A143621 a(n) = (-1)^binomial(n,4): Periodic sequence 1,1,1,1,-1,-1,-1,-1,... .

1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1

Offset: 0

Author

Peter Bala, Aug 30 2008

Keywords

• easy
• sign

Comments

Periodic sequence with period 8. More generally, 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), A057077 (r = 1) and A143622 (r = 3).

Nonsimple continued fraction expansion of A188943 = 1.767591879243... - R. J. Mathar, Mar 08 2012

Examples

			G.f. = 1 + x + x^2 + x^3 - x^4 - x^5 - x^6 - x^7 + x^8 + x^9 + x^10 + ...
		

Links

• Maciej Gawron, and Maciej Ulas, On formal inverse of the Prouhet-Thue-Morse sequence, Discrete Mathematics 339.5 (2016): 1459-1470. Also arXiv preprintarXiv:1601.04840 [math.CO], 2016. The sequence appears on page 1464, prefixed by three 0's.

Crossrefs

Cf. A033999, A057077, A130151, A143622.

Programs

Maple

with(combinat):
a := n -> (-1)^binomial(n,4):
seq(a(n),n = 0..103);

Mathematica

Table[(-1)^Binomial[n, 4], {n, 0, 100}] (* Wesley Ivan Hurt, May 20 2014 *)
a[ n_] := (-1)^Quotient[n, 4]; (* Michael Somos, May 05 2015 *)
PadRight[{},120,{1,1,1,1,-1,-1,-1,-1}] (* Harvey P. Dale, Nov 29 2024 *)

PARI

{a(n) = (-1)^(n \ 4)}; /* Michael Somos, Sep 30 2011 */

PARI

x='x+O('x^99); Vec((1-x^4)^2/((1-x)*(1-x^8))) \\ Altug Alkan, Apr 15 2016

Python

def A143621(n): return -1 if n&4 else 1 # Chai Wah Wu, Jan 18 2023

Formula

a(n) = (-1)^binomial(n,4) = (-1)^floor(n/4), since Sum_{k = 1..n-3} k*(k+1)(k+2)/3! = binomial(n,4) == floor(n/4) (mod 2) for n = 0,1,...,7 by calculation and both sides increase by an even number if we substitute n+8 for n.

a(n) = (1/4)*((n+4) mod 8 - n mod 8).

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

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), E(2) and E(3) (a Dobinski-type relation).

a(n) = (-1)^A180969(2,n), where the first index in A180969(.,.) is the row index. - Adriano Caroli, Nov 18 2010

Euler transform of length 8 sequence [ 1, 0, 0, -2, 0, 0, 0, 1]. - Michael Somos, Sep 30 2011

G.f.: (1 - x^4)^2 / ((1 - x) * (1 - x^8)). a(n) = -a(-1 - n) for all n in Z. - Michael Somos, Sep 30 2011

E.g.f.: sin(x/sqrt(2))*sinh(x/sqrt(2)) + (sqrt(2)*sin(x/sqrt(2)) + cos(x/sqrt(2)))*cosh(x/sqrt(2)). - Ilya Gutkovskiy, Apr 15 2016