Displaying 1-10 of 21 results found.
a(n) = largest-nth-power(n, 3) * radical(n) = A053150(n) * A007947(n), where the largest-nth-power(n, e) is the largest positive integer b such that b^e divides n.
+20
2
1, 2, 3, 2, 5, 6, 7, 4, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, 21, 22, 23, 12, 5, 26, 9, 14, 29, 30, 31, 4, 33, 34, 35, 6, 37, 38, 39, 20, 41, 42, 43, 22, 15, 46, 47, 12, 7, 10, 51, 26, 53, 18, 55, 28, 57, 58, 59, 30, 61, 62, 21, 8, 65, 66, 67, 34, 69
FORMULA
Multiplicative with a(p^e) = p^(1 + floor(e/3)). - Amiram Eldar, Jul 13 2022 Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(5)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 2/p^5 + 1/p^6) = 0.3643121583... . - Amiram Eldar, Nov 13 2022
MAPLE
with(NumberTheory): seq(LargestNthPower(n, 3)*Radical(n), n=1..69);
MATHEMATICA
f[p_, e_] := p^(1 + Floor[e/3]); a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jul 13 2022 *)
PROG
(Python)
from math import prod
from sympy import factorint
def A355263(n): return prod(p**(e//3+1) for p, e in factorint(n).items()) # Chai Wah Wu, Jul 13 2022
(1) Number of solutions to x^2 == 0 (mod n). (2) Also square root of largest square dividing n. (3) Also max_{ d divides n } gcd(d, n/d).
+10
147
1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 2, 5, 1, 3, 2, 1, 1, 1, 4, 1, 1, 1, 6, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 4, 7, 5, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 3, 8, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 5, 2, 1, 1, 1, 4, 9, 1, 1, 2, 1, 1, 1, 2, 1, 3
COMMENTS
Labos Elemer and Henry Bottomley independently proved that (2) and (3) define the same sequence. Bottomley also showed that (1) and (2) define the same sequence.
Proof that (2) = (3): Let max{gcd(d, n/d)} = K, then d = Kx, n/d = Ky so n = KKxy where xy is the squarefree part of n, otherwise K is not maximal. Observe also that g = gcd(K, xy) is not necessarily 1. Thus K is also the "maximal square-root factor" of n. - Labos Elemer, July 2000 We can write sqrt(n) = b*sqrt(c) where c is squarefree. Then b = A000188(n) is the "inner square root" of n, c = A007913(n), lcm(b,c) = A007947(n) = "squarefree kernel" of n and bc = A019554(n) = "outer square root" of n. [The relation with LCM is wrong if b is not squarefree. One must, e.g., replace b with A007947(b). - M. F. Hasler, Mar 03 2018]
LINKS
Andrew Reiter, On (mod n) spirals (2014), and posting to Number Theory Mailing List, Mar 23 2014.
FORMULA
a(n) = Sum_{d^2|n} phi(d), where phi is the Euler totient function A000010. Dirichlet series: Sum_{n >= 1} a(n)/n^s = zeta(2*s - 1)*zeta(s)/zeta(2*s), (Re(s) > 1).
Sum_{n>=1} lambda(n)*a(n)*x^n/(1-x^n) = Sum_{n>=1} n*x^(n^2), where lambda() is the Liouville function A008836 (cf. A205801). - Mamuka Jibladze, Feb 15 2015 a(2*n) = a(n)*( A096268(n-1) + 1). - observed by Velin Yanev, Jul 14 2017, The formula says that a(2n) = 2*a(n) only when 2-adic valuation of n ( A007814(n)) is odd, otherwise a(2n) = a(n). This follows easily from the definition (2). - Antti Karttunen, Nov 28 2017 Sum_{k=1..n} a(k) ~ 3*n*((log(n) + 3*gamma - 1)/Pi^2 - 12*zeta'(2)/Pi^4), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Dec 01 2020 G.f.: Sum_{k>=1} phi(k) * x^(k^2) / (1 - x^(k^2)). - Ilya Gutkovskiy, Aug 20 2021
EXAMPLE
a(8) = 2 because the largest square dividing 8 is 4, the square root of which is 2.
a(9) = 3 because 9 is a perfect square and its square root is 3.
a(10) = 1 because 10 is squarefree.
MAPLE
with(numtheory): A000188 := proc(n) local i: RETURN(op(mul(i, i=map(x->x[1]^floor(x[2]/2), ifactors(n)[2])))); end;
MATHEMATICA
Array[Function[n, Count[Array[PowerMod[#, 2, n ] &, n, 0 ], 0 ] ], 100]
(* Second program: *)
nMax = 90; sList = Range[Floor[Sqrt[nMax]]]^2; Sqrt[#] &/@ Table[ Last[ Select[ sList, Divisible[n, #] &]], {n, nMax}] (* Harvey P. Dale, May 11 2011 *) a[n_] := With[{d = Divisors[n]}, Max[GCD[d, Reverse[d]]]] (* Mamuka Jibladze, Feb 15 2015 *) f[p_, e_] := p^Floor[e/2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)
PROG
(PARI) a(n)=if(n<1, 0, sum(i=1, n, i*i%n==0))
(PARI) a(n)=sqrtint(n/core(n)) \\ Zak Seidov, Apr 07 2009 (Haskell)
a000188 n = product $ zipWith (^)
(a027748_row n) $ map (`div` 2) (a124010_row n)
(Python)
from sympy.ntheory.factor_ import core
from sympy import integer_nthroot
Smallest number whose square is divisible by n.
+10
40
1, 2, 3, 2, 5, 6, 7, 4, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, 21, 22, 23, 12, 5, 26, 9, 14, 29, 30, 31, 8, 33, 34, 35, 6, 37, 38, 39, 20, 41, 42, 43, 22, 15, 46, 47, 12, 7, 10, 51, 26, 53, 18, 55, 28, 57, 58, 59, 30, 61, 62, 21, 8, 65, 66, 67, 34, 69, 70, 71, 12, 73, 74, 15, 38, 77
COMMENTS
A note on square roots of numbers: we can write sqrt(n) = b*sqrt(c) where c is squarefree. Then b = A000188(n) is the "inner square root" of n, c = A007913(n), lcm(b,c) = A007947(n) = "squarefree kernel" of n and bc = A019554(n) = "outer square root" of n. [The relation with LCM is wrong if b is not squarefree. One must, e.g., replace b with A007947(b). - M. F. Hasler, Mar 03 2018] Instead of the terms "inner square root" and "outer square root", we may use the terms "lower square root" and "upper square root", respectively. Upper k-th roots have been studied by Broughan (2002, 2003, 2006). - Petros Hadjicostas, Sep 15 2019 The number of times each number k appears in this sequence is A034444(k). The first time k appears is at position A102631(k). - N. J. A. Sloane, Jul 28 2021
FORMULA
Replace any square factors in n by their square roots.
Multiplicative with a(p^e) = p^ceiling(e/2).
Dirichlet series:
Sum_{n>=1} a(n)/n^s = zeta(2*s-1)*zeta(s-1)/zeta(2*s-2), (Re(s) > 2);
Sum_{n>=1} (1/a(n))/n^s = zeta(2*s+1)*zeta(s+1)/zeta(2*s+2), (Re(s) > 0).
Sum_{k=1..n} 1/a(k) ~ 3*log(n)^2/(2*Pi^2) + (9*gamma/Pi^2 - 36*zeta'(2)/Pi^4)*log(n) + 6*gamma^2/Pi^2 - 108*gamma*zeta'(2)/Pi^4 + 432*zeta'(2)^2/Pi^6 - 36*zeta''(2)/Pi^4 - 15*sg1/Pi^2, where gamma is the Euler-Mascheroni constant A001620 and sg1 is the first Stieltjes constant (see A082633). - Vaclav Kotesovec, Jul 27 2021
MAPLE
with(numtheory): A019554 := proc(n) local i: RETURN(op(mul(i, i=map(x->x[1]^ceil(x[2]/2), ifactors(n)[2])))); end;
MATHEMATICA
Flatten[Table[Select[Range[n], Divisible[#^2, n]&, 1], {n, 100}]] (* Harvey P. Dale, Oct 17 2011 *) f[p_, e_] := p^Ceiling[e/2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2020 *)
PROG
(Haskell)
a019554 n = product $ zipWith (^)
(a027748_row n) (map ((`div` 2) . (+ 1)) $ a124010_row n)
(Python 3.8+)
from math import prod
from sympy import factorint
def A019554(n): return n//prod(p**(q//2) for p, q in factorint(n).items()) # Chai Wah Wu, Aug 18 2021
4th root of largest 4th power dividing n.
+10
23
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, 2, 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, 2, 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, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2
COMMENTS
Multiplicative with a(p^e) = p^[e/4]. - Mitch Harris, Apr 19 2005
FORMULA
Multiplicative with a(p^e) = p^[e/4].
Dirichlet g.f.: zeta(4s-1)*zeta(s)/zeta(4s). - R. J. Mathar, Apr 09 2011 Sum_{k=1..n} a(k) ~ 90*zeta(3)*n/Pi^4 + 3*zeta(1/2)*sqrt(n)/Pi^2. - Vaclav Kotesovec, Dec 01 2020 G.f.: Sum_{k>=1} phi(k) * x^(k^4) / (1 - x^(k^4)). - Ilya Gutkovskiy, Aug 20 2021
EXAMPLE
a(32) = 2 since 2 = 16^(1/4) and 16 is the largest 4th power dividing 32.
MAPLE
A053164 := proc(n) local a, f, e, p ; for f in ifactors(n)[2] do e:= op(2, f) ; p := op(1, f) ; a := a*p^floor(e/4) ; end do ; a ; end proc: # R. J. Mathar, Jan 11 2012
MATHEMATICA
f[list_] := list[[1]]^Quotient[list[[2]], 4]; Table[Apply[Times, Map[f, FactorInteger[n]]], {n, 1, 81}] (* Geoffrey Critzer, Jan 21 2015 *)
PROG
(Scheme, with memoization macro definec)
(define ( A002265 n) (floor->exact (/ n 4))) ;; For MIT/GNU Scheme
Number of cubes dividing n.
+10
23
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, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1
FORMULA
Multiplicative with a(p^e) = floor(e/3) + 1. - Mitch Harris, Apr 19 2005 G.f.: Sum_{n>=1} x^(n^3)/(1-x^(n^3)). - Joerg Arndt, Jan 30 2011 Sum_{k=1..n} a(k) ~ zeta(3)*n + zeta(1/3)*n^(1/3). - Vaclav Kotesovec, Dec 01 2020 a(n) = Sum_{k=1..n} (1 - ceiling(n/k^3) + floor(n/k^3)). - Wesley Ivan Hurt, Jan 28 2021
EXAMPLE
a(128) = 3 since 128 is divisible by 1^3 = 1, 2^3 = 8 and 4^3 = 64.
MAPLE
N:= 1000: # to get a(1)..a(N)
G:= add(x^(n^3)/(1-x^(n^3)), n=1..floor(N^(1/3))):
S:= series(G, x, N+1):
# alternative
local a, pe ;
a := 1 ;
for pe in ifactors(n)[2] do
op(2, pe) ;
a := a*(1+floor(%/3)) ;
end do:
a ;
end proc:
MATHEMATICA
nn = 100; f[list_, i_]:= list[[i]]; Table[ DirichletConvolve[ f[ Boole[ Map[ IntegerQ[#] &, Map[#^(1/3) &, Range[nn]]]], n], f[Table[1, {nn}], n], n, m], {m, 1, nn}] (* Geoffrey Critzer, Feb 07 2015 *) Table[DivisorSum[n, 1 &, IntegerQ[#^(1/3)] &], {n, 105}] (* Michael De Vlieger, Jul 28 2017 *) f[p_, e_] := 1 + Floor[e/3]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 15 2020 *)
PROG
(PARI) a(n) = sumdiv(n, d, ispower(d, 3)); \\ Michel Marcus, Jan 31 2015
1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 27, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 27, 1, 8, 1, 1, 1, 1, 1, 1, 1, 64, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 27
LINKS
Eric Weisstein's World of Mathematics, Cubic Part.
FORMULA
Multiplicative with a(p^e) = p^(3[e/3]). - Mitch Harris, Apr 19 2005 Dirichlet g.f.: zeta(s)*zeta(3s-3)/zeta(3s). The Dirichlet convolution of this sequence with A050985 generates A000203. - R. J. Mathar, Apr 05 2011 Sum_{k=1..n} a(k) ~ 45 * zeta(4/3) * n^(4/3) / (2*Pi^4). - Vaclav Kotesovec, Jan 31 2019
MAPLE
with(numtheory): [ seq( expand(nthpow(i, 3)), i=1..200) ];
# alternative:
local p;
a := 1 ;
for p in ifactors(n)[2] do
e := floor(op(2, p)/3) ;
a := a*op(1, p)^(3*e) ;
end do:
a ;
end proc:
MATHEMATICA
a[n_] := Times @@ (#[[1]]^(#[[2]] - Mod[#[[2]], 3]) & ) /@ FactorInteger[n]; Table[a[n], {n, 1, 81}]
upto=1000; Flatten[With[{c=Range[Floor[Surd[upto, 3]], 1, -1]^3}, Table[ Select[ c, Divisible[n, #]&, 1], {n, upto}]]](* Harvey P. Dale, Apr 07 2013 *)
PROG
(Python)
from math import prod
from sympy import factorint
def A008834(n): return prod(p**(e-e%3) for p, e in factorint(n).items()) # Chai Wah Wu, Aug 08 2024
Smallest number whose cube is divisible by n.
+10
16
1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 4, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 12, 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
COMMENTS
This can be thought as an "upper 3rd root" of a positive integer. Upper k-th roots were studied by Broughan (2002, 2003, 2006). The sequence of "lower 3rd root" of positive integers is given by A053150. - Petros Hadjicostas, Sep 15 2019
FORMULA
Replace any cubic factors in n by their cube roots.
Multiplicative with a(p^e) = p^ceiling(e/3). - R. J. Mathar, May 29 2011 Dirichlet g.f.: zeta(3*s-1) * Product_{p prime} (1 + p^(1 - s) + p^(1 - 2*s)).
Dirichlet g.f.: zeta(3*s-1) * zeta(s-1) * Product_{p prime} (1 - p^(2 - 3*s) + p^(1 - 2*s) - p^(2 - 2*s)).
Sum_{k=1..n} a(k) ~ c * zeta(5) * n^2 / 2, where c = Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4) = 0.684286924186862318141968725791218083472312736723163777284618226290055... (End)
MAPLE
f:= n -> mul(t[1]^ceil(t[2]/3), t = ifactors(n)[2]):
MATHEMATICA
cubes=Range[85]^3; Table[Position[Divisible[cubes, i], True, 1, 1][[1, 1]], {i, 85}] (* Harvey P. Dale, Jan 12 2011 *) f[p_, e_] := p^Ceiling[e/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jan 06 2024 *)
PROG
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + p*X + p*X^2)/(1 - p*X^3))[n], ", ")) \\ Vaclav Kotesovec, Aug 30 2021 (PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i, 1]^ceil(f[i, 2]/3)); } \\ Amiram Eldar, Jan 06 2024 (Sage) [prod([t[0]^(ceil(t[1]/3)) for t in factor(n)]) for n in range(1, 79)] # Danny Rorabaugh, Sep 22 2015 (Python 3.8+)
from math import prod
from sympy import factorint
def A019555(n): return prod(p**((q%3 != 0)+(q//3)) for p, q in factorint(n).items()) # Chai Wah Wu, Aug 18 2021
Smallest positive integer for which n divides a(n)^4.
+10
15
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, 4, 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
COMMENTS
According to Broughan (2002, 2003, 2006), a(n) is the "upper 4th root of n". The "lower 4th root of n" is sequence A053164. - Petros Hadjicostas, Sep 15 2019
FORMULA
If n is 5th-power-free (i.e., not 32, 64, 128, 243, ...) then a(n) = A007947(n). Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(7)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6) = 0.3528057925... . - Amiram Eldar, Oct 27 2022
MATHEMATICA
f[p_, e_] := p^Ceiling[e/4]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 08 2020 *)
PROG
(PARI) a(n) = my(f=factor(n)); for (i=1, #f~, f[i, 2] = ceil(f[i, 2]/4)); factorback(f); \\ Michel Marcus, Jun 09 2014
Number of exponents larger than 2 in the prime factorization of n.
+10
13
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
FORMULA
Additive with a(p^e) = 1 if e>2, 0 otherwise.
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
EXAMPLE
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.
MATHEMATICA
Array[Count[FactorInteger[#][[All, -1]], _?(# > 2 &)] &, 105] (* Michael De Vlieger, Nov 28 2017 *)
PROG
(PARI) a(n) = { my(v = factor(n)[, 2], i=0); for(x=1, length(v), if(v[x]>2, i++)); i; } \\ Iain Fox, Nov 29 2017
CROSSREFS
Cf. A001221, A003557, A053150, A056169, A056170, A085541, A162642, A295657, A295662, A295883, A295884.
n divided by largest cubefree factor of n.
+10
10
1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 9, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 2
FORMULA
Multiplicative with a(p^e) = p^max(e-2, 0). - Amiram Eldar, Sep 07 2020 Dirichlet g.f.: zeta(s-1) * Product_{p prime} (1 - 1/p^(s-1) + 1/p^s - 1/p^(2*s-1) + 1/p^(2*s)). - Amiram Eldar, Dec 07 2023
MATHEMATICA
f[p_, e_] := p^Max[e-2, 0]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 07 2020 *)
CROSSREFS
Cf. A000189, A000578, A007948, A008834, A019555, A048798, A050985, A053149, A053150, A056551, A056552. See A003557 for squares and A062379 for 4th powers. Differs from A073753 for the first time at n=90, where a(90) = 1, while A073753(90) = 3.
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