A143622 a(n) = (-1)^binomial(n,8): 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, -1, -1, -1, 1, 1, 1, 1, 1, 1

Offset

0,1

Comments

Periodic sequence with period 16. 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 A143621 (r = 2).

Nonsimple continued fraction expansion of (47+sqrt(445))/42 = 1.62131007404... - R. J. Mathar, Mar 08 2012

Links

Table of n, a(n) for n=0..85.

Formula

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).

Maple

with(combinat):
a := n -> (-1)^binomial(n, 8):
seq(a(n), n = 0..95);

Crossrefs

Cf. A033999, A057077, A130151, A143621.

Sequence in context: A143621 A292117 A098417 * A246016 A306638 A076479

Adjacent sequences: A143619 A143620 A143621 * A143623 A143624 A143625

Keyword

easy,sign

Author

Peter Bala, Aug 30 2008

Status

approved