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Number of trailing 2-digits in ternary representation of A254103(n).
5

%I #7 Sep 12 2017 20:38:04

%S 0,0,1,0,1,0,2,0,1,0,1,0,1,0,2,0,1,0,3,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,

%T 1,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,1,0,1,0,3,0,

%U 1,0,1,0,1,0,1,0,1,0,2,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,2,0,1,0,1,0,1,0,2,0,1

%N Number of trailing 2-digits in ternary representation of A254103(n).

%H Antti Karttunen, <a href="/A292242/b292242.txt">Table of n, a(n) for n = 0..16384</a>

%F a(n) = A007949(1+A254103(n)).

%F a(n) = A007814(1+A291760(n)).

%F a(0) = 0, after which, a(2n) = 1 + A292241(n/2), a(2n+1) = 0.

%o (Scheme)

%o (define (A292242 n) (A007949 (+ 1 (A254103 n))))

%o (define (A292242 n) (A007814 (+ 1 (A291760 n))))

%o (define (A292242 n) (cond ((zero? n) n) ((odd? n) 0) (else (+ 1 (A292241 (/ n 2))))))

%Y Cf. A007814, A007949, A254103, A291760, A292241 (even bisection subtracted by one).

%Y Cf. also A292252, A292262.

%K nonn,base

%O 0,7

%A _Antti Karttunen_, Sep 12 2017