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%I #32 May 15 2020 02:34:05
%S 0,1,0,1,1,1,1,2,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,2,2,2,
%T 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,
%U 3,3,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2
%N Number of bits in binary expansion of n minus the number of digits of n when written in base 3.
%C Record highs are at n = 2^A054414. All n=2^k >= 2 are increases, all n=3^j are decreases, and there is either one or none 3^j between 2^(k-1) and 2^k. When one, a(2^k) = a(2^(k-1)) so not a record high. When none, a(2^k) = a(2^(k-1)) + 1 which is a record high. If 2^k and 2^(k-1) are the same length in ternary then there is no 3^j between them. This is when 2^k has most significant ternary digit 2 since 2^(k-1) >= 3^j is 2^k >= 2*3^j. These k are A054414. Non-record increases are at its complement n = 2^A020914 >= 2. - _Kevin Ryde_, Apr 04 2020
%F a(n) = A000523(n) - A062153(n) = floor(log_2(n)) - floor(log_3(n)).
%F a(n) = length(A007088(n)) - length(A007089(n)).
%e a(8) = 2 = 4 - 2 for binary 1000 and ternary 22.
%e a(64) = 3 = 7 - 4 for binary 1000000 and ternary 2101.
%p a:= n-> ilog[2](n)-ilog[3](n):
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Mar 15 2020
%t a[n_]: = Floor @ Log[2, n] - Floor @ Log[3, n]; Array[a, 100] (* _Amiram Eldar_, Mar 16 2020 *)
%o (Rexx)
%o L = 1 ; M = 1 ; B = 2 ; T = 3 ; S = 0
%o do N = 2 while length( S ) < 258
%o if B = N then do ; B = B * 2 ; L = L + 1 ; end
%o if T = N then do ; T = T * 3 ; M = M + 1 ; end
%o S = S || ',' L - M
%o end N
%o say S ; return S
%o (PARI) a(n) = logint(n,2) - logint(n,3); \\ _Kevin Ryde_, May 15 2020
%Y Cf. A007088 ( binary), A000523 (floor(log_2(n)), A029837.
%Y Cf. A007089 (ternary), A062153 (floor(log_3(n)), A117966.
%K nonn,base,easy
%O 1,8
%A _Frank Ellermann_, Mar 15 2020
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