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Least k such that the ternary representation of 2^k has exactly 2*n 1's, or -1 if no such k exists.
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%I #13 Aug 17 2024 22:51:34

%S 1,2,8,14,24,26,42,45,50,53,70,74,96,76,124,98,116,121,143,141,179,

%T 150,187,181,192,215,209,233,220,257,245,264,243,278,260,310,297,303,

%U 315,339,329,387,341,357,354,366,403,420,350,400,411,415,474,455,466,442

%N Least k such that the ternary representation of 2^k has exactly 2*n 1's, or -1 if no such k exists.

%F Conjecture: a(n) ~ 6*log_2(3)*n = 6*A020857*n.

%e For n = 3, the smallest power of 2 with exactly 2*3 = 6 1's in its ternary representation is 2^14 = 211110211_3, so a(3) = 14.

%o (PARI) a(n) = my(k=1); while (#select(x->(x==1), digits(2^k, 3)) != 2*n, k++); k; \\ _Michel Marcus_, Aug 17 2024

%Y Cf. A020857, A060035, A104321, A365214.

%K nonn,base

%O 0,2

%A _Pontus von Brömssen_, Aug 17 2024