A330904 - OEIS

%I #68 Aug 04 2020 19:30:40

%S 0,1,10,12,13,34,36,37,66,67,120,121,192,193,202,264,265,272,273,282,

%T 283,354,355,360,361,514,516,517,520,526,544,576,577,688,840,841,848,

%U 849,904,928,1026,1027,1028,1029,1032,1033,1038,1039,1062,1063,1074,1075

%N Numbers m such that the number of 1's in the binary expansion of m equals the sum of the balanced ternary trits of m.

%C If a(n) mod 6 = 0, then a(n+1) = a(n)+1.

%C a(41) = 1026, a(42) = 1027, a(43) = 1028 and a(44) = 1029 is the first time that four consecutive numbers appear in a(n).  Conjecture: There is no occurrence of five or more consecutive numbers in a(n).  Tested by exhaustive search up to 3^30. - _Thomas König_, Jul 19 2020

%H Thomas König, <a href="/A330904/b330904.txt">Table of n, a(n) for n = 1..20000</a>

%H Thomas König, <a href="/A330904/a330904.c.txt">C program</a>

%H Thomas König, <a href="/A330904/a330904.f90.txt">Fortran program to test the conjecture that there are at most four consecutive numbers in sequence</a>

%F Integers m such that A065363(m) = A000120(m).

%e 34_10 = 11T1_bt = 10010_2, the sum of the digits is 1+1-1+1 = 2 for balanced ternary and 1+1 = 2 for base 2, so 34 is a term.

%o (PARI) bt(n)= if (n==0, return (0)); my(d=digits(n, 3), c=1); while(c, if(d[1]==2, d=concat(0, d)); c=0; for(i=2, #d, if(d[i]==2, d[i]=-1; d[i-1]+=1; c=1))); vecsum(d); \\ A065363

%o isok(m) = bt(m) == hammingweight(m); \\ _Michel Marcus_, Jun 07 2020

%Y Aside from the first term, subsequence of A174659.

%Y Cf. A000120, A037301, A065363.

%K nonn,base

%O 1,3

%A _Thomas König_, May 02 2020

%E Offset corrected by _Thomas König_, Jul 09 2020