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a(n) = 2*3^(s-1) - n, where s is the number of trits of n in balanced ternary form.
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%I #43 Nov 19 2024 00:42:38

%S 1,4,3,2,13,12,11,10,9,8,7,6,5,40,39,38,37,36,35,34,33,32,31,30,29,28,

%T 27,26,25,24,23,22,21,20,19,18,17,16,15,14,121,120,119,118,117,116,

%U 115,114,113,112,111,110,109,108,107,106,105,104,103,102,101,100

%N a(n) = 2*3^(s-1) - n, where s is the number of trits of n in balanced ternary form.

%C Analogous to a bit, a ternary digit is a trit (trinary digit).

%C Per the definition, n + a(n) = 2*3^(s-1), where s is the number of trits of n and a(n), n and a(n) form a decomposition of 2*3^(s-1).

%H Lei Zhou, <a href="/A274601/b274601.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = 2*3^(floor(log_3(2*n-1))) - n. [corrected by _Jason Yuen_, Nov 18 2024]

%e For n=1 the balanced ternary form of 1 is 1, which has 1 trits. 2*3^(1-1)-1 = 1, so a(1) = 1.

%e For n=2 the balanced ternary form of 2 is 1T, which has 2 trits. 2*3^(2-1)-2 = 4, so a(2) = 4.

%e For n=3 the balanced ternary form of 3 is 10, which has 2 trits. 2*3^(2-1)-3 = 3, so a(2) = 3.

%e ...

%e For n=62 the balanced ternary form of 62 is 1T10T, which has 5 trits. 2*3^(5-1)-62 = 100, so a(62) = 100.

%t Table[2*3^(Floor[Log[3, 2*n - 1]]) - n, {n, 1, 62}]

%o (Python)

%o from sympy import integer_log

%o def a(n): return 2*3**(integer_log(2*n - 1, 3)[0]) - n

%o print([a(n) for n in range(1, 101)]) # _Indranil Ghosh_, Jun 10 2017

%o (PARI) a(n) = 2*3^logint(2*n-1,3)-n \\ _Jason Yuen_, Nov 18 2024

%Y Cf. A117966, A134021, A140267, A134028.

%K nonn,base,easy,changed

%O 1,2

%A _Lei Zhou_, Nov 10 2016