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A133872
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Period 4: repeat 1,1,0,0.
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28
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1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| Partial sums of A056594.
Let i=sqrt(-1) and S(n)=Sum_{k=0..n-1} exp(2*pi*i*k^2/n) for n>=1 the famous Gauss sum. Then S(n)=(a(n)+a(n+1)*i)*sqrt(n). - Franz Vrabec (franz.vrabec(AT)aon.at), Nov 08 2007
For any n>=1 the sequence gives the minimum value m>=0 we can get using addition and subtraction among all the numbers from 1 to n. E.g.: n=1 -> m=1; n=2 -> m=2-1=1; n=3 -> m=3-2-1=0; n=4 -> m=4-3-2+1=0; n=5 -> m=5-4+3-2-1=1; n=6 -> m=6+5-4-3-2-1=6-5+4-3-2+1=1; n=7 -> m=7-6+5-4-3+2-1=7+6-5-4-3-2+1=0; etc. - Paolo P. Lava (paoloplava(AT)gmail.com), Feb 29 2008
a(A042948(n)) = 1; a(A042964(n)) = 0. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 03 2008]
a(n) is also the real part of partial sum of powers of the complex unit i [From Barbarel Tres Mil (barbarel3000(AT)yahoo.es), Aug 16 2009]
Periodic sequences having a period of 2k and composed of k ones followed by k zeros have a closed formula of floor(((n+k) mod 2k)/k). Listed sequences of this form are k=1..A000035(n+1),k=2..A133872(n),k=3..A088911,k=4..A131078(n),k=5..A112713(n-1). [From Gary Detlefs, May 17 2011]
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LINKS
| Index entries for characteristic functions
Psychedelic Geometry Blogspot, Curious Series-001 [From Barbarel Tres Mil (barbarel3000(AT)yahoo.es), Aug 16 2009]
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FORMULA
| a(n) = (1+floor(n/2)) mod 2.
a(n) = A004526(A000035(n+2)).
a(n) = 1+floor(n/2)-2*floor((n+2)/4).
a(n) = (((n+2) mod 4)-(n mod 2))/2.
a(n) = ((n+2-(n mod 2))/2) mod 2.
a(n) = ((2n+3+(-1)^n)/4) mod 2.
a(n) = (1+(-1)^((2n-1+(-1)^n)/4))/2.
a(n) = binomial(n+2,n) mod 2 =binomial(n+2,2) mod 2.
a(n) = A000217(n+1) mod 2.
G.f.: (1+x)/(1-x^4) = 1/((1-x)(1+x^2)).
a(n) = 1/2+(1/2)*cos(Pi*n/2)+(1/2)*sin(Pi*n/2). a(n) = A021913(n+2). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 15 2007
a(n)=(1/12)*(-2*(n mod 4)+((n+1) mod 4)+4*((n+2) mod 4)+((n+3) mod 4)), with n>=0 [From Paolo P. Lava (paoloplava(AT)gmail.com), Oct 06 2008]
Contribution from Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Dec 05 2008: (Start)
a(n) = 1/2+sin((2n+1)pi/4)/sqrt(2)
a(n) = 1/2+cos((2n-1)pi/4)/sqrt(2) (End)
a(n) = Re(sum(k=0,n,i^k)), where i=sqrt(-1) and Re is the real part of a complex number. a(n) = (1/2)*((sum(k=0,n,i^k))+sum(k=0,n,i^-k)) = Re((1/2)*(1+i)*(1-i^(n+1))). [From Barbarel Tres Mil (barbarel3000(AT)yahoo.es), Aug 16 2009]
a(n) = (1+i^(n*(n-1)))/2, where i=sqrt(-1). - Bruno Berselli, May 18 2011
a(n)= sum(k^j,k=1..n) mod 2, for any j.[From Gary Detlefs, Dec 28 2011]
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PROG
| (PARI) a(n)=n%4<2 [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Mar 17 2009]
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CROSSREFS
| Cf. A056594, A133620-A133625, A133630, A038509, A133634-A133636, A021913, A000217, A133882, A133880, A133890, A133900, A133910, A000035, A088911, A131078, A112713.
Sequence in context: A125999 A073784 A128130 * A068434 A127015 A068432
Adjacent sequences: A133869 A133870 A133871 * A133873 A133874 A133875
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KEYWORD
| nonn,easy
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AUTHOR
| Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Oct 10 2007
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EXTENSIONS
| Definition rewritten by N. J. A. Sloane, Apr 30 2009.
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