A053150 -id:A053150 - cp's OEIS frontend

A015052 a(n) is the smallest positive integer m such that m^5 is divisible by n.

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 2, 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, 6, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 4, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38, 77, 78

Offset: 1

R. Muller

A multiplicative companion function n/a(n) = 1,1,1,2,1,1,1,4,3,1,1,2,1,1,1,8,1,... can be defined using the 5th instead of the 4th power in A000190, which differs from A000190 and also from A003557. - R. J. Mathar, Jul 14 2012

Amiram Eldar, Table of n, a(n) for n = 1..10000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
Henry Ibstedt, Surfing on the Ocean of Numbers, Erhus Univ. Press, Vail, 1997.
Florentin Smarandache, Collected Papers, Vol. II, Tempus Publ. Hse, Bucharest, 1996.
Eric Weisstein's World of Mathematics, Smarandache Ceil Function.

Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A019555 (outer 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015053 (outer 6th root).

Cf. A013667.

f[p_, e_] := p^Ceiling[e/5]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

a(n) = my(f=factor(n)); for (i=1, #f~, f[i,2] = ceil(f[i,2]/5)); factorback(f); \\ Michel Marcus, Feb 15 2015

Multiplicative with a(p^e) = p^(ceiling(e/5)). - Christian G. Bower, May 16 2005

Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(9)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6 + 1/p^7 - 1/p^8) = 0.3523622369... . - Amiram Eldar, Oct 27 2022

Corrected by David W. Wilson, Jun 04 2002

Name reworded by Jon E. Schoenfield, Oct 28 2022

A015053 Smallest positive integer for which n divides a(n)^6.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 2, 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, 6, 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, 77, 78

Offset: 1

R. Muller (Research37(AT)aol.com)

Differs from A007947 as follows: A007947(128)=2, a(128)=4; A007947(256)=2, a(256)=4; A007947(384)=6, a(384)=12; A007947(512)=2, a(512)=4; A007947(640)=10, a(640)=20, etc. - R. J. Mathar, Oct 28 2008

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Henry Bottomley, Some Smarandache-type multiplicative sequences.
Kevin A. Broughan, Restricted divisor sums, Acta Arithmetica, 101(2) (2002), 105-114.
Kevin A. Broughan, Relationship between the integer conductor and k-th root functions, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.
Kevin A. Broughan, Relaxations of the ABC Conjecture using integer k'th roots, New Zealand J. Math. 35(2) (2006), 121-136.
Henry Ibstedt, Surfing on the Ocean of Numbers, Erhus Univ. Press, Vail, 1997.
Eric Weisstein's World of Mathematics, Smarandache Ceil Function.

Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A019555 (outer 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (5th outer root).

Cf. A013669.

spi[n_]:=Module[{k=1},While[PowerMod[k,6,n]!=0,k++];k]; Array[spi,80] (* Harvey P. Dale, Feb 29 2020 *)
f[p_, e_] := p^Ceiling[e/6]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)

a(n) = my(f=factor(n)); for (i=1, #f~, f[i,2] = ceil(f[i,2]/6)); factorback(f); \\ Michel Marcus, Feb 15 2015

Multiplicative with a(p^e) = p^ceiling(e/6). - Christian G. Bower, May 16 2005

Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(11)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6 + 1/p^7 - 1/p^8 + 1/p^9 - 1/p^10) = 0.3522558764... . - Amiram Eldar, Oct 27 2022

Corrected by David W. Wilson, Jun 04 2002

A056552 Powerfree kernel of cubefree part of n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 1, 3, 10, 11, 6, 13, 14, 15, 2, 17, 6, 19, 10, 21, 22, 23, 3, 5, 26, 1, 14, 29, 30, 31, 2, 33, 34, 35, 6, 37, 38, 39, 5, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, 51, 26, 53, 2, 55, 7, 57, 58, 59, 30, 61, 62, 21, 1, 65, 66, 67, 34, 69, 70, 71, 3, 73, 74, 15, 38, 77

Offset: 1

Henry Bottomley, Jun 25 2000

			a(32) = 2 because cubefree part of 32 is 4 and powerfree kernel of 4 is 2.

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

Cf. A000189, A000578, A007947, A008834, A013664, A019555, A048798, A050985, A053149, A053150, A056551.

f[p_, e_] :=  p^If[Divisible[e, 3], 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]/3), f[k,2] = 1, f[k,2] = 0)); factorback(f); \\ Michel Marcus, Feb 28 2019

a(n) = A007947(A050985(n)) = A019555(A050985(n)) = n/(A053150(n)*A000189(n)) = A019555(n)/A053150(n) = A056551(n)^(1/3).
If n = Product_{j} Pj^Ej then a(n) = Product_{j} Pj^Fj, where Fj = 0 if Ej is 0 or a multiple of 3 and Fj = 1 otherwise.
Multiplicative with a(p^e) = p^(if 3|e, then 0, else 1). - Mitch Harris, Apr 19 2005
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(6)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4) = 0.3480772773... . - Amiram Eldar, Oct 28 2022
Dirichlet g.f.: zeta(3*s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-1)). - Amiram Eldar, Sep 16 2023

A333843 Expansion of g.f.: Sum_{k>=1} k * x^(k^3) / (1 - x^(k^3)).

Original entry on oeis.org

1, 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, 4, 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, 4, 1, 3, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 4, 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, 1, 4

Offset: 1

Ilya Gutkovskiy, Apr 07 2020

Sum of cube roots of cube divisors of n.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Atul Dixit, Bibekananda Maji, and Akshaa Vatwani, Voronoi summation formula for the generalized divisor function sigma_z^k(n), arXiv:2303.09937 [math.NT], 2023, sigma(z=3,k=3,n).

Cf. A000203, A053150, A061704, A069290, A113061, A333844.

nmax = 108; CoefficientList[Series[Sum[k x^(k^3)/(1 - x^(k^3)), {k, 1, Floor[nmax^(1/3)] + 1}], {x, 0, nmax}], x] // Rest
Table[DivisorSum[n, #^(1/3) &, IntegerQ[#^(1/3)] &], {n, 108}]
f[p_, e_] := (p^(Floor[e/3] + 1) - 1)/(p - 1); a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Dec 01 2020 *)

a(n) = {my(f = factor(n)); prod(i=1, #f~, (f[i,1]^(f[i,2]\3 + 1) - 1)/(f[i,1] - 1));} \\ Amiram Eldar, Sep 05 2023

Dirichlet g.f.: zeta(s) * zeta(3*s-1).
If n = Product (p_j^k_j) then a(n) = Product ((p_j^(floor(k_j/3) + 1) - 1)/(p_j - 1)).
Sum_{k=1..n} a(k) ~ Pi^2*n/6 + zeta(2/3)*n^(2/3)/2. - Vaclav Kotesovec, Dec 01 2020
a(n) = A000203(A053150(n)) (the sum of divisors of the cube root of largest cube dividing n). - Amiram Eldar, Sep 05 2023

A056551 Smallest cube divisible by n divided by largest cube which divides n.

Original entry on oeis.org

1, 8, 27, 8, 125, 216, 343, 1, 27, 1000, 1331, 216, 2197, 2744, 3375, 8, 4913, 216, 6859, 1000, 9261, 10648, 12167, 27, 125, 17576, 1, 2744, 24389, 27000, 29791, 8, 35937, 39304, 42875, 216, 50653, 54872, 59319, 125, 68921, 74088, 79507, 10648

Offset: 1

Henry Bottomley, Jun 25 2000

			a(16) = 8 since smallest cube divisible by 16 is 64 and smallest cube which divides 16 is 8 and 64/8 = 8.

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

Cf. A000189, A000578, A008834, A019555, A048798, A050985, A053149, A053150, A056552.

Cf. A002117, A013670, A330596.

f[p_, e_] := p^If[Divisible[e, 3], 0, 1]; a[n_] := (Times @@ (f @@@ FactorInteger[ n]))^3; Array[a, 100] (* Amiram Eldar, Aug 29 2019*)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%3, f[i,1], 1))^3; } \\ Amiram Eldar, Oct 28 2022

a(n) = A053149(n)/A008834(n) = A048798(n)*A050985(n) = A056552(n)^3.
From Amiram Eldar, Oct 28 2022: (Start)
Multiplicative with a(p^e) = 1 if e is divisible by 3, and a(p^e) = p^3 otherwise.
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(12)/(4*zeta(3))) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = A013670 * A330596 / (4*A002117) = 0.1557163105... . (End)
Dirichlet g.f.: zeta(3*s) * Product_{p prime} (1 + 1/p^(s-3) + 1/p^(2*s-3)). - Amiram Eldar, Sep 16 2023

A248763 Greatest k such that k^3 divides n!

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 2, 4, 12, 12, 12, 24, 24, 24, 360, 1440, 1440, 1440, 1440, 2880, 60480, 60480, 60480, 120960, 604800, 604800, 1814400, 3628800, 3628800, 3628800, 3628800, 14515200, 479001600, 479001600, 479001600, 958003200, 958003200, 958003200

Offset: 1

Clark Kimberling, Oct 14 2014

Every term divides all its successors.

			a(4) = 2 because 2^3 divides 24 and if k > 2, then k^3 > 8 does not divide 24.

Clark Kimberling, Table of n, a(n) for n = 1..1000
Rafael Jakimczuk, On the h-th free part of the factorial, International Mathematical Forum, Vol. 12, No. 13 (2017), pp. 629-634.

Cf. A000142, A053150, A145642, A248762.

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 = 3; Table[p[m, n], {n, 1, z}]  (* A248762 *)
Table[p[m, n]^(1/m), {n, 1, z}]   (* A248763 *)
Table[n!/p[m, n], {n, 1, z}]      (* A145642 *)
f[p_, e_] := p^Floor[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 40] (* Amiram Eldar, Sep 01 2024 *)

a(n) = {my(f = factor(n!)); prod(i = 1, #f~, f[i, 1]^(f[i, 2]\3));} \\ Amiram Eldar, Sep 01 2024

From Amiram Eldar, Sep 01 2024: (Start)
a(n) = A053150(n!).
a(n) = (n! / A145642(n))^(1/3) = A248762(n)^(1/3).
log(a(n)) = (1/3)*n*log(n) - (log(3)+1)*n/3 + o(n) (Jakimczuk, 2017). (End)

A295657 Multiplicative with a(p^e) = p^floor((e-1)/2).

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Nov 28 2017

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A000188, A003557, A004526, A020639, A028234, A067029, A295658, A295659.
Cf. A004709 (positions of ones), A082020, A212793.
Cf. also A008834, A053150, A061704.

Array[Apply[Times, FactorInteger[#] /. {p_, e_} /; p > 0 :> p^Floor[(e - 1)/2]] &, 105] (* Michael De Vlieger, Nov 28 2017 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^floor((f[i,2]-1)/2));} \\ Amiram Eldar, Nov 30 2022

a(1) = 1; for n > 1, a(n) = A020639(n)^A004526(A067029(n)-1) * a(A028234(n)).
a(n) = A000188(A003557(n)).
a(n) = 1 iff A212793(n) = 1.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 15/Pi^2 = 1.519817... (A082020). - Amiram Eldar, Nov 30 2022

A334215 T(n, k) is the greatest positive integer m such that m^k divides n; square array T(n, k), n, k > 0 read by antidiagonals downwards.

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 2, 5, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 2, 9, 1, 1, 1, 1, 1, 1, 1, 2, 3, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Rémy Sigrist, Apr 19 2020

			Square array starts:
  n\k|   1  2  3  4  5  6  7  8  9 10
  ---+-------------------------------
    1|   1  1  1  1  1  1  1  1  1  1
    2|   2  1  1  1  1  1  1  1  1  1
    3|   3  1  1  1  1  1  1  1  1  1
    4|   4  2  1  1  1  1  1  1  1  1
    5|   5  1  1  1  1  1  1  1  1  1
    6|   6  1  1  1  1  1  1  1  1  1
    7|   7  1  1  1  1  1  1  1  1  1
    8|   8  2  2  1  1  1  1  1  1  1
    9|   9  3  1  1  1  1  1  1  1  1
   10|  10  1  1  1  1  1  1  1  1  1
   11|  11  1  1  1  1  1  1  1  1  1
   12|  12  2  1  1  1  1  1  1  1  1
   13|  13  1  1  1  1  1  1  1  1  1
   14|  14  1  1  1  1  1  1  1  1  1
   15|  15  1  1  1  1  1  1  1  1  1
   16|  16  4  2  2  1  1  1  1  1  1

Cf. A000188, A051903, A053150, A053164, A060175, A261969, A334215, A334216.

T(n,k) = { my (f=factor(n)); prod (i=1, #f~, f[i,1]^(f[i,2]\k)) }

T(n, 1) = n.
T(n, 2) = A000188(n).
T(n, 3) = A053150(n).
T(n, 4) = A053164(n).
T(n, A051903(n)) = A261969(n).
T(n, k) = 1 for any k > A051903(n).
T(n^k, k) = n.

A355265 Bicubeful numbers.

Original entry on oeis.org

64, 128, 192, 256, 320, 384, 448, 512, 576, 640, 704, 729, 768, 832, 896, 960, 1024, 1088, 1152, 1216, 1280, 1344, 1408, 1458, 1472, 1536, 1600, 1664, 1728, 1792, 1856, 1920, 1984, 2048, 2112, 2176, 2187, 2240, 2304, 2368, 2432, 2496, 2560, 2624, 2688, 2752

Offset: 1

Peter Luschny, Jul 12 2022

nonn

Let lp(n, e) denote the largest positive integer b such that b^e divides n. For example for e = 1, 2, 3, 4 the sequences (lp(n, e), n >= 1) are A000027, A000188, A053150, and A053164. Let rad(n) = A007947(n) be the squarefree kernel of n. k is in this sequence if lp(n, 3) does not divide rad(n). The case e = 1 gives A013929, and the case e = 2 is A046101.

The asymptotic density of this sequence is 1 - 1/zeta(6) = 1 - 945/Pi^6 = 0.017047... . - Amiram Eldar, Jul 13 2022

			n = 512 = 2^9, rad(n) = 2, lp(n, 3) = 8 since n/8^3 = 1. But 8 does not divide 2.
n = 704 = 2^6*11, rad(n) = 22, lp(n, 3) = 4 since n/4^3 = 11. But 4 does not divide 22.

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

Cf. A007947, A000188, A053150, A053164, A013929, A046101 (biquadrateful).

Cf. A013664, A343359.

with(NumberTheory):
isBicubeful := n -> irem(Radical(n), LargestNthPower(n, 3)) <> 0:
select(isBicubeful, [`$`(1..2752)]);

bicubQ[n_] := AnyTrue[FactorInteger[n][[;; , 2]], # > 5 &]; Select[Range[3000], bicubQ] (* Amiram Eldar, Jul 13 2022 *)

from itertools import count, islice
from sympy import factorint
def A355265_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:any(map(lambda m:m>5,factorint(n).values())),count(max(startvalue,1)))
A355265_list = list(islice(A355265_gen(),30)) # Chai Wah Wu, Jul 12 2022

A number k is bicubeful iff it is divisible by the 6th power of an integer > 1.

A365487 The number of divisors of the largest cube dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 4, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 4, 1, 4, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 4, 1, 1, 1, 1, 1, 1

Offset: 1

Amiram Eldar, Sep 05 2023

The number of divisors of the cube root of the largest cube dividing n, A053150(n), is A061704(n).

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

Cf. A000005, A000578, A004709, A008834, A053150, A061704.

f[p_, e_] := 3*Floor[e/3] + 1; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100]

a(n) = vecprod(apply(x -> 3*(x\3) + 1, factor(n)[, 2]));

a(n) = A000005(A008834(n)).
Multiplicative with a(p^e) = 3*floor(e/3) + 1.
a(n) = 1 if and only if n is cubefree (A004709).
a(n) <= A000005(n) with equality if and only if n is a cube (A000578).
Dirichlet g.f.: zeta(s) * zeta(3*s) * Product_{p prime} (1 + 2/p^(3*s)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(3) * Product_{p prime} (1 + 2/p^3) = 1.6552343865608... .

cp's OEIS Frontend

A015052 a(n) is the smallest positive integer m such that m^5 is divisible by n.

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A015053 Smallest positive integer for which n divides a(n)^6.

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A056552 Powerfree kernel of cubefree part of n.

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A333843 Expansion of g.f.: Sum_{k>=1} k * x^(k^3) / (1 - x^(k^3)).

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A056551 Smallest cube divisible by n divided by largest cube which divides n.

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A248763 Greatest k such that k^3 divides n!

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A295657 Multiplicative with a(p^e) = p^floor((e-1)/2).

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