190, A000583, A008835, A053164, A053165, A053166, A053167, A056553, A056554, A056555, A058035. See A003557 for squares and A062378 for cubes.

f[p_, e_] := p^Max[e-3, 0]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 07 2020 *)

(define (A062379 n) (/ n (A058035 n)))
(define (A058035 n) (if (= 1 n) n (* (expt (A020639 n) (min 3 (A067029 n))) (A058035 (A028234 n)))))
;; Antti Karttunen, Sep 13 2017

a(n) = n/A058035(n).

Multiplicative with a(p^e) = p^max(e-3, 0). - Amiram Eldar, Sep 07 2020

A248764 Greatest 4th power integer that divides n!

Original entry on oeis.org

1, 1, 1, 1, 1, 16, 16, 16, 1296, 20736, 20736, 20736, 20736, 20736, 20736, 331776, 331776, 429981696, 429981696, 268738560000, 268738560000, 268738560000, 268738560000, 4299816960000, 4299816960000, 4299816960000, 348285173760000, 13379723235164160000

Offset: 1

Clark Kimberling, Oct 14 2014

Every term divides all its successors.

			a(6) = 16 because 16 divides 6! and if k > 2 then k^4 does not divide 6!.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A000142, A008835, A248762, A248765, A248766.

z = 40; f[n_] := f[n] = FactorInteger[n!]; r[m_, x_] := r[m, x] = m*Floor[x/m];
u[n_] := Table[f[n][[i, 1]], {i, 1, Length[f[n]]}];
v[n_] := Table[f[n][[i, 2]], {i, 1, Length[f[n]]}];
p[m_, n_] := p[m, n] = Product[u[n][[i]]^r[m, v[n]][[i]], {i, 1, Length[f[n]]}];
m = 4; Table[p[m, n], {n, 1, z}]  (* A248764 *)
Table[p[m, n]^(1/m), {n, 1, z}]   (* A248765 *)
Table[n!/p[m, n], {n, 1, z}]      (* A248766 *)
f[p_, e_] := p^(4*Floor[e/4]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 30] (* Amiram Eldar, Sep 01 2024 *)

a(n) = {my(f = factor(n!)); prod(i = 1, #f~, f[i, 1]^(4*(f[i, 2]\4)));} \\ Amiram Eldar, Sep 01 2024

a(n) = n!/A248766(n).
From Amiram Eldar, Sep 01 2024: (Start)
a(n) = A008835(n!).
a(n) = A248765(n)^4. (End)

A365490 The number of divisors of the largest 4th power dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 5, 1, 1, 1, 1, 1, 1

Offset: 1

Amiram Eldar, Sep 05 2023

The number of divisors of the 4th root of the largest 4th power dividing n, A053164(n), is A063775(n).

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

Cf. A000005, A000583, A046100, A008835, A053164, A063775.

f[p_, e_] := 4*Floor[e/4] + 1; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100]

a(n) = vecprod(apply(x -> 4*(x\4) + 1, factor(n)[, 2]));

from math import prod
from sympy import factorint
def A365490(n): return prod(e&-4|1 for e in factorint(n).values()) # Chai Wah Wu, Aug 08 2024

a(n) = A000005(A008835(n)).
Multiplicative with a(p^e) = 4*floor(e/4) + 1.
a(n) = 1 if and only if n is a biquadratefree number (A046100).
a(n) <= A000005(n) with equality if and only if n is a fourth power (A000583).
Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 + 3/p^(4*s)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(4) * Product_{p prime} (1 + 3/p^4) = 1.3414590511076... . In general, the asymptotic mean of the number of divisors of the largest k-th power dividing n is zeta(k) * Product_{p prime} (1 + (k-1)/p^k).

A128251 n^4 - 1 divided by its largest fourth power divisor.

Original entry on oeis.org

15, 5, 255, 39, 1295, 150, 4095, 410, 9999, 915, 20735, 1785, 38415, 3164, 65535, 5220, 104975, 8145, 159999, 12155, 234255, 17490, 331775, 24414, 456975, 33215, 614655, 44205, 809999, 57720, 1048575, 74120, 1336335, 93789, 1679615, 117135

Offset: 2

Jonathan Vos Post, May 03 2007

In other words, biquadratefree part of n^4-1, or biquadratefree kernel of n^4-1. Fourth power analog of what A128972 is to cubes and A068310 is to squares. A046100 Biquadratefree numbers. A008835 Largest 4th power dividing n.

			a(3) = 5 because (3^4 - 1)/16 = 80/16 = (2^4 * 5)/(2^4) = 5.
a(5) = 39 because (5^4 - 1)/16 = 624/16 = (2^4 * 3 * 13)/(2^4) = 39.
a(7) = 150 because (7^4 - 1)/16 = 2400/16 = (2^5 * 3 * 5^2)/(2^4) = 150.
a(9) = 410 because (9^4 - 1)/16 = 6560/16 = (2^5 * 5 * 41)/(2^4) = 410.
a(63) = 61535 because (63^4 - 1)/256 = 15752960/256 = (2^8 * 5 * 31 * 397)/(2^8) = 61535.

Eric Weisstein's World of Mathematics, Biquadratefree.

Cf. A000188, A000583, A002350, A004709, A007948, A008835, A062378, A067872, A033314, A068310, A128972.

a(n) = (n^4 - 1)/A008835(n^4 - 1) = (A000583(n)-1)/A008835((A000583(n)-1)).

cp's OEIS Frontend

A058035 Largest 4th-power-free number dividing n.

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A295884 Number of exponents larger than 3 in the prime factorization of n.

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A335324 Square part of 4th-power-free part of n.

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A056553 Smallest 4th-power divisible by n divided by largest 4th-power which divides n.

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A056554 Powerfree kernel of 4th-powerfree part of n.

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A056555 Smallest number k (k>0) such that n*k is a perfect 4th power.

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A062379 n divided by largest 4th-power-free factor of n.

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