cp's OEIS Frontend

A365459 a(n) = n - the largest power of 3 that is less than or equal to n.

%I A365459 #16 Sep 17 2023 18:24:25
%S A365459 0,1,0,1,2,3,4,5,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,0,1,2,3,
%T A365459 4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,
%U A365459 29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,0,1,2,3,4,5,6,7
%N A365459 a(n) = n - the largest power of 3 that is less than or equal to n.
%H A365459 Michael De Vlieger, <a href="/A365459/b365459.txt">Table of n, a(n) for n = 1..19683</a> (3^9 = 19683).
%F A365459 a(n) = n - A365458(n).
%F A365459 a(n) = n - 3^floor((log n)/(log 3)). - _Michael De Vlieger_, Sep 17 2023
%t A365459 Array[# - 3^Floor@ Log[3, #] &, 88] (* _Michael De Vlieger_, Sep 17 2023 *)
%o A365459 (PARI) A365459(n) = if(1==n,0,my(k=0); while((3^k) < n, k++); if((3^k) > n,k--); n-(3^k));
%o A365459 (Python)
%o A365459 def A365459(n):
%o A365459     kmin, kmax = 0, 1
%o A365459     while 3**kmax <= n:
%o A365459         kmax <<= 1
%o A365459     while True:
%o A365459         kmid = kmax+kmin>>1
%o A365459         if 3**kmid > n:
%o A365459             kmax = kmid
%o A365459         else:
%o A365459             kmin = kmid
%o A365459         if kmax-kmin <= 1:
%o A365459             break
%o A365459     return n-3**kmin # _Chai Wah Wu_, Sep 17 2023
%Y A365459 Cf. A000244, A053735 (gives the number of iterations needed to reach 0), A365458.
%K A365459 nonn
%O A365459 1,5
%A A365459 _Antti Karttunen_, Sep 17 2023