cp's OEIS Frontend

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,... .

%I A143622 #18 Jun 15 2021 01:41:19
%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,
%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,
%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,-1,1,1,1,1,1,1
%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 A143621 (r = 2).
%C A143622 Nonsimple continued fraction expansion of (47+sqrt(445))/42 = 1.62131007404... - _R. J. Mathar_, Mar 08 2012
%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).
%F A143622 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} 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).
%F A143622 a(n) = (-1)^A180969(3,n).
%p A143622 with(combinat):
%p A143622 a := n -> (-1)^binomial(n,8):
%p A143622 seq(a(n),n = 0..95);
%Y A143622 Cf. A033999, A057077, A130151, A143621.
%K A143622 easy,sign
%O A143622 0,1
%A A143622 _Peter Bala_, Aug 30 2008