|
Search: id:A133872
|
|
|
| A133872 |
|
Period 4: repeat 1,1,0,0. |
|
+0 23
|
|
| 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; listen)
|
|
|
OFFSET
|
0,1
|
|
|
COMMENT
|
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 famouse 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 (ppl(AT)spl.at), 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 the finite sum of consecutive powers of the complex unit i [From Barbarel Tres Mil (barbarel3000(AT)yahoo.es), Aug 16 2009]
|
|
LINKS
|
Index entries for characteristic functions
Psychedelic Geometry Blogspot, Curious Series-001 [From Barbarel Tres Mil (barbarel3000(AT)yahoo.es), Aug 16 2009]
|
|
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. g(x)=(1+x)/(1-x^4).
G.f. g(x)=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 (ppl(AT)spl.at), 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)) a(n)=Re((1/2)*(1+i)*(1-i^(n+1))) [From Barbarel Tres Mil (barbarel3000(AT)yahoo.es), Aug 16 2009]
|
|
MATHEMATICA
|
Re[ Sum[ \[ ImaginaryI ]^k, {k, 0, n} ] ] [ From Barbarel Tres Mil (barbarel3000(AT)yahoo.es), Aug 16 2009 ]
|
|
PROGRAM
|
(PARI) a(n)=n%4<2 [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Mar 17 2009]
|
|
CROSSREFS
|
Cf. A056594. A133620-A133625, A133630, A133633-A133636. A021913. A000217.
Cf. A133882, A133880, A133890, A133900, A133910.
Sequence in context: A125999 A073784 A128130 this_sequence A071026 A068434 A127015
Adjacent sequences: A133869 A133870 A133871 this_sequence A133873 A133874 A133875
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Oct 10 2007
|
|
EXTENSIONS
|
Definition rewritten by N. J. A. Sloane, Apr 30 2009.
|
|
|
Search completed in 0.002 seconds
|