Search: id:A049347 Results 1-1 of 1 results found. %I A049347 %S A049347 1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0, %T A049347 1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0, %U A049347 1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0 %V A049347 1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0, %W A049347 1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0, %X A049347 1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0 %N A049347 Periodic sequence with period {1,-1,0}. %C A049347 (G.f.)^(-1)= cyclotomic(3,x) (cyclotomic polynomial). %C A049347 Self-convolution yields (-1)^n*A099254(n). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 06 2008 %C A049347 Hankel transform of A099324. [From Paul Barry (pbarry(AT)wit.ie), Aug 10 2009] %D A049347 A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 175. %H A049347 Index entries for sequences related to linear recurrences with constant coefficients %H A049347 Index entries for sequences related to Chebyshev polynomials. %F A049347 a(n) = +1 if n mod 3 = 0, a(n) = -1 if n mod 3 = 1 else 0. a(n) = S(n, -1) = U(n, -1/2) (Chebyshev's U(n, x) polynomials.) G.f.: 1/(1+x+x^2). %F A049347 a(n) = (1/2)((-1)^floor((2n+2)/3) + (-1)^floor((2n+1)/3)). - Mario Catalani, Jan 16 2003 %F A049347 a(n)=(1/2)((-1)^(Floor[(2n)/3]) + 1). - Mario Catalani (mario.catalani(AT)unito.it), Oct 22 2003 %F A049347 a(n)=2sqrt(3)cos(2*pi*n/3+pi/6)/3. - Paul Barry (pbarry(AT)wit.ie), Mar 15 2004 %F A049347 a(n) = Sum[k>=0, (-1)^(n-k)*C(n-k, k) ]. %F A049347 Given g.f. A(x), then B(x)=x*A(x) satisfies 0=f(B(x), B(x^2)) where f(u, v)= u^2 -v +2*u*v . - Michael Somos Oct 03 2006 %F A049347 Euler transform of length 3 sequence [ -1, 0, 1]. - Michael Somos Oct 03 2006 %F A049347 a(n)=b(n+1) where b(n) is multiplicative with b(3^e) = 0^e, b(p^e) = 1 if p == 1 (mod 3), b(p^e) = (-1)^e if p == 2 (mod 3). - Michael Somos Oct 03 2006 %F A049347 G.f.: (1-x)/(1-x^3). a(n)=-a(1-n)=-a(n-1)-a(n-2)=a(n-3). - Michael Somos Oct 03 2006 %F A049347 a(n)= -(1/3)*[n mod 3+(n+1) mod 3-2*((n+2) mod 3)] - Paolo P. Lava (ppl(AT)spl.at), Oct 09 2006 %e A049347 1 - x + x^3 - x^4 + x^6 - x^7 + x^9 - x^10 + x^12 - x^13 + x^15 + ... %o A049347 (PARI) {a(n)=n++; kronecker(-3,n)} /* Michael Somos Oct 03 2006 */ %o A049347 (PARI) {a(n) = [1, -1, 0][n%3 + 1]} /* Michael Somos Oct 15 2008 */ %o A049347 (PARI) a(n)=(n+2)%3-1 [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Mar 24 2009] %Y A049347 Cf. A010892, A057078. %Y A049347 A057078(n) = a(-n). A106510(n+1) = a(n) unless n=0. %Y A049347 Sequence in context: A011646 A016350 A117441 this_sequence A010892 A091338 A016345 %Y A049347 Adjacent sequences: A049344 A049345 A049346 this_sequence A049348 A049349 A049350 %K A049347 easy,sign,mult %O A049347 0,1 %A A049347 Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de) Search completed in 0.002 seconds