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%I #7 Aug 09 2012 06:25:48
%S 1,1,0,1,0,0,0,1,1,0,0,0,1,0,1,1,1,1,1,0,1,0,0,0,1,1,0,0,0,1,0,1,1,1,
%T 1,1,1,1,0,0,0,1,0,0,0,0,0,0,1,1,0,1,1,0,1,0,0,0,0,1,0,0,0,1,0,1,0,1,
%U 1,1,1,1,0,1,0,1,1,0,1,0,1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,1,1,0,1,0,0
%N a(n)=1 if (n modulo 65536)+1 is a quadratic residue of 65537, 0 otherwise.
%C This sequence gives essentially the same information as A165471, but in contrast to it (and A165472), the period of this sequence is explicitly defined as 65536 (instead of 65537), so that in essence the zeros at A165471(k*65537) are silently skipped. Several derived sequences to be computed.
%H A. Karttunen, <a href="/A179416/b179416.txt">Table of n, a(n) for n = 0..65535</a>
%F a(n) = 1 if A165471(1+(n%65536))=+1, otherwise 0. Period 65536.
%o (MIT Scheme:)
%o (define (A179416 n) (1+halved (A165471 (1+ (modulo n 65536)))))
%o (define (1+halved n) (floor->exact (/ (1+ n) 2)))
%o (Sage)
%o def A179416_list(n) : # for n <= 65536
%o Q = quadratic_residues(65537)
%o return [int(i in Q) for i in (1..n)]
%o A179416_list(102) # _Peter Luschny_, Aug 08 2012
%K nonn
%O 0,1
%A _Antti Karttunen_, Jul 27 2010
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