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%I #10 Dec 29 2015 03:06:58
%S 0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,6,0,1,
%T 0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,0,1,0,2,
%U 0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,6,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,3
%N Zero-based row index to A265345; 2-adic valuation of bijective base-3 reversal of n: a(n) = A007814(A263273(n)).
%H Antti Karttunen, <a href="/A265330/b265330.txt">Table of n, a(n) for n = 1..6561</a>
%F a(n) = A007814(A263273(n)).
%F a(2n+1) = 0, a(2n) = 1 + a(A265352(n)).
%e For n = 32, in base-3 "1012" [= A007089(32)], when we reverse it, we get "2101" [= A007089(64)], and 2-adic valuation of 64 [= "1000000" = A007088(64)] is 6, thus a(32) = 6.
%o (Scheme, two variants)
%o (definec (A265330 n) (if (odd? n) 0 (+ 1 (A265330 (A265352 (/ n 2))))))
%o (define (A265330 n) (A007814 (A263273 n)))
%Y One less than A265331.
%Y Cf. A007088, A007089, A263273, A265345, A265352.
%Y Cf. A265910 (corresponding other index).
%Y Cf. also A265336, A265337, A265340.
%Y Differs from A007814 for the first time at n=32, where a(32) = 6, while A007814(32) = 5.
%K nonn,base
%O 1,4
%A _Antti Karttunen_, Dec 18 2015
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