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 A057079 Periodic sequence: repeat [1,2,1,-1,-2,-1]; expansion of (1+x)/(1-x+x^2). 54
 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, -1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Inverse binomial transform of A057083. Binomial transform of A061347. The sums of consecutive pairs of elements give A084103. - Paul Barry, May 15 2003 Hexaperiodic sequence identical to its third differences. - Paul Curtz, Dec 13 2007 a(n+1) is the Hankel transform of A001700(n+1)-A001700(n). - Paul Barry, Apr 21 2009 Non-simple continued fraction expansion of 1 = 1+1/(2+1/(1+1/(-1+...))). - R. J. Mathar, Mar 08 2012 Pisano period lengths: 1, 3, 2, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, ... - R. J. Mathar, Aug 10 2012 Alternating row sums of Riordan triangle A111125. - Wolfdieter Lang, Oct 18 2012 Periodic sequences of this type can be also calculated by a(n) = c + floor(q/(p^m-1)*p^n) mod p, where c is a constant, q is the number representing the periodic digit pattern and m is the period length. c, p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D, min = minimum value of elements in D. Than c := min, p := max - min + 1 and q := p^m*sum_{i=1..m} (D(i)-min)/p^i. Example: D = (1, 2, 1, -1, -2, -1), c = -2, m = 6, p = 5 and q = 12276 for this sequence. - Hieronymus Fischer, Jan 04 2013 REFERENCES Fink, Alex, Richard Guy, and Mark Krusemeyer. "Partitions with parts occurring at most thrice." Contributions to Discrete Mathematics 3.2 (2008), 76-114. See Section 13. LINKS T.-X. He, L. W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic. 532 (2017) 25-41, theorem 2.5, k=3. Tanya Khovanova, Recursive Sequences Index entries for linear recurrences with constant coefficients, signature (1,-1). FORMULA a(n) = S(n, 1)+S(n-1, 1) = S(2*n, sqrt(3)); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 1) = A010892(n). a(n) = 2*cos((n-1)*Pi/3) = a(n-1)-a(n-2) = -a(n-3) = a(n-6) = (A022003(n+1)+1)*(-1)^floor(n/3). Unsigned a(n) = 4-a(n-1)-a(n-2). - Henry Bottomley, Mar 29 2001 a(n) = (-1)^floor(n/3)+((-1)^floor((n-1)/3)+(-1)^floor((n+1)/3))/2. - Mario Catalani (mario.catalani(AT)unito.it), Jan 07 2003 a(n) = (1/2-sqrt(3)i/2)^(n-1)+(1/2+sqrt(3)i/2)^(n-1) = cos(Pi*n/3)+sqrt(3)sin(Pi*n/3). - Paul Barry, Mar 15 2004 The period 3 sequence (2, -1, -1, ...) has a(n) = 2*cos(2*Pi*n/3) = (-1/2-sqrt(3)i/2)^n+(-1/2+sqrt(3)i/2)^n. - Paul Barry, Mar 15 2004 Euler transform of length 6 sequence [2, -2, -1, 0, 0, 1]. - Michael Somos, Jul 14 2006 G.f.: (1 + x) / (1 -x + x^2) = (1 - x^2)^2 * (1 - x^3) / ((1 - x)^2 * (1 - x^6)). a(n) = a(2-n) for all n in Z. - Michael Somos, Jul 14 2006 a(n) = -1/6*(2*(n mod 6)+((n+1) mod 6)-((n+2) mod 6)-2*((n+3) mod 6)-((n+4) mod 6)+((n+5) mod 6)). - Paolo P. Lava, Nov 20 2006 a(n) = A033999(A002264(n))*(A000035(A010872(n))+1). - Hieronymus Fischer, Jun 20 2007 a(n) = (3*A033999(A002264(n))-A033999(n))/2. - Hieronymus Fischer, Jun 20 2007 a(n) = (-1)^floor(n/3)*((n mod 3) mod 2 + 1). - Hieronymus Fischer, Jun 20 2007 a(n) = (3*(-1)^floor(n/3)-(-1)^n)/2. - Hieronymus Fischer, Jun 20 2007 a(n) = (-1)^((n-1)/3)+(-1)^((1-n)/3). - Jaume Oliver Lafont, May 13 2010 E.g.f.: E(x) = S(0), S(k) = 1 + 2*x/(6*k+1 - x*(6*k+1)/(4*(3*k+1) + x + 4*x*(3*k+1)/(6*k + 3 - x - x*(6*k+3)/(3*k + 2 + x - x*(3*k+2)/(12*k + 10 + x - x*(12*k+10)/(x - (6*k+6)/S(k+1))))))); (continued fraction). - Sergei N. Gladkovskii, Dec 14 2011 a(n) = -2 + floor(281/819*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 04 2013 a(n) = -2 + floor(11/14*5^(n+1)) mod 5. - Hieronymus Fischer, Jan 04 2013 a(n) = A010892(n) + A010892(n-1). a(n) = ( (1+i*sqrt(3))^(n-1)+(1-i*sqrt(3))^(n-1) )/2^(n-1), where i=sqrt(-1). - Bruno Berselli, Dec 01 2014 a(n) = 2*sin((2n+1)*Pi/6). - Wesley Ivan Hurt, Apr 04 2015 a(n) = hypergeom([-n/2-2, -n/2-5/2], [-n-4], 4). - Peter Luschny, Dec 17 2016 G.f.: 1 / (1 - 2*x / (1 + 3*x / (2 - x))). - Michael Somos, Dec 29 2016 a(n) = (2*n+1)*(Sum_{k=0..n} ((-1)^k/(2*k+1))*binomial(n+k,2*k)) for n >= 0. - Werner Schulte, Jul 10 2017 Sum_{n>=0} (a(n)/(2*n+1))*x^(2*n+1) = arctan(x/(1-x^2)) for -1 < x < 1. - Werner Schulte, Jul 10 2017 EXAMPLE G.f. = 1 + 2*x + x^2 - x^3 - 2*x^4 - x^5 + x^6 + 2*x^7 + x^8 - x^9 - 2*x^10 + x^11 + ... MAPLE A057079:=n->[1, 2, 1, -1, -2, -1][(n mod 6)+1]: seq(A057079(n), n=0..100); # Wesley Ivan Hurt, Mar 10 2015 MATHEMATICA a[n_] := {1, 2, 1, -1, -2, -1}[[Mod[n, 6] + 1]]; Array[a, 100, 0] (* Jean-François Alcover, Jul 05 2013 *) CoefficientList[Series[(1 + x)/(1 - x + x^2), {x, 0, 71}], x] (* Michael De Vlieger, Jul 10 2017 *) PROG (PARI) {a(n) = [1, 2, 1, -1, -2, -1][n%6 + 1]}; /* Michael Somos, Jul 14 2006 */ (PARI) {a(n) = if( n<0, n = 2-n); polcoeff( (1 + x) / (1 - x + x^2) + x * O(x^n), n)}; /* Michael Somos, Jul 14 2006 */ (PARI) a(n)=2^(n%3%2)*(-1)^(n\3) \\ Tani Akinari, Aug 15 2013 CROSSREFS Cf. A049310. Apart from signs, same as A061347. Cf. A002264, A010872. Sequence in context: A122876 A131713 A100063 * A132419 A131556 A107751 Adjacent sequences:  A057076 A057077 A057078 * A057080 A057081 A057082 KEYWORD easy,sign AUTHOR Wolfdieter Lang, Aug 04 2000 STATUS approved

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Last modified December 13 22:07 EST 2018. Contains 318087 sequences. (Running on oeis4.)