A013776 -id:A013776 - cp's OEIS frontend

A004171 a(n) = 2^(2n+1).

2, 8, 32, 128, 512, 2048, 8192, 32768, 131072, 524288, 2097152, 8388608, 33554432, 134217728, 536870912, 2147483648, 8589934592, 34359738368, 137438953472, 549755813888, 2199023255552, 8796093022208, 35184372088832, 140737488355328, 562949953421312

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(2, 8), L(2, 8), P(2, 8), T(2, 8). See A008776 for definitions of Pisot sequences.
In the Chebyshev polynomial of degree 2n, a(n) is the coefficient of x^2n. - Benoit Cloitre, Mar 13 2002
1/2 - 1/8 + 1/32 - 1/128 + ... = 2/5. - Gary W. Adamson, Mar 03 2009
From Adi Dani, May 15 2011: (Start)
Number of ways of placing an even number of indistinguishable objects in n + 1 distinguishable boxes with at most 3 objects in box.
Number of compositions of even natural numbers into n + 1 parts less than or equal to 3 (0 is counted as part). (End)
Also the number of maximal cliques in the (n+1)-Sierpinski tetrahedron graph for n > 0. - Eric W. Weisstein, Dec 01 2017
Assuming the Collatz conjecture is true, any starting number eventually leads to a power of 2. A number in this sequence can never be the first power of 2 in a Collatz sequence except of course for the Collatz sequence starting with that number. For example, except for 8, 4, 2, 1, any Collatz sequence that includes 8 must also include 16 (e.g., 5, 16, 8, 4, 2, 1). - Alonso del Arte, Oct 01 2019
First differences of A020988, and thus the "wavelengths" of the local maxima in A020986. See the Brillhart and Morton link, pp. 855-856. - John Keith, Mar 04 2021

			G.f. = 2 + 8*x + 32*x^2 + 128*x^3 + 512*x^4 + 2048*x^5 + 8192*x^6 + 32768*x^7 + ...
From _Adi Dani_, May 15 2011: (Start)
a(1) = 8 because all compositions of even natural numbers into 2 parts less than or equal to 3 are:
  for 0: (0, 0)
  for 2: (0, 2), (2, 0), (1, 1)
  for 4: (1, 3), (3, 1), (2, 2)
  for 6: (3, 3).
a(2) = 32 because all compositions of even natural numbers into 3 parts less than or equal to 3 are:
  for 0: (0, 0, 0)
  for 2: (0, 0, 2), (0, 2, 0), (2, 0, 0), (0, 1, 1), (1, 0, 1) , (1, 1, 0)
  for 4: (0, 1, 3), (0, 3, 1), (1, 0, 3), (1, 3, 0), (3, 0, 1), (3, 1, 0), (0, 2, 2), (2, 0, 2), (2, 2, 0), (1, 1, 2), (1, 2, 1), (2, 1, 1)
  for 6: (0, 3, 3), (3, 0, 3), (3, 3, 0), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1), (2, 2, 2)
  for 8: (2, 3, 3), (3, 2, 3), (3, 3, 2).
(End)

Adi Dani, Quasicompositions of natural numbers, Proceedings of III congress of mathematicians of Macedonia, Struga Macedonia 29 IX -2 X 2005 pages 225-238.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
J. Brillhart and P. Morton, A case study in mathematical research: the Golay-Rudin-Shapiro sequence, Amer. Math. Monthly, 103 (1996) 854-869.
Ling Gao, Graph assembly for spider and tadpole graphs, Master's Thesis, Cal. State Poly. Univ. (2023).
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Tanya Khovanova, Recursive Sequences.
Mitchell Paukner, Lucy Pepin, Manda Riehl, and Jarred Wieser, Pattern Avoidance in Task-Precedence Posets, arXiv:1511.00080 [math.CO], 2015-2016.
Eric Weisstein's World of Mathematics, Maximal Clique.
Eric Weisstein's World of Mathematics, Sierpinski Tetrahedron Graph.
Index entries for linear recurrences with constant coefficients, signature (4).
Index to divisibility sequences.

Cf. A013708-A013729.
Absolute value of A009117. Essentially the same as A081294.
Cf. A132020, A164632. Equals A000980(n) + 2*A181765(n). Cf. A013776.

List([0..30],n->2^(2*n+1)); # Muniru A Asiru, Mar 12 2019

a004171 = (* 2) . a000302
a004171_list = iterate (* 4) 2  -- Reinhard Zumkeller, Jan 09 2013

[2^(2*n+1): n in [0..30]]; // Vincenzo Librandi, May 16 2011

seq(2^(2*n+1),n=0..24); # Nathaniel Johnston, Jun 25 2011

Table[2^(2 n + 1), {n, 0, 24}]
2^(2 Range[20] - 1) (* Eric W. Weisstein, Dec 01 2017 *)
LinearRecurrence[{4}, {2}, 20] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[2/(1 - 4 x), {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)

a(n)=2<<(2*n) \\ Charles R Greathouse IV, Apr 07 2012

a(n) = 2^(2*n+1) \\ Michel Marcus, Aug 12 2014

[2**(2*n+1) for n in range(0,25)] # Stefano Spezia, Jul 23 2025

((List.fill(20)(4: BigInt)).scanLeft(1: BigInt)( * )).map(2 * ) // _Alonso del Arte, Sep 12 2019

a(n) = 2*4^n.
a(n) = 4*a(n-1).
1 = 1/2 + Sum_{n >= 1} 3/a(n) = 3/6 + 3/8 + 3/32 + 3/128 + 3/512 + 3/2048 + ...; with partial sums: 1/2, 31/32, 127/128, 511/512, 2047/2048, ... - Gary W. Adamson, Jun 16 2003
From Philippe Deléham, Nov 23 2008: (Start)
a(n) = 2*A000302(n).
G.f.: 2/(1-4*x). (End)
a(n) = A081294(n+1) = A028403(n+1) - A000079(n+1) for n >= 1. a(n-1) = A028403(n) - A000079(n). - Jaroslav Krizek, Jul 27 2009
E.g.f.: 2*exp(4*x). - Ilya Gutkovskiy, Nov 01 2016
a(n) = A002063(n)/3 - A000302(n). - Zhandos Mambetaliyev, Nov 19 2016
a(n) = Sum_{k = 0..2*n} (-1)^(k+n)*binomial(4*n + 2, 2*k + 1); a(2*n) = Sum_{k = 0..2*n} binomial(4*n + 2, 2*k + 1) = A013776(n). - Peter Bala, Nov 25 2016
Product_{n>=0} (1 - 1/a(n)) = A132020. - Amiram Eldar, May 08 2023

A078164 Numbers k such that phi(k) is a perfect biquadrate.

Original entry on oeis.org

1, 2, 17, 32, 34, 40, 48, 60, 257, 512, 514, 544, 640, 680, 768, 816, 960, 1020, 1297, 1387, 1417, 1729, 1971, 2109, 2223, 2289, 2331, 2445, 2457, 2565, 2594, 2608, 2774, 2812, 2834, 2835, 3052, 3260, 3458, 3888, 3912, 3924, 3942, 3996, 4104, 4212, 4218

Offset: 1

Labos Elemer, Nov 27 2002

nonn

Corresponding values of phi include 1, 16, 256, 1296, 4096, ... and these arise several times each.
a(3) = A053576(4).
A013776 is a subsequence since phi(2^(4*n+1)) = (2^n)^4. - Bernard Schott, Sep 22 2022
Subsequence of primes is A037896 since in this case: phi(k^4+1) = k^4. - Bernard Schott, Mar 05 2023

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

Subsequence of A039770. A037896 is a subsequence.
Sequences where phi(k) is a perfect power: A039770 (square), A039771 (cube), this sequence (4th), A078165 (5th), A078166 (6th), A078167 (7th), A078168 (8th), A078169 (9th), A078170 (10th).
Cf. A001317, A053576, A037896, A045544, A000010, A013776.

k=4; Do[s=EulerPhi[n]^(1/k); If[IntegerQ[s], Print[n]], {n, 1, 5000}]
Select[Range[5000],IntegerQ[Surd[EulerPhi[#],4]]&] (* Harvey P. Dale, Apr 30 2015 *)

is(n)=ispower(eulerphi(n),4) \\ Charles R Greathouse IV, Apr 24 2020

from itertools import count, islice
from sympy import totient, integer_nthroot
def A078164_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:integer_nthroot(totient(n),4)[1], count(max(1,startvalue)))
A078164_list = list(islice(A078164_gen(),20)) # Chai Wah Wu, Feb 28 2023

A144864 a(n) = (4*16^(n-1)-1)/3.

Original entry on oeis.org

1, 21, 341, 5461, 87381, 1398101, 22369621, 357913941, 5726623061, 91625968981, 1466015503701, 23456248059221, 375299968947541, 6004799503160661, 96076792050570581, 1537228672809129301, 24595658764946068821, 393530540239137101141, 6296488643826193618261, 100743818301219097892181

Offset: 1

Artur Jasinski, Sep 23 2008

Old name was: A144863, read as binary numbers, converted to base 10.
All numbers in this sequence for n>1 are congruent to 5 mod 16. - Artur Jasinski, Sep 25 2008
From Omar E. Pol, Sep 10 2011: (Start)
It appears that this is a bisection of A002450.
It appears that this is a bisection of A084241.
It appears that this is a bisection of A153497.
It appears that this is a bisection of A088556, if n>=2.
(End)
All of the above is trivially true. - Joerg Arndt, Aug 19 2014
The aerated sequence (b(n))n>=1 = [1, 0, 21, 0, 341, 0, 5461, 0, 87381, ...] is a fourth-order linear divisibility sequence; that is, a(n) divides a(m) whenever n divides m. It is the case P1 = 0, P2 = -9, Q = -4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Aug 26 2022

Vincenzo Librandi, Table of n, a(n) for n = 1..500
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A001025, A002450, A013776, A056830, A084241, A088556, A094028, A098704, A131865, A135576, A144863, A153497.

Third quadrisection of Jacobsthal numbers A001045; the other quadrisections are A195156 (first), A139792 (second), and A141060 (fourth).

[16^n/12-1/3: n in [1..20]]; // Vincenzo Librandi, Aug 03 2011

Table[1/3 (-1 + 16^(n - 1)) + 16^(n - 1), {n, 1, 17}] (* Artur Jasinski, Sep 25 2008 *)
LinearRecurrence[{17,-16},{1,21},20] (* Harvey P. Dale, Jun 29 2022 *)

vector(66,n,(4*16^(n-1)-1)/3) \\ Joerg Arndt, Aug 19 2014

a(n) = 16^n/12 - 1/3; a(n) = 16*a(n-1) + 5, a(1)=1. - Artur Jasinski, Sep 25 2008
G.f.: x*(1+4*x) / ( (16*x-1)*(x-1) ). - R. J. Mathar, Jan 06 2011
a(n)=b such that Integral_{x=-Pi/2..Pi/2} (-1)^(n+1)*2^(2*n-3)*(cos((2*n-1)*x))/(5/4+sin(x)) dx = c+b*log(3). - Francesco Daddi, Aug 02 2011
a(n) = (2^(4*n-2)-1)/3. - Klaus Purath, Jan 31 2021
From Jianing Song, Aug 30 2022: (Start)
a(n) = A001045(4*n-2).
a(n+1) - a(n) = 10*A013776(n-1) = 20*A001025(n-1) for n >= 1.
a(n) = 10*A098704(n) + 1 = 20*A131865(n-2) + 1 for n >= 2. (End)
E.g.f.: (exp(16*x) - 4*exp(x) + 3)/12. - Stefano Spezia, Apr 18 2024

New name from Joerg Arndt, Aug 19 2014

A141060 Fourth quadrisection of Jacobsthal numbers A001045: a(n)=16a(n-1)-5.

Original entry on oeis.org

3, 43, 683, 10923, 174763, 2796203, 44739243, 715827883, 11453246123, 183251937963, 2932031007403, 46912496118443, 750599937895083, 12009599006321323, 192153584101141163, 3074457345618258603, 49191317529892137643

Offset: 0

Paul Curtz, Jul 30 2008

Jacobsthal numbers ending with the decimal digit 3. - Jianing Song, Aug 30 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (17,-16).

The other quadrisections of A001045 are A195156 (first), A139792 (second), and A144864 (third).

[(1/3)*(1+8*16^n): n in [0..25]]; // Vincenzo Librandi, May 25 2011

LinearRecurrence[{17,-16},{3,43},30] (* Harvey P. Dale, Mar 16 2015 *)

a(n)=8*16^n\3+1 \\ Charles R Greathouse IV, May 25 2011

def A141060(n): return ((1<<(n<<2)+3)|1)//3 # Chai Wah Wu, Apr 18 2025

a(n) = A139792(n) + A013776(n).
a(n+1) - a(n) = 10*A013709(n) = 40*A001025(n).
G.f.: (3-8*x)/((1-x)*(1-16*x)). [Colin Barker, Apr 05 2012]
a(0)=3, a(1)=43, a(n)=17*a(n-1)-16*a(n-2). - Harvey P. Dale, Mar 16 2015
From Jianing Song, Aug 30 2022: (Start)
a(n) = A001045(4*n+3).
a(n) = 10*A141032(n) + 3 = 20*A098704(n+1) + 1 = 40*A131865(n-1) + 1 for n >= 1. (End)

A264960 Half-convolution of the central binomial coefficients A000984 with itself.

Original entry on oeis.org

1, 2, 10, 32, 146, 512, 2248, 8192, 35218, 131072, 556040, 2097152, 8815496, 33554432, 140107040, 536870912, 2230302098, 8589934592, 35541690568, 137438953472, 566823203656, 2199023255552, 9044910175520, 35184372088832, 144393718191496

Offset: 0

Peter Bala, Nov 29 2015

The half-convolution of a sequence {s(n)}n>=0 with itself is defined by r(n) := Sum_{k = 0..floor(n/2)} s(k)*s(n-k). See A201204.

Muniru A Asiru, Table of n, a(n) for n = 0..1520

Cf. A002894, A013776, A112830.

List([0..24],n->Sum([0..Int(n/2)],k->Binomial(2*k,k)*Binomial(2*n-2*k,n-k))); # Muniru A Asiru, Nov 25 2018

[(&+[Binomial(2*k,k)*Binomial(2*n-2*k, n-k): k in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, Nov 26 2018

A264960:= n-> add(binomial(2*k,k)*binomial(2*n - 2*k, n - k),k = 0..floor(n/2)):
seq(A264960(n),n = 0..24);

a[n_] := Sum[Binomial[2k, k]*Binomial[2n - 2k, n - k], {k, 0, Floor[n/2]}]; Array[a, 30, 0] (* Amiram Eldar, Nov 25 2018 *)

a(n) = sum(k = 0, n\2, binomial(2*k,k)*binomial(2*n - 2*k, n - k)); \\ Michel Marcus, Nov 30 2015

[sum(binomial(2*k,k)*binomial(2*n-2*k, n-k) for k in (0..floor(n/2))) for n in range(30)] # G. C. Greubel, Nov 26 2018

a(n) = Sum_{k = 0..floor(n/2)} binomial(2*k,k)*binomial(2*n - 2*k, n - k).
a(2*n + 1) = 2^(4*n + 1) = A013776(n).
a(2*n) = (1/2)*(binomial(2*n,n)^2 + 16^n) = A112830(2*n,n).
O.g.f.: (1/2)*( 2/Pi*EllipticK(4*x) + 1/(1 - 4*x) ).
E.g.f.: (1/2)*( cosh(4*x) + sinh(4*x) + (BesselI(0,2*x))^2 ).
D-finite with recurrence: - (2*n-3)*n^2*a(n) + 4*(2*n-1)*(n-1)^2*a(n-1) + 16*(2*n-3)*(n-1)^2*a(n-2) - 64*(2*n-1)*(n-2)^2*a(n-3) = 0. - Georg Fischer, Nov 25 2022

A307690 Integers with only one prime factor and whose Euler's totient is a perfect biquadrate.

Original entry on oeis.org

2, 17, 32, 257, 512, 1297, 8192, 65537, 131072, 160001, 331777, 614657, 1336337, 1419857, 2097152, 4477457, 5308417, 8503057, 9834497, 29986577, 33554432, 40960001, 45212177, 59969537, 65610001, 126247697, 193877777, 303595777, 384160001, 406586897, 536870912, 562448657, 655360001

Offset: 1

Bernard Schott, Apr 22 2019

nonn

An integer q is a term iff q = p^(4*m+1), when p is prime of the form k^4 + 1 and m >= 0, then phi(q) = (k * (k^4+1)^m)^4. The primitive terms of this sequence are the primes of the form p = k^4 + 1, which are exactly in A037896.

			a(14) = 1419857 = 17^5 and phi(1419857) = 34^4.

Subsequences: A013776 (2^(4*m+1)), A013806 (17^(4*m+1)), A037896 (primes of the form k^4 + 1).
Intersection of A078164 and A246655.
Cf. A054755 (idem with Euler's totient is square).

[n:n in [1..10000000]| #PrimeDivisors(n) eq 1 and IsPower(EulerPhi(n),4)]; // Marius A. Burtea, May 09 2019

isok(n) = isprimepower(n) && ispower(eulerphi(n), 4); \\ Michel Marcus, Apr 23 2019

A013792 a(n) = 10^(4*n + 1).

Original entry on oeis.org

10, 100000, 1000000000, 10000000000000, 100000000000000000, 1000000000000000000000, 10000000000000000000000000, 100000000000000000000000000000, 1000000000000000000000000000000000, 10000000000000000000000000000000000000, 100000000000000000000000000000000000000000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (10000).

Subsequence of A011557.

Cf. A013715, A013776, A013782, A016813, A098608.

[10^(4*n+1): n in [0..10]]; // Vincenzo Librandi, Jun 28 2011

a(n) = 10000*a(n-1). - Wesley Ivan Hurt, Apr 19 2023
From Elmo R. Oliveira, Aug 30 2024: (Start)
G.f.: 10/(1 - 10000*x).
E.g.f.: 10*exp(10000*x).
a(n) = A011557(A016813(n)) = A098608(n)*A013715(n) = A013776(n)*A013782(n). (End)

A013806 a(n) = 17^(4*n+1).

Original entry on oeis.org

17, 1419857, 118587876497, 9904578032905937, 827240261886336764177, 69091933913008732880827217, 5770627412348402378939569991057, 481968572106750915091411825223071697, 40254497110927943179349807054456171205137

Offset: 0

N. J. A. Sloane

As phi(a(n)) = (2*17^n)^4 is a perfect biquadrate (where phi is the Euler totient A000010), this is a subsequence of A078164 and A307690. - Bernard Schott, Mar 29 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (83521).

Cf. A000010, A013776, A307690.

Intersection of A001026 and A078164.

[17^(4*n+1): n in [0..15]]; // Vincenzo Librandi, Jul 06 2011

17^(4Range[0,10]+1) (* or *) NestList[83521#&,17,20] (* Harvey P. Dale, May 21 2013 *)

a(0)=17, a(n)=83521*a(n-1). - Harvey P. Dale, May 21 2013
Sum_{n>=0} 1/a(n) = 4913/83520. - Bernard Schott, Mar 29 2022
Sum_{n>=0} (-1)^n/a(n) = 4913/83522. - Bernard Schott, Apr 08 2022

A241888 a(n) = 2^(4*n + 1) - 1.

Original entry on oeis.org

1, 31, 511, 8191, 131071, 2097151, 33554431, 536870911, 8589934591, 137438953471, 2199023255551, 35184372088831, 562949953421311, 9007199254740991, 144115188075855871, 2305843009213693951, 36893488147419103231, 590295810358705651711, 9444732965739290427391

Offset: 0

Wassan Letourneur, Aug 09 2014

Jens Kruse Andersen, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A000225, A004171, A013776.

List([0..20],n->2^(4*n+1)-1); # Muniru A Asiru, Mar 12 2019

seq(coeff(series((14*x+1)/((x-1)*(16*x-1)),x,n+1), x, n), n = 0 .. 20); # Muniru A Asiru, Mar 12 2019

Table[2^(4n + 1) - 1, {n, 0, 29}]
CoefficientList[ Series[(14x + 1)/((x - 1) (16x - 1)), {x, 0, 18}], x] (* Robert G. Wilson v, Jan 28 2015 *)
LinearRecurrence[{17,-16},{1,31},30] (* Harvey P. Dale, Mar 13 2016 *)

vector(40, n, 2^(4*n-3)-1) \\ Derek Orr, Aug 11 2014

Vec((14*x+1)/((x-1)*(16*x-1)) + O(x^100)) \\ Colin Barker, Aug 31 2014

a(n) = 2^(4*n + 1) - 1 = A000225(4*n + 1) = A013776(n) - 1 = 4*A000225(4*n - 1) + 3.
From Colin Barker, Aug 31 2014: (Start)
a(n) = 17*a(n-1) - 16*a(n-2).
G.f.: (14*x+1)/((x-1)*(16*x-1)). (End)
E.g.f.: exp(x)*(2*exp(15*x) - 1). - Elmo R. Oliveira, Feb 20 2025

A013812 a(n) = 20^(4*n + 1).

Original entry on oeis.org

20, 3200000, 512000000000, 81920000000000000, 13107200000000000000000, 2097152000000000000000000000, 335544320000000000000000000000000, 53687091200000000000000000000000000000, 8589934592000000000000000000000000000000000, 1374389534720000000000000000000000000000000000000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (160000).

Subsequence of A009964.

Cf. A013776, A013792, A016813.

[20^(4*n+1): n in [0..15]]; // Vincenzo Librandi, Jul 06 2011

A013812:=n->20^(4*n+1): seq(A013812(n), n=0..10); # Wesley Ivan Hurt, Mar 31 2017

20^(4*Range[0,10]+1) (* or *) NestList[160000#&,20,10] (* Harvey P. Dale, Dec 30 2012 *)

a(n) = 160000*a(n-1). - Harvey P. Dale, Dec 30 2012
From Elmo R. Oliveira, Jul 13 2025: (Start)
G.f.: 20/(1-160000*x).
E.g.f.: 20*exp(160000*x).
a(n) = A013776(n)*A013792(n) = A009964(A016813(n)). (End)

cp's OEIS Frontend

A004171 a(n) = 2^(2n+1).

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A078164 Numbers k such that phi(k) is a perfect biquadrate.

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A144864 a(n) = (4*16^(n-1)-1)/3.

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A141060 Fourth quadrisection of Jacobsthal numbers A001045: a(n)=16a(n-1)-5.

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A264960 Half-convolution of the central binomial coefficients A000984 with itself.

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A307690 Integers with only one prime factor and whose Euler's totient is a perfect biquadrate.

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