%I A143622
%S A143622 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,
%T A143622 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,
%U A143622 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
%V A143622 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,
%W A143622 -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,
%X A143622 -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
%N 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,... .
%C A143622 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 A1436221 (r = 2).
%F A143622 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)).
%F A143622 Define E(k) = sum {n = 0..inf} 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).
%p A143622 with(combinat):
%p A143622 a := n -> (-1)^binomial(n,8):
%p A143622 seq(a(n),n = 0..95);
%Y A143622 A033999, A057077, A130151, A143621.
%Y A143622 Sequence in context: A165326 A143621 A098417 this_sequence A076479 A155040 
               A033999
%Y A143622 Adjacent sequences: A143619 A143620 A143621 this_sequence A143623 A143624 
               A143625
%K A143622 easy,sign
%O A143622 0,1
%A A143622 Peter Bala (pbala(AT)toucansurf.com), Aug 30 2008