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%I #14 Jul 27 2017 14:27:27
%S 1,-1,-2,-1,1,2,-2,2,4,-1,1,2,1,-1,-2,2,-2,-4,-2,2,4,2,-2,-4,4,-4,-8,
%T -1,1,2,1,-1,-2,2,-2,-4,1,-1,-2,-1,1,2,-2,2,4,2,-2,-4,-2,2,4,-4,4,8,
%U -2,2,4,2,-2,-4,4,-4,-8,2,-2,-4,-2,2,4,-4,4,8,4,-4,-8,-4,4,8,-8,8,16,-1,1,2,1,-1,-2,2,-2,-4,1,-1
%N a(n) = a(3n) = -a(3n+1) = -a(3n+2)/2.
%C a(n) = a(3n)/a(0) = a(3n+1)/a(1) = a(3n+2)/a(2).
%C Row sums of A117941.
%H Antti Karttunen, <a href="/A117942/b117942.txt">Table of n, a(n) for n = 0..19683</a>
%F a(n) = 2^A081603(n) * (-1)^(A062756(n)+A081603(n)). - _Antti Karttunen_, Jul 26 2017
%o (Scheme)
%o ;; A stand-alone recurrence:
%o (define (A117942 n) (cond ((zero? n) 1) ((zero? (modulo n 3)) (A117942 (/ n 3))) (else (let ((d (modulo n 3))) (- (* d (A117942 (/ (- n d) 3))))))))
%o ;; An implementation based on a new formula:
%o (Scheme) (define (A117942 n) (* (A000079 (A081603 n)) (expt -1 (+ (A062756 n) (A081603 n)))))
%o ;; _Antti Karttunen_, Jul 26 2017
%Y Cf. A117592 (gives the absolute values).
%Y Cf. A062756, A081603.
%K sign
%O 0,3
%A _Paul Barry_, Apr 05 2006
%E More terms from _Antti Karttunen_, Jul 26 2017
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