cp's OEIS Frontend

G.f.: (1+x)/(1+x^2).

a(n) = S(n, 0) + S(n-1, 0) = S(2*n, sqrt(2)); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 0)=A056594.

a(n) = (-1)^binomial(n,2) = (-1)^floor(n/2) = 1/2*((n+2) mod 4 - n mod 4). For fixed r = 0,1,2,..., it appears that (-1)^binomial(n,2^r) gives a periodic sequence of period 2^(r+1), the period consisting of a block of 2^r plus ones followed by a block of 2^r minus ones. See A033999 (r = 0), A143621 (r = 2) and A143622 (r = 3). Define E(k) = sum {n = 0..inf} a(n)*n^k/n! for k = 0,1,2,... . Then E(0) = cos(1) + sin(1), E(1) = cos(1) - sin(1) and E(k) is an integral linear combination of E(0) and E(1) (a Dobinski-type relation). Precisely, E(k) = A121867(k) * E(0) - A121868(k) * E(1). See A143623 and A143624 for the decimal expansions of E(0) and E(1) respectively. For a fixed value of r, similar relations hold between the values of the sums E_r(k) := Sum_{n>=0} (-1)^floor(n/r)*n^k/n!, k = 0,1,2,... . For particular cases see A000587 (r = 1) and A143628 (r = 3). - Peter Bala, Aug 28 2008

Sum_{k>=0} a(k)/(k+1) = Sum_{k>=0} 1/((a(k)*(k+1))) = log(2)/2 + Pi/4. - Jaume Oliver Lafont, Apr 30 2010

a(n) = (-1)^A180969(1,n), where the first index in A180969(.,.) is the row index. - Adriano Caroli, Nov 18 2010

a(n) = (-1)^((2*n+(-1)^n-1)/4) = i^((n-1)*n), with i=sqrt(-1). - Bruno Berselli, Dec 27 2010 - Aug 26 2011

Non-simple continued fraction expansion of (3+sqrt(5))/2 = A104457. - R. J. Mathar, Mar 08 2012

E.g.f.: cos(x)*(1 + tan(x)). - Arkadiusz Wesolowski, Aug 31 2012

From Ricardo Soares Vieira, Oct 15 2019: (Start)

E.g.f.: sin(x) + cos(x) = sqrt(2)*sin(x + Pi/4).

a(n) = sqrt(2)*(d^n/dx^n) sin(x)|_x=Pi/4, i.e., a(n) equals sqrt(2) times the n-th derivative of sin(x) evaluated at x=Pi/4. (End)

a(n) = 4*floor(n/4) - 2*floor(n/2) + 1. - Ridouane Oudra, Mar 23 2024

a(n+6) = a(n), a(0)=a(1)=a(2)=-a(3)=-a(4)=-a(5)=1.

a(n) = ((-1)^n * (4 * (cos((2*n + 1)*Pi/3) + cos(n*Pi)) + 1) - 4) / 3. - Federico Acha Neckar (f0383864(AT)hotmail.com), Sep 01 2007

a(n) = (-1)^n * (4 * cos((2*n + 1) * Pi/3) + 1) / 3. - Federico Acha Neckar (f0383864(AT)hotmail.com), Sep 02 2007

G.f.: (1+x+x^2)/((1+x)*(x^2-x+1)). - R. J. Mathar, Nov 14 2007

a(n) = 3*a(n-1) - a(n-3) + 3*a(n-4) for n>3. - Paul Curtz, Nov 22 2007

a(n) = (-1)^floor(n/3). Compare with A057077, A143621 and A143622. Define E(k) = Sum_{n >= 0} a(n)*n^k/n! for k = 0,1,2,... . Then E(k) is an integral linear combination of E(0), E(1) and E(2) (a Dobinski-type relation). Precisely, E(k) = A143628(k)*E(0) + A143629(k)*E(1) + A143630(k)*E(2). - Peter Bala, Aug 28 2008

Euler transform of length 6 sequence [1, 0, -2, 0, 0, 1]. - Michael Somos, Feb 26 2011

a(n) = b(2*n + 1) where b(n) is multiplicative with b(2^e) = 0^e, b(3^e) = -(-1)^e if e>0, b(p^e) = 1 if p == 1 (mod 4), b(p^e) = (-1)^e if p == 3 (mod 4) and p>3. - Michael Somos, Feb 26 2011

a(n + 3) = a(-1 - n) = -a(n) for all n in Z. - Michael Somos, Feb 26 2011

a(n) = (-1)^n * A257075(n) for all n in Z. - Michael Somos, Apr 15 2015

G.f.: 1 / (1 - x / (1 + 2*x^2 / (1 + x / (1 + x / (1 - x))))). - Michael Somos, Apr 15 2015

From Wesley Ivan Hurt, Jul 05 2016: (Start)

a(n) + a(n-3) = 0 for n>2.

a(n) = (cos(n*Pi) + 2*cos(n*Pi/3) + 2*sqrt(3)*sin(n*Pi/3)) / 3. (End)

a(n)*a(n-4) = a(n-1)*a(n-3) for all n in Z. - Michael Somos, Feb 25 2020

a(n) = (-1)^binomial(n,8) = (-1)^floor(n/8), since sum {k = 1..n-7} k*(k+1)*...*(k+6)/7! = binomial(n,8) == floor(n/8) (mod 2) for n = 0,1,...,15 by calculation and both sides increase by an even number if we substitute n+16 for n. a(n) = (1/8)*((n+8) mod 16 - n mod 16).

O.g.f.: (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7)/(1+x^8) = (1+x)*(1+x^2)*(1+x^4)/(1+x^8) = (1-x^8)/((1-x)*(1+x^8)).

Define E(k) = Sum_{n>=0} a(n)*n^k/n! for k = 0,1,2,... . Then E(k) is a an integral linear combination of E(0),E(1),...,E(7) (a Dobinski-type relation).

a(n) = (-1)^A180969(3,n).