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Number of powers of 7 between 2^n and 2^(n+1).
0

%I #15 Sep 08 2022 08:44:47

%S 0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,1,0,0,1,

%T 0,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,1,

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

%N Number of powers of 7 between 2^n and 2^(n+1).

%F a(n) = floor(log_7 2^(n + 1)) - floor(log_7 2^n). - _Alonso del Arte_, Nov 04 2018

%e Between 2^2 and 2^3 there is only one power of 7, which is 7 itself. Hence a(2) = 1.

%e Between 2^3 and 2^4 there are no powers of 7, so a(3) = 0.

%t Table[Floor[Log[7, 2^(n + 1)]] - Floor[Log[7, 2^n]], {n, 0, 127}] (* _Alonso del Arte_, Nov 04 2018 *)

%o (PARI) logint(2^(n+1),7)-logint(2^n,7) \\ _Charles R Greathouse IV_, Jan 16 2017

%o (Magma) [Floor(Log(7, 2^(n+1))) - Floor(Log(7, 2^n)): n in [0..100]]; // _Vincenzo Librandi_, Nov 05 2018

%K nonn

%O 0,1

%A _Clark Kimberling_

%E Definition clarified by _Alonso del Arte_, Nov 04 2018